I've came across some similar problems to this one in the past, and I still haven't got good idea how to solve this problem. Problem goes like this:
You are given an positive integer array with size n <= 1000 and k <= n which is the number of contiguous subarrays that you will have to split your array into. You have to output minimum m, where m = max{s[1],..., s[k]}, and s[i] is the sum of the i-th subarray. All integers in the array are between 1 and 100. Example :
Input: Output:
5 3 >> n = 5 k = 3 3
2 1 1 2 3
Splitting array into 2+1 | 1+2 | 3 will minimize the m.
My brute force idea was to make first subarray end at position i (for all possible i) and then try to split the rest of the array in k-1 subarrays in the best way possible. However, this is exponential solution and will never work.
So I'm looking for good ideas to solve it. If you have one please tell me.
Thanks for your help.
You can use dynamic programming to solve this problem, but you can actually solve with greedy and binary search on the answer. This algorithm's complexity is O(n log d), where d is the output answer. (An upper bound would be the sum of all the elements in the array.) (or O( n d ) in the size of the output bits)
The idea is to binary search on what your m would be - and then greedily move forward on the array, adding the current element to the partition unless adding the current element pushes it over the current m -- in that case you start a new partition. The current m is a success (and thus adjust your upper bound) if the numbers of partition used is less than or equal to your given input k. Otherwise, you used too many partitions, and raise your lower bound on m.
Some pseudocode:
// binary search
binary_search ( array, N, k ) {
lower = max( array ), upper = sum( array )
while lower < upper {
mid = ( lower + upper ) / 2
// if the greedy is good
if partitions( array, mid ) <= k
upper = mid
else
lower = mid
}
}
partitions( array, m ) {
count = 0
running_sum = 0
for x in array {
if running_sum + x > m
running_sum = 0
count++
running_sum += x
}
if running_sum > 0
count++
return count
}
This should be easier to come up with conceptually. Also note that because of the monotonic nature of the partitions function, you can actually skip the binary search and do a linear search, if you are sure that the output d is not too big:
for i = 0 to infinity
if partitions( array, i ) <= k
return i
Dynamic programming. Make an array
int best[k+1][n+1];
where best[i][j] is the best you can achieve splitting the first j elements of the array int i subarrays. best[1][j] is simply the sum of the first j array elements. Having row i, you calculate row i+1 as follows:
for(j = i+1; j <= n; ++j){
temp = min(best[i][i], arraysum[i+1 .. j]);
for(h = i+1; h < j; ++h){
if (min(best[i][h], arraysum[h+1 .. j]) < temp){
temp = min(best[i][h], arraysum[h+1 .. j]);
}
}
best[i+1][j] = temp;
}
best[m][n] will contain the solution. The algorithm is O(n^2*k), probably something better is possible.
Edit: a combination of the ideas of ChingPing, toto2, Coffee on Mars and rds (in the order they appear as I currently see this page).
Set A = ceiling(sum/k). This is a lower bound for the minimum. To find a good upper bound for the minimum, create a good partition by any of the mentioned methods, moving borders until you don't find any simple move that still decreases the maximum subsum. That gives you an upper bound B, not much larger than the lower bound (if it were much larger, you'd find an easy improvement by moving a border, I think).
Now proceed with ChingPing's algorithm, with the known upper bound reducing the number of possible branches. This last phase is O((B-A)*n), finding B unknown, but I guess better than O(n^2).
I have a sucky branch and bound algorithm ( please dont downvote me )
First take the sum of array and dvide by k, which gives you the best case bound for you answer i.e. the average A. Also we will keep a best solution seen so far for any branch GO ( global optimal ).Lets consider we put a divider( logical ) as a partition unit after some array element and we have to put k-1 partitions. Now we will put the partitions greedily this way,
Traverse the array elements summing them up until you see that at the next position we will exceed A, now make two branches one where you put the divider at this position and other where you put at next position, Do this recursiely and set GO = min (GO, answer for a branch ).
If at any point in any branch we have a partition greater then GO or the no of position are less then the partitions left to be put we bound. In the end you should have GO as you answer.
EDIT:
As suggested by Daniel, we could modify the divider placing strategy a little to place it until you reach sum of elements as A or the remaining positions left are less then the dividers.
This is just a sketch of an idea... I'm not sure that it works, but it's very easy (and probably fast too).
You start say by putting the separations evenly distributed (it does not actually matter how you start).
Make the sum of each subarray.
Find the subarray with the largest sum.
Look at the right and left neighbor subarrays and move the separation on the left by one if the subarray on the left has a lower sum than the one on the right (and vice-versa).
Redo for the subarray with the current largest sum.
You'll reach some situation where you'll keep bouncing the separation between the same two positions which will probably mean that you have the solution.
EDIT: see the comment by #rds. You'll have to think harder about bouncing solutions and the end condition.
My idea, which unfortunately does not work:
Split the array in N subarrays
Locate the two contiguous subarrays whose sum is the least
Merge the subarrays found in step 2 to form a new contiguous subarray
If the total number of subarrays is greater than k, iterate from step 2, else finish.
If your array has random numbers, you can hope that a partition where each subarray has n/k is a good starting point.
From there
Evaluate this candidate solution, by computing the sums
Store this candidate solution. For instance with:
an array of the indexes of every sub-arrays
the corresponding maximum of sum over sub-arrays
Reduce the size of the max sub-array: create two new candidates: one with the sub-array starting at index+1 ; one with sub-array ending at index-1
Evaluate the new candidates.
If their maximum is higher, discard
If their maximum is lower, iterate on 2, except if this candidate was already evaluated, in which case it is the solution.
Related
I have tried solving this for so long but I can't seem to be able to.
The question is as follows:
Given an array n numbers where all of the numbers in it occur twice except for one, which occurs only once, find the number that occurs only once.
Now, I have found many solutions online for this, but none of them satisfy the additional constraints of the question.
The solution should:
Run in linear time (aka O(n)).
Not use hash tables.
Assume that computer supports only comparison and the arithmetic (addition, subtraction, multiplication, division).
The number of bits in each number in the array is about O(log(n)).
Therefore, trying something like this https://stackoverflow.com/a/4772568/7774315 using the XOR operator isn't possible, since we don't have the XOR operator. Since the number of bits in each number is about O(log(n)), trying to implement the XOR operator using normal arithmetic (bit by bit) will take about O(log(n)) actions, which will give us an overall solution of O(nlog(n)).
The closest I have come to solving it is if I had a way to get the sum of all unique values in the array in linear time, I could subtract twice that sum from the overall sum to get (negative) the element that occurs only once, because if the numbers that appear twice are {a1,a2,....,ak} and the number that appears once is x, then the overall sum is
sum=2(a1+...+ak)+x
As far as I know, sets are implemented using hash tables, so using them to find the sum of all unique values is no good.
Let's imagine we had a way to find the exact median in linear time and partition the array so all greater elements are on one side and smaller elements on the other. By the parity of expected number of elements, we could identify which side the target element is in. Now perform this routine recursively in the section we identified. Since the section is halved in size each time, the total number of elements traversed cannot exceed O(2n) = O(n).
The key element in the question seems to be this one:
The number of bits in each number in the array is about O(log(n)).
The issue is that this clue is vague a little bit.
A first approach is to consider that the maximum value is O(n). Then a counting sort can be performed in O(n) operations and O(n) memory.
It will consists in finding the maximum value MAX, setting an integer array C[MAX] and performing directly a classical counting sort thanks to it
C[a[i]]++;
Looking for an odd value in array C[] will provide the solution.
A second approach, I guess more efficient, would be to set an array of size n, each element consisting of an array of unknown size. Then, a kind of almost counting sort would consists in :
C[a[i]%n].append (a[i]);
To find the unique element, we then have to find a sub-array of odd size, and then to examine the elements in this sub-array.
The maximum size k of each sub-array will be about 2*(MAX/n). According to the clue, this value should be very low. Dealing with this sub-array has a complexity O(k), for example by performing a counting sort on the b[j]/n, all the elements being equal modulo n.
We can note that practically, this is equivalent to perform a kind of ad-hoc hashing.
Global complexity is O(n + MAX/n).
This should do the trick as long as your a dealing with integers of size O(log n). It is a Python implementation of the algorithm sketched #גלעד ברקן answer (including #OneLyner comments), where the median is replaced by a mean or mid-value.
def mean(items):
result = 0
for i, item in enumerate(items, 1):
result = (result * (i - 1) + item) / i
return result
def midval(items):
min_val = max_val = items[0]
for item in items:
if item < min_val:
min_val = item
elif item > max_val:
max_val = item
return (max_val - min_val) / 2
def find_singleton(items, pivoting=mean):
n = len(items)
if n == 1:
return items[0]
else:
# find pivot - O(n)
pivot = pivoting(items)
# partition the items - O(n)
j = 0
for i, item in enumerate(items):
if item > pivot:
items[j], items[i] = items[i], items[j]
j += 1
# recursion on the partition with odd number of elements
if j % 2:
return find_singleton(items[:j])
else:
return find_singleton(items[j:])
The following code is just for some sanity-checking on random inputs:
def gen_input(n, randomize=True):
"""Generate inputs with unique pairs except one, with size (2 * n + 1)."""
items = sorted(set(random.randint(-n, n) for _ in range(n)))[:n]
singleton = items[-1]
items = items + items[:-1]
if randomize:
random.shuffle(items)
return items, singleton
items, singleton = gen_input(100)
print(singleton, len(items), items.index(singleton), items)
print(find_singleton(items, mean))
print(find_singleton(items, midval))
For a symmetric distribution the median and the mean or mid-value coincide.
With the log(n) requirement on the number of bits for the entries, one
can show that any arbitrary sub-sampling cannot be skewed enough to provide more than log(n) recursions.
For example, considering the case of k = log(n) bits with k = 4 and only positive numbers, the worst case is: [0, 1, 1, 2, 2, 4, 4, 8, 8, 16, 16]. Here pivoting by the mean will reduce the input by 2 at time, resulting in k + 1 recursive calls, but adding any other couple to the input will not increase the number of recursive calls, while it will increase the input size.
(EDITED to provide a better explanation.)
Here is an (unoptimized) implementation of the idea sketched by גלעד ברקן .
I'm using Median_of_medians to get a value close enough to the median to ensure the linear time in the worst case.
NB: this in fact uses only comparisons, and is O(n) whatever the size of the integers as long as comparisons and copies are counted as O(1).
def median_small(L):
return sorted(L)[len(L)//2]
def median_of_medians(L):
if len(L) < 20:
return median_small(L)
return median_of_medians([median_small(L[i:i+5]) for i in range(0, len(L), 5)])
def find_single(L):
if len(L) == 1:
return L[0]
pivot = median_of_medians(L)
smaller = [i for i in L if i <= pivot]
bigger = [i for i in L if i > pivot]
if len(smaller) % 2:
return find_single(smaller)
else:
return find_single(bigger)
This version needs O(n) additional space, but could be implemented with O(1).
I am trying to find the maximum sum of two elements in an array minus the distance between them.
Specifically I am trying to calculate max{ a[i]+a[j]-|i-j| }
I am currently stuck. I have obviously considered the naive approach (O(n^2)). However ,I am pretty sure there is a better ,more efficient approach (O(nlogn)) or even O(n).
Can someone please help me on how to approach the problem. I would be grateful if anyone threw some hints or a simple idea to have something to start from. Sorting the array first? Maybe using a dynamic programming approach?
Edit:
I think I have found an O(n) solution
Let's assume that our max sum comes from a[i] and a[j] , a[i] contributes to that sum with : a[i]+i . a[j] contributes to that sum with a[j]-j. (Because our sum is a[i]+a[j]-|j-i|= a[i]+a[j]+i-j. )
Approach: for convenience we compute the matrices A_plus_index=a[i]+i and A_minus_index=a[i]-i.
Then we use two helping arrays:
i) The first one has for every i ,the max value of A_plus_index array considering only the elements from 0 to i.
ii) The second has for every i, the max value of A_minus_index array considering only the elements from N to i ,where N is the length of array a.
Now we traverse the arrays once and find the max: A_plus_index[i]+ A_minus_index[i+1].
Total complexity O(n).
#JeffersonWhite your idea works and you could post it as an answer and accept it.
But I am going to improve upon your idea a little bit:
You could build only one array instead of 2, which contains the maximum of A[j] - j so far for each j from N-1 to 1.
And then traverse the array forward each time computing the max( A[i] + i + max_so_far-_reverse[i+1])
//Building the reverse array
max_so_far_reverse = array of length N
max_reverse = A[N-1]-(N-1)
max_so_far_reverse[N-1] = max_reverse
for j = N-2 to 1:
max_reverse = max(max_reverse, A[j]-j)
max_so_far_reverse[j] = max_reverse
//Computing maximum value by traversing forward
max = 0
for i = 0 to N-2:
max = max(max, A[i] + i + max_so_far_reverse[i+1])
return max
Given series of integers having relation where a number is equal to sum of previous 2 numbers and starting integer is 1
Series ->1,2,3,5,8,13,21,34,55
find the number of ways such that sum of k elements equal to p.We can use an element any number of times.
p=8
k=4.
So,number of ways would be 4.Those are,
1,1,1,5
1,1,3,3
1,2,2,3
2,2,2,2
I am able to sove this question through recursion.I sense dynamic programming here but i am not getting how to do it.Can it be done in much lesser time???
EDIT I forgot to mention that the sequence of the numbers does not matter and will be counted once. for ex=3->(1,2)and(2,1).here number of ways would be 1 only.
EDIT: Poster has changed the original problem since this was posted. My algorithm still works, but maybe can be improved upon. Original problem had n arbitrary input numbers (he has now modified it to be a Fibonacci series). To apply my algorithm to the modified post, truncate the series by taking only elements less than p (assume there are n of them).
Here's an n^(k/2) algorithm. (n is the number of elements in the series)
Use a table of length p, such that table[i] contains all combinations of k/2 elements that sum to i. For example, in the example data that you provided, table[4] contains {1,3} and {2,2}.
EDIT: If the space is prohibitive, this same algorithm can be done with an ordered linked lists, where you only store the non-empty table entries. The linked list has to be both directions: forward and backwards, which makes the final step of the algorithm cleaner.
Once this table is computed, then we get all solutions by combining every table[j] with every table[p-j], whenever both are non-empty.
To get the table, initialize the entire thing to empty. Then:
For i_1 = 0 to n-1:
For i_2 = i_1 to n-1:
...
For i_k/2 = i_k/2-1 to n-1:
sum = series[i_1] + ... + series[i_k/2]
if sum <= p:
store {i_1, i_2, ... , i_k/2 } in table[sum]
This "variable number of loops" looks impossible to implement, but actually it can be done with an array of length k/2 that keeps track of where each i_` is.
Let's go back to your data and see how our table would look:
table[2] = {1,1}
table[3] = {1,2}
table[4] = {1,3} and {2,2}
table[5] = {2,3}
table[6] = {1,5}
table[7] = {2,5}
table[8] = {3,5}
Solutions are found by combining table[2] with table[6], table[3] with table[5], and table[4] with table[4]. Thus, solutions are: {1,1,1,5} {1,2,2,3}, {1,1,3,3}, {2,2,2,2}, {1,3,2,2}.
You can use dynamic programming. Let C(p, k) be the number of ways that sum k element equal to p and a be the array of elements. Then
C(p, k) = C(p - a[0], k - 1) + C(p - a[1], k - 1) + .... + C(p - a[n-1], k - 1)
Then, you can use memorization to speed up your code.
Hint:
Your problem is well-known. It is the sum set problem, a variation of knapsack problem. Check this pretty good explanation. sum-set problem
Here is the problem:
Given is an array A of n integers, a seperate integer M, and an integer d. Find a consecutive subarray S of A, such that the size of the subarray is less than or equal to d and the sum of all the elements in S is M. Return the indexes of A that make the left and right index the subarray S. All numbers are positive.
If there is more than one result, give the rightmost result.
We have to make the algorithm run in better time than: O(n^2) or O(n*d). So basically it has to be O(nlog(n)), and divide and conquer I'm assuming is the way to go. I know how to do maximum continuous subarray problem, but that is made easier because when you divide and conquer you can look for max subarrays, with this one you don't really know what you are looking for in the subarrays, if that makes sense, since the solution could come from combinations of subarrays with small numbers and subarrays with big
Any help to lead me to the solution would be greatly appreciated!
I'm about 80% sure at this point that this is not possible... I keep looking it over and I can not think of a single way to make this work, is it possible this is a massive trick question?
This is relatively easy if the integers in A are >= 0, because you can just maintain a couple of pointers that define an interval with sum close to M and slide this along the array from right to left. Is it possible that you have missed some extra information like this in the question?
OK - here is some expansion. You have a left pointer and a right pointer. Move the right hand pointer from right to left, maintaining the invariant that the left hand pointer is no more than d places from the right hand pointer, and the sum of elements enclosed by the two pointers is the greatest possible number <= M. Repeatedly move the right hand pointer one step to the left and move the left hand pointer to the left until either you reach the limit of d or moving it further would produce a sum > M. Each time you move a pointer you can increment or decrement to maintain a running total of the sum enclosed by the two pointers.
Because the numbers are >= 0 every time you move the right hand pointer you decrease the sum or it stays the same so you always want to leave the left hand pointer the same or move it to the left. Because the numbers are >=0 you know that if there is an answer starting at the right hand pointer position you will find it with the left hand pointer position - anything that doesn't extend as far as the left hand pointer is too small, and anything that extends further is too large, except in the case where there are zeros and in that case you will find a solution, it's just that there are other solutions.
Each pointer is moved only in one direction so the maximum number of pointer movements is O(n) and the cost per pointer move is fixed so the complexity is O(n)
If all numbers are non-negative, this has a straightforward O(N) solution. The requirement of length<=d doesn't add any complexity, just add a check current_length<=d. I assume there are negative numbers in the array. We need additional O(N) space.
Compute prefix-sum of each index of S: p(i) = sum(S,0,i). Place p(i) in an additional array P: P[i]=p(i).
Make a copy of P: PSorted = P. Sort PSorted with a stable sort algorithm. We use it as a map prefix-sum -> index, with the index being a tie-breaker.
For each index k of S, starting from the largest:
Set p = P[k].
Look up p-M in PSorted using binary search, biased to the right. Say the resulting index is q.
If found, and q-k<d, return the answer (k,q).
This has overall O(n log n) complexity.
Expected running time can be reduced to O(N) if one uses a hash table instead of a sorted array, but one needs to be careful to always return the rightmost index which is smaller than the current index.
Correctly working algorithm, time complexity is O(n), if you count number operations closely.
public void SubArraySum(int[] arr, int d, int sum)
{
int n = arr.Length-1;
int curr_sum = arr[0], start = 0, i;
/* Add elements one by one to curr_sum and if the curr_sum exceeds the
sum, then remove starting element */
for (i = 1; i <= n; i++)
{
// If curr_sum exceeds the sum, then remove the starting elements
while (curr_sum > sum && start < i - 1)
{
curr_sum = curr_sum - arr[start];
start++;
}
// If curr_sum becomes equal to sum, then return true
if (curr_sum == sum && Math.Abs(start - i - 1) <= d)
{
Console.WriteLine("Sum found between indexes {0} and {1}", start, i - 1);
return;
}
// Add this element to curr_sum
if (i < n)
curr_sum = curr_sum + arr[i];
}
// If we reach here, then no subarray
Console.WriteLine("No subarray found");
}
Hope this help :)
I have a question and I tried to think over it again and again... but got nothing so posting the question here. Maybe I could get some view-point of others, to try and make it work...
The question is: we are given a SORTED array, which consists of a collection of values occurring an EVEN number of times, except one, which occurs ODD number of times. We need to find the solution in log n time.
It is easy to find the solution in O(n) time, but it looks pretty tricky to perform in log n time.
Theorem: Every deterministic algorithm for this problem probes Ω(log2 n) memory locations in the worst case.
Proof (completely rewritten in a more formal style):
Let k > 0 be an odd integer and let n = k2. We describe an adversary that forces (log2 (k + 1))2 = Ω(log2 n) probes.
We call the maximal subsequences of identical elements groups. The adversary's possible inputs consist of k length-k segments x1 x2 … xk. For each segment xj, there exists an integer bj ∈ [0, k] such that xj consists of bj copies of j - 1 followed by k - bj copies of j. Each group overlaps at most two segments, and each segment overlaps at most two groups.
Group boundaries
| | | | |
0 0 1 1 1 2 2 3 3
| | | |
Segment boundaries
Wherever there is an increase of two, we assume a double boundary by convention.
Group boundaries
| || | |
0 0 0 2 2 2 2 3 3
Claim: The location of the jth group boundary (1 ≤ j ≤ k) is uniquely determined by the segment xj.
Proof: It's just after the ((j - 1) k + bj)th memory location, and xj uniquely determines bj. //
We say that the algorithm has observed the jth group boundary in case the results of its probes of xj uniquely determine xj. By convention, the beginning and the end of the input are always observed. It is possible for the algorithm to uniquely determine the location of a group boundary without observing it.
Group boundaries
| X | | |
0 0 ? 1 2 2 3 3 3
| | | |
Segment boundaries
Given only 0 0 ?, the algorithm cannot tell for sure whether ? is a 0 or a 1. In context, however, ? must be a 1, as otherwise there would be three odd groups, and the group boundary at X can be inferred. These inferences could be problematic for the adversary, but it turns out that they can be made only after the group boundary in question is "irrelevant".
Claim: At any given point during the algorithm's execution, consider the set of group boundaries that it has observed. Exactly one consecutive pair is at odd distance, and the odd group lies between them.
Proof: Every other consecutive pair bounds only even groups. //
Define the odd-length subsequence bounded by the special consecutive pair to be the relevant subsequence.
Claim: No group boundary in the interior of the relevant subsequence is uniquely determined. If there is at least one such boundary, then the identity of the odd group is not uniquely determined.
Proof: Without loss of generality, assume that each memory location not in the relevant subsequence has been probed and that each segment contained in the relevant subsequence has exactly one location that has not been probed. Suppose that the jth group boundary (call it B) lies in the interior of the relevant subsequence. By hypothesis, the probes to xj determine B's location up to two consecutive possibilities. We call the one at odd distance from the left observed boundary odd-left and the other odd-right. For both possibilities, we work left to right and fix the location of every remaining interior group boundary so that the group to its left is even. (We can do this because they each have two consecutive possibilities as well.) If B is at odd-left, then the group to its left is the unique odd group. If B is at odd-right, then the last group in the relevant subsequence is the unique odd group. Both are valid inputs, so the algorithm has uniquely determined neither the location of B nor the odd group. //
Example:
Observed group boundaries; relevant subsequence marked by […]
[ ] |
0 0 Y 1 1 Z 2 3 3
| | | |
Segment boundaries
Possibility #1: Y=0, Z=2
Possibility #2: Y=1, Z=2
Possibility #3: Y=1, Z=1
As a consequence of this claim, the algorithm, regardless of how it works, must narrow the relevant subsequence to one group. By definition, it therefore must observe some group boundaries. The adversary now has the simple task of keeping open as many possibilities as it can.
At any given point during the algorithm's execution, the adversary is internally committed to one possibility for each memory location outside of the relevant subsequence. At the beginning, the relevant subsequence is the entire input, so there are no initial commitments. Whenever the algorithm probes an uncommitted location of xj, the adversary must commit to one of two values: j - 1, or j. If it can avoid letting the jth boundary be observed, it chooses a value that leaves at least half of the remaining possibilities (with respect to observation). Otherwise, it chooses so as to keep at least half of the groups in the relevant interval and commits values for the others.
In this way, the adversary forces the algorithm to observe at least log2 (k + 1) group boundaries, and in observing the jth group boundary, the algorithm is forced to make at least log2 (k + 1) probes.
Extensions:
This result extends straightforwardly to randomized algorithms by randomizing the input, replacing "at best halved" (from the algorithm's point of view) with "at best halved in expectation", and applying standard concentration inequalities.
It also extends to the case where no group can be larger than s copies; in this case the lower bound is Ω(log n log s).
A sorted array suggests a binary search. We have to redefine equality and comparison. Equality simple means an odd number of elements. We can do comparison by observing the index of the first or last element of the group. The first element will be an even index (0-based) before the odd group, and an odd index after the odd group. We can find the first and last elements of a group using binary search. The total cost is O((log N)²).
PROOF OF O((log N)²)
T(2) = 1 //to make the summation nice
T(N) = log(N) + T(N/2) //log(N) is finding the first/last elements
For some N=2^k,
T(2^k) = (log 2^k) + T(2^(k-1))
= (log 2^k) + (log 2^(k-1)) + T(2^(k-2))
= (log 2^k) + (log 2^(k-1)) + (log 2^(k-2)) + ... + (log 2^2) + 1
= k + (k-1) + (k-2) + ... + 1
= k(k+1)/2
= (k² + k)/2
= (log(N)² + log(N))/ 2
= O(log(N)²)
Look at the middle element of the array. With a couple of appropriate binary searches, you can find the first and its last appearance in the array. E.g., if the middle element is 'a', you need to find i and j as shown below:
[* * * * a a a a * * *]
^ ^
| |
| |
i j
Is j - i an even number? You are done! Otherwise (and this is the key here), the question to ask is i an even or an odd number? Do you see what this piece of knowledge implies? Then the rest is easy.
This answer is in support of the answer posted by "throwawayacct". He deserves the bounty. I spent some time on this question and I'm totally convinced that his proof is correct that you need Ω(log(n)^2) queries to find the number that occurs an odd number of times. I'm convinced because I ended up recreating the exact same argument after only skimming his solution.
In the solution, an adversary creates an input to make life hard for the algorithm, but also simple for a human analyzer. The input consists of k pages that each have k entries. The total number of entries is n = k^2, and it is important that O(log(k)) = O(log(n)) and Ω(log(k)) = Ω(log(n)). To make the input, the adversary makes a string of length k of the form 00...011...1, with the transition in an arbitrary position. Then each symbol in the string is expanded into a page of length k of the form aa...abb...b, where on the ith page, a=i and b=i+1. The transition on each page is also in an arbitrary position, except that the parity agrees with the symbol that the page was expanded from.
It is important to understand the "adversary method" of analyzing an algorithm's worst case. The adversary answers queries about the algorithm's input, without committing to future answers. The answers have to be consistent, and the game is over when the adversary has been pinned down enough for the algorithm to reach a conclusion.
With that background, here are some observations:
1) If you want to learn the parity of a transition in a page by making queries in that page, you have to learn the exact position of the transition and you need Ω(log(k)) queries. Any collection of queries restricts the transition point to an interval, and any interval of length more than 1 has both parities. The most efficient search for the transition in that page is a binary search.
2) The most subtle and most important point: There are two ways to determine the parity of a transition inside a specific page. You can either make enough queries in that page to find the transition, or you can infer the parity if you find the same parity in both an earlier and a later page. There is no escape from this either-or. Any set of queries restricts the transition point in each page to some interval. The only restriction on parities comes from intervals of length 1. Otherwise the transition points are free to wiggle to have any consistent parities.
3) In the adversary method, there are no lucky strikes. For instance, suppose that your first query in some page is toward one end instead of in the middle. Since the adversary hasn't committed to an answer, he's free to put the transition on the long side.
4) The end result is that you are forced to directly probe the parities in Ω(log(k)) pages, and the work for each of these subproblems is also Ω(log(k)).
5) Things are not much better with random choices than with adversarial choices. The math is more complicated, because now you can get partial statistical information, rather than a strict yes you know a parity or no you don't know it. But it makes little difference. For instance, you can give each page length k^2, so that with high probability, the first log(k) queries in each page tell you almost nothing about the parity in that page. The adversary can make random choices at the beginning and it still works.
Start at the middle of the array and walk backward until you get to a value that's different from the one at the center. Check whether the number above that boundary is at an odd or even index. If it's odd, then the number occurring an odd number of times is to the left, so repeat your search between the beginning and the boundary you found. If it's even, then the number occurring an odd number of times must be later in the array, so repeat the search in the right half.
As stated, this has both a logarithmic and a linear component. If you want to keep the whole thing logarithmic, instead of just walking backward through the array to a different value, you want to use a binary search instead. Unless you expect many repetitions of the same numbers, the binary search may not be worthwhile though.
I have an algorithm which works in log(N/C)*log(K), where K is the length of maximum same-value range, and C is the length of range being searched for.
The main difference of this algorithm from most posted before is that it takes advantage of the case where all same-value ranges are short. It finds boundaries not by binary-searching the entire array, but by first quickly finding a rough estimate by jumping back by 1, 2, 4, 8, ... (log(K) iterations) steps, and then binary-searching the resulting range (log(K) again).
The algorithm is as follows (written in C#):
// Finds the start of the range of equal numbers containing the index "index",
// which is assumed to be inside the array
//
// Complexity is O(log(K)) with K being the length of range
static int findRangeStart (int[] arr, int index)
{
int candidate = index;
int value = arr[index];
int step = 1;
// find the boundary for binary search:
while(candidate>=0 && arr[candidate] == value)
{
candidate -= step;
step *= 2;
}
// binary search:
int a = Math.Max(0,candidate);
int b = candidate+step/2;
while(a+1!=b)
{
int c = (a+b)/2;
if(arr[c] == value)
b = c;
else
a = c;
}
return b;
}
// Finds the index after the only "odd" range of equal numbers in the array.
// The result should be in the range (start; end]
// The "end" is considered to always be the end of some equal number range.
static int search(int[] arr, int start, int end)
{
if(arr[start] == arr[end-1])
return end;
int middle = (start+end)/2;
int rangeStart = findRangeStart(arr,middle);
if((rangeStart & 1) == 0)
return search(arr, middle, end);
return search(arr, start, rangeStart);
}
// Finds the index after the only "odd" range of equal numbers in the array
static int search(int[] arr)
{
return search(arr, 0, arr.Length);
}
Take the middle element e. Use binary search to find the first and last occurrence. O(log(n))
If it is odd return e.
Otherwise, recurse onto the side that has an odd number of elements [....]eeee[....]
Runtime will be log(n) + log(n/2) + log(n/4).... = O(log(n)^2).
AHhh. There is an answer.
Do a binary search and as you search, for each value, move backwards until you find the first entry with that same value. If its index is even, it is before the oddball, so move to the right.
If its array index is odd, it is after the oddball, so move to the left.
In pseudocode (this is the general idea, not tested...):
private static int FindOddBall(int[] ary)
{
int l = 0,
r = ary.Length - 1;
int n = (l+r)/2;
while (r > l+2)
{
n = (l + r) / 2;
while (ary[n] == ary[n-1])
n = FindBreakIndex(ary, l, n);
if (n % 2 == 0) // even index we are on or to the left of the oddball
l = n;
else // odd index we are to the right of the oddball
r = n-1;
}
return ary[l];
}
private static int FindBreakIndex(int[] ary, int l, int n)
{
var t = ary[n];
var r = n;
while(ary[n] != t || ary[n] == ary[n-1])
if(ary[n] == t)
{
r = n;
n = (l + r)/2;
}
else
{
l = n;
n = (l + r)/2;
}
return n;
}
You can use this algorithm:
int GetSpecialOne(int[] array, int length)
{
int specialOne = array[0];
for(int i=1; i < length; i++)
{
specialOne ^= array[i];
}
return specialOne;
}
Solved with the help of a similar question which can be found here on http://www.technicalinterviewquestions.net
We don't have any information about the distribution of lenghts inside the array, and of the array as a whole, right?
So the arraylength might be 1, 11, 101, 1001 or something, 1 at least with no upper bound, and must contain at least 1 type of elements ('number') up to (length-1)/2 + 1 elements, for total sizes of 1, 11, 101: 1, 1 to 6, 1 to 51 elements and so on.
Shall we assume every possible size of equal probability? This would lead to a middle length of subarrays of size/4, wouldn't it?
An array of size 5 could be divided into 1, 2 or 3 sublists.
What seems to be obvious is not that obvious, if we go into details.
An array of size 5 can be 'divided' into one sublist in just one way, with arguable right to call it 'dividing'. It's just a list of 5 elements (aaaaa). To avoid confusion let's assume the elements inside the list to be ordered characters, not numbers (a,b,c, ...).
Divided into two sublist, they might be (1, 4), (2, 3), (3, 2), (4, 1). (abbbb, aabbb, aaabb, aaaab).
Now let's look back at the claim made before: Shall the 'division' (5) be assumed the same probability as those 4 divisions into 2 sublists? Or shall we mix them together, and assume every partition as evenly probable, (1/5)?
Or can we calculate the solution without knowing the probability of the length of the sublists?
The clue is you're looking for log(n). That's less than n.
Stepping through the entire array, one at a time? That's n. That's not going to work.
We know the first two indexes in the array (0 and 1) should be the same number. Same with 50 and 51, if the odd number in the array is after them.
So find the middle element in the array, compare it to the element right after it. If the change in numbers happens on the wrong index, we know the odd number in the array is before it; otherwise, it's after. With one set of comparisons, we figure out which half of the array the target is in.
Keep going from there.
Use a hash table
For each element E in the input set
if E is set in the hash table
increment it's value
else
set E in the hash table and initialize it to 0
For each key K in hash table
if K % 2 = 1
return K
As this algorithm is 2n it belongs to O(n)
Try this:
int getOddOccurrence(int ar[], int ar_size)
{
int i;
int xor = 0;
for (i=0; i < ar_size; i++)
xor = xor ^ ar[i];
return res;
}
XOR will cancel out everytime you XOR with the same number so 1^1=0 but 1^1^1=1 so every pair should cancel out leaving the odd number out.
Assume indexing start at 0. Binary search for the smallest even i such that x[i] != x[i+1]; your answer is x[i].
edit: due to public demand, here is the code
int f(int *x, int min, int max) {
int size = max;
min /= 2;
max /= 2;
while (min < max) {
int i = (min + max)/2;
if (i==0 || x[2*i-1] == x[2*i])
min = i+1;
else
max = i-1;
}
if (2*max == size || x[2*max] != x[2*max+1])
return x[2*max];
return x[2*min];
}