We have an strictly increasing array of length n ( 1 < n < 500) . We sum the digits of each element to create a new array with each elements values is in range 1 to 500.The task is to rebuild the old array from the new one. since there might be more than one answer, we want the answers with the minimum value of the last element.
Example:
3 11 23 37 45 123 =>3 2 5 10 9 6
now from the second array, we can rebuild the original array in many different ways for instance:
12 20 23 37 54 60
from all the possible combinations, we need the one we minimum last element.
My Thoughts so far:
The brute force way is to find all possible permutations to create each number and then create all combinations possible of all numbers of the second array and find the combination with minimum last element. It is obvious that this is not a good choice.
Using this algorithm(with exponential time!) we can create all possible permutations of digits that sum to a number in the second arrays. Note that we know the original elements were less than 500 so we can limit the death of search of the algorithm.
One way I thought of that might find the answer faster is to:
start from the last element in the new arrays and find all possible
numbers that their digit sum resulted this element.
Then try to use the smallest amount in the last step for this element.
Now try to do the same with the second to last element. If the
minimum permutation value found for the second to last element is bigger
than the one found for the last element, backtrack to the last
element and try a larger permutation.
Do this until you get to the first element.
I think this is a greed solution but I'm not very sure about the time complexity. Also I want to know is there a better solution for this problem? like using dp?
For simplicity, let's have our sequence 1-based and the input sequence is called x.
We will also use an utility function, which returns the sum of the digits of a given number:
int sum(int x) {
int result = 0;
while (x > 0) {
result += x % 10;
x /= 10;
}
return result;
}
Let's assume that we are at index idx and try to set there some number called value (given that the sum of digits of value is x[idx]). If we do so, then what could we say about the previous number in the sequence? It should be strictly less than value.
So we already have a state for a potential dp approach - [idx, value], where idx is the index where we are currently at and value denotes the value we are trying to set on this index.
If the dp table holds boolean values, we will know we have found an answer if we have found a suitable number for the first number in the sequence. Therefore, if there is a path starting from the last row in the dp table and ends at row 0 then we'll know we have found an answer and we could then simply restore it.
Our recurrence function will be something like this:
f(idx, value) = OR {dp[idx - 1][value'], where sumOfDigits(value) = x[idx] and value' < value}
f(0, *) = true
Also, in order to restore the answer, we need to track the path. Once we set any dp[idx][value] cell to be true, then we can safe the value' to which we would like to jump in the previous table row.
Now let's code that one. I hope the code is self-explanatory:
boolean[][] dp = new boolean[n + 1][501];
int[][] prev = new int[n + 1][501];
for (int i = 0; i <= 500; i++) {
dp[0][i] = true;
}
for (int idx = 1; idx <= n; idx++) {
for (int value = 1; value <= 500; value++) {
if (sum(value) == x[idx]) {
for (int smaller = 0; smaller < value; smaller++) {
dp[idx][value] |= dp[idx - 1][smaller];
if (dp[idx][value]) {
prev[idx][value] = smaller;
break;
}
}
}
}
}
The prev table only keeps information about which is the smallest value', which we can use as previous to our idx in the resulting sequence.
Now, in order to restore the sequence, we can start from the last element. We would like it to be minimal, so we can find the first one that has dp[n][value] = true. Once we have such element, we then use the prev table to track down the values up to the first one:
int[] result = new int[n];
int idx = n - 1;
for (int i = 0; i <= 500; i++) {
if (dp[n][i]) {
int row = n, col = i;
while (row > 0) {
result[idx--] = col;
col = prev[row][col];
row--;
}
break;
}
}
for (int i = 0; i < n; i++) {
out.print(result[i]);
out.print(' ');
}
If we apply this on an input sequence:
3 2 5 10 9 6
we get
3 11 14 19 27 33
The time complexity is O(n * m * m), where n is the number of elements we have and m is the maximum possible value that an element could hold.
The space complexity is O(n * m) as this is dominated by the size of the dp and prev tables.
We can use a greedy algorithm: proceed through the array in order, setting each element to the least value that is greater than the previous element and has digits with the appropriate sum. (We can just iterate over the possible values and check the sums of their digits.) There's no need to consider any greater value than that, because increasing a given element will never make it possible to decrease a later element. So we don't need dynamic programming here.
We can calculate the sum of the digits of an integer m in O(log m) time, so the whole solution takes O(b log b) time, where b is the upper bound (500 in your example).
I was given the following question in an interview, and couldn't find the solution.
Given is an array of chars length n, and "important section" (all chars in this section must be saved) length m where n >= m >= 0 as follows:
Without extra space, perform the following process:
Remove all occurrences of A and duplicate all occurrences of B, return a sub array of the mutated array. For example, for the above array [C,A,X,B,B,F,Q] n=7, m=5 ,output will be [C,X,B,B,B,B]. Note that the mutated array length is 6, since Q was in the redundant section and B was duplicated.
Return -1 if the operation can't be performed.
Examples:
n=2, m=2 , [A,B] => [B,B]
n=2, m=2 , [B,B] => -1 (since the result [B,B,B,B] is larger then the array)
n=3, m=2 , [A,B,C] => [B,B]
n=3, m=3 , [A,B,C] => [B,B,C]
n=3, m=2 , [Z,B,A] => [Z,B,B] (since A was in the redundant section)
Looking for a code example, Could this be done in O(n) time complexity?
Scan array to determine if is it possible to store mutated array in available space -- count As and B, and check N-M >= numB-numA
Walk array left to right: Shift elements to the left by the number of As so far (filling places of A)
Walk array right to left: Shift elements to the right by numB-B_so_far, inserting additional Bs
Start from the end of the input array. We will figure out from the back to the front what to fill in.
Look at the last significant character in the input (position m). If it is a, ignore it. Otherwise, add the symbol. Repeat until you read all the input.
This removes as. Now we will duplicate bs.
Start from the beginning of the array. Find the last value you wrote during the above steps. If it is a b, write two bs. If it is something else, just write one of them. Repeat. NOTE: if you ever "catch up", needing to write where you need to read, you don't have enough room and you output -1. Otherwise, return the part of the array from position 1 to the last read position.
Example:
Phase 1: removing A
CAXBBFQ
CAXBBFB
CAXBBBB
CAXBXBB
CAXCXBB
Phase 2: duplicating B
CAXCXBB
CXXCXBB
CXBBXBB
CXBBBBB
^^^^^^
Phase 1 is linear (we read m symbols and write no more than m).
Phase 2 is linear (we read fewer than m symbols and write no more than 2m).
m is less than n so everything is O(m) and O(n).
The code, with some optimizations, would look something like this, O(n):
// returns length of the relevant part of the mutated array or -1
public static int mutate(char[] a, int m) {
// delete As and count Bs in the relevant part
int bCount = 0, position = 0;
for (int i = 0; i < m; i++) {
if (a[i] != 'A') {
if (a[i] == 'B')
bCount++;
a[position++] = a[i];
}
}
// check if it is possible
int n = bCount + position;
if (n > a.length)
return -1;
// duplicate the Bs in the relevant part
for (int i = position - 1, index = n - 1; i >= 0; i--) {
if (a[i] != 'B') {
a[index--] = a[i];
} else {
a[index--] = 'B';
a[index--] = 'B';
}
}
return n;
}
Hey guys I have been working on this for 3 days and have come up with nothing from everywere I have looked.
I am trying to take an Array of around 250 floats and find the Kth largest value without changing the array in anyway or making a new array.
I can change it or create a new one because other functions need the placing of the data in the correct order and my Arduino cant hold any more values in its memory space so the 2 easiest options are out.
The values in the Array can ( and probably will ) have duplicates in them.
As an EG : if you have the array ::: 1,36,2,54,11,9,22,9,1,36,0,11;
from Max to min would be ::
1) 54
2) 36
3) 36
4) 22
5) 11
6) 11
7) 9
8) 9
9) 2
10) 1
11) 1
12) 0
Any help would be great.
It may be to much to ask for a function that would do this nicely for me :) hahaha
here is the code I have so far but I have not even tried to get the duplicates working yet
and it for some reason only gives me one answer for some reason that's 2 ,,, no clue why though
void setup()
{
Serial.begin(9600);
}
void loop ()
{
int Array[] = {1,2,3,4,5,6,7,8,9,10};
int Kth = 6; //// just for testing putting the value as a constant
int tr = 0; /// traking threw the array to find the MAX
for (int y=0;y<10;y++) //////////// finding the MAX first so I have somewhere to start
{
if (Array[y]>Array[tr])
{
tr = y;
}
}
Serial.print("The max number is ");
int F = Array[tr];
Serial.println(F); // Prints the MAX ,,, mostly just for error checking this is done
///////////////////////////////////////////////////////// got MAX
for ( int x = 1; x<Kth;x++) //// run the below Kth times and each time lowering the "Max" making the loop run Kth times
{
for(int P=0;P<10;P++) // run threw every element
{
if (Array[P]<F)
{
for(int r=0;r<10;r++) //and then test that element against every other element to make sure
//its is bigger then all the rest but small then MAX
{
Serial.println(r);
if(r=tr) /////////////////// done so the max dosent clash with the number being tested
{
r++;
Serial.println("Max's Placeing !!!!");
}
if(Array[P]>Array[r])
{
F=Array[P]; ////// if its bigger then all others and smaller then the MAx then make that the Max
Serial.print(F);
Serial.println(" on the ");
}
}}}}
Serial.println(F); /// ment to give me the Kth largest number
delay(1000);
}
If speed isn't an issue you can take this approach (pseudocode):
current=inf,0
for i in [0,k):
max=-inf,0
for j in [0,n):
item=x[j],j
if item<current and item>max:
max=item
current=max
current will then contain the kth largest item, where an item is a pair of value and index.
The idea is simple. To find the first largest item, you just find the largest item. To find the second largest item, you find the largest item that isn't greater than your first largest item. To find the third largest item, you find the largest item that isn't greater than your second largest item. etc.
The only trick here is that since there can be duplicates, the items need to include both a value and an index to make them unique.
Here is how it might be implemented in C:
void loop()
{
int array[] = {1,2,3,4,5,6,7,8,9,10};
int n = 10;
int k = 6; //// just for testing putting the value as a constant
int c = n; // start with current index being past the end of the array
// to indicate that there is no current index.
for (int x = 1; x<=k; x++) {
int m = -1; // start with the max index being before the beginning of
// the array to indicate there is no max index
for (int p=0; p<n; p++) {
int ap = array[p];
// if this item is less than current
if (c==n || ap<array[c] || (ap==array[c] && p<c)) {
// if this item is greater than max
if (m<0 || ap>array[m] || (ap==array[m] && p>m)) {
// make this item be the new max
m = p;
}
}
}
// update current to be the max
c = m;
}
Serial.println(array[c]); /// ment to give me the Kth largest number
delay(1000);
}
In the C version, I just keep track of the current and max indices, since I can always get the current and max values by looking in the array.
An interesting interview question that a colleague of mine uses:
Suppose that you are given a very long, unsorted list of unsigned 64-bit integers. How would you find the smallest non-negative integer that does not occur in the list?
FOLLOW-UP: Now that the obvious solution by sorting has been proposed, can you do it faster than O(n log n)?
FOLLOW-UP: Your algorithm has to run on a computer with, say, 1GB of memory
CLARIFICATION: The list is in RAM, though it might consume a large amount of it. You are given the size of the list, say N, in advance.
If the datastructure can be mutated in place and supports random access then you can do it in O(N) time and O(1) additional space. Just go through the array sequentially and for every index write the value at the index to the index specified by value, recursively placing any value at that location to its place and throwing away values > N. Then go again through the array looking for the spot where value doesn't match the index - that's the smallest value not in the array. This results in at most 3N comparisons and only uses a few values worth of temporary space.
# Pass 1, move every value to the position of its value
for cursor in range(N):
target = array[cursor]
while target < N and target != array[target]:
new_target = array[target]
array[target] = target
target = new_target
# Pass 2, find first location where the index doesn't match the value
for cursor in range(N):
if array[cursor] != cursor:
return cursor
return N
Here's a simple O(N) solution that uses O(N) space. I'm assuming that we are restricting the input list to non-negative numbers and that we want to find the first non-negative number that is not in the list.
Find the length of the list; lets say it is N.
Allocate an array of N booleans, initialized to all false.
For each number X in the list, if X is less than N, set the X'th element of the array to true.
Scan the array starting from index 0, looking for the first element that is false. If you find the first false at index I, then I is the answer. Otherwise (i.e. when all elements are true) the answer is N.
In practice, the "array of N booleans" would probably be encoded as a "bitmap" or "bitset" represented as a byte or int array. This typically uses less space (depending on the programming language) and allows the scan for the first false to be done more quickly.
This is how / why the algorithm works.
Suppose that the N numbers in the list are not distinct, or that one or more of them is greater than N. This means that there must be at least one number in the range 0 .. N - 1 that is not in the list. So the problem of find the smallest missing number must therefore reduce to the problem of finding the smallest missing number less than N. This means that we don't need to keep track of numbers that are greater or equal to N ... because they won't be the answer.
The alternative to the previous paragraph is that the list is a permutation of the numbers from 0 .. N - 1. In this case, step 3 sets all elements of the array to true, and step 4 tells us that the first "missing" number is N.
The computational complexity of the algorithm is O(N) with a relatively small constant of proportionality. It makes two linear passes through the list, or just one pass if the list length is known to start with. There is no need to represent the hold the entire list in memory, so the algorithm's asymptotic memory usage is just what is needed to represent the array of booleans; i.e. O(N) bits.
(By contrast, algorithms that rely on in-memory sorting or partitioning assume that you can represent the entire list in memory. In the form the question was asked, this would require O(N) 64-bit words.)
#Jorn comments that steps 1 through 3 are a variation on counting sort. In a sense he is right, but the differences are significant:
A counting sort requires an array of (at least) Xmax - Xmin counters where Xmax is the largest number in the list and Xmin is the smallest number in the list. Each counter has to be able to represent N states; i.e. assuming a binary representation it has to have an integer type (at least) ceiling(log2(N)) bits.
To determine the array size, a counting sort needs to make an initial pass through the list to determine Xmax and Xmin.
The minimum worst-case space requirement is therefore ceiling(log2(N)) * (Xmax - Xmin) bits.
By contrast, the algorithm presented above simply requires N bits in the worst and best cases.
However, this analysis leads to the intuition that if the algorithm made an initial pass through the list looking for a zero (and counting the list elements if required), it would give a quicker answer using no space at all if it found the zero. It is definitely worth doing this if there is a high probability of finding at least one zero in the list. And this extra pass doesn't change the overall complexity.
EDIT: I've changed the description of the algorithm to use "array of booleans" since people apparently found my original description using bits and bitmaps to be confusing.
Since the OP has now specified that the original list is held in RAM and that the computer has only, say, 1GB of memory, I'm going to go out on a limb and predict that the answer is zero.
1GB of RAM means the list can have at most 134,217,728 numbers in it. But there are 264 = 18,446,744,073,709,551,616 possible numbers. So the probability that zero is in the list is 1 in 137,438,953,472.
In contrast, my odds of being struck by lightning this year are 1 in 700,000. And my odds of getting hit by a meteorite are about 1 in 10 trillion. So I'm about ten times more likely to be written up in a scientific journal due to my untimely death by a celestial object than the answer not being zero.
As pointed out in other answers you can do a sort, and then simply scan up until you find a gap.
You can improve the algorithmic complexity to O(N) and keep O(N) space by using a modified QuickSort where you eliminate partitions which are not potential candidates for containing the gap.
On the first partition phase, remove duplicates.
Once the partitioning is complete look at the number of items in the lower partition
Is this value equal to the value used for creating the partition?
If so then it implies that the gap is in the higher partition.
Continue with the quicksort, ignoring the lower partition
Otherwise the gap is in the lower partition
Continue with the quicksort, ignoring the higher partition
This saves a large number of computations.
To illustrate one of the pitfalls of O(N) thinking, here is an O(N) algorithm that uses O(1) space.
for i in [0..2^64):
if i not in list: return i
print "no 64-bit integers are missing"
Since the numbers are all 64 bits long, we can use radix sort on them, which is O(n). Sort 'em, then scan 'em until you find what you're looking for.
if the smallest number is zero, scan forward until you find a gap. If the smallest number is not zero, the answer is zero.
For a space efficient method and all values are distinct you can do it in space O( k ) and time O( k*log(N)*N ). It's space efficient and there's no data moving and all operations are elementary (adding subtracting).
set U = N; L=0
First partition the number space in k regions. Like this:
0->(1/k)*(U-L) + L, 0->(2/k)*(U-L) + L, 0->(3/k)*(U-L) + L ... 0->(U-L) + L
Find how many numbers (count{i}) are in each region. (N*k steps)
Find the first region (h) that isn't full. That means count{h} < upper_limit{h}. (k steps)
if h - count{h-1} = 1 you've got your answer
set U = count{h}; L = count{h-1}
goto 2
this can be improved using hashing (thanks for Nic this idea).
same
First partition the number space in k regions. Like this:
L + (i/k)->L + (i+1/k)*(U-L)
inc count{j} using j = (number - L)/k (if L < number < U)
find first region (h) that doesn't have k elements in it
if count{h} = 1 h is your answer
set U = maximum value in region h L = minimum value in region h
This will run in O(log(N)*N).
I'd just sort them then run through the sequence until I find a gap (including the gap at the start between zero and the first number).
In terms of an algorithm, something like this would do it:
def smallest_not_in_list(list):
sort(list)
if list[0] != 0:
return 0
for i = 1 to list.last:
if list[i] != list[i-1] + 1:
return list[i-1] + 1
if list[list.last] == 2^64 - 1:
assert ("No gaps")
return list[list.last] + 1
Of course, if you have a lot more memory than CPU grunt, you could create a bitmask of all possible 64-bit values and just set the bits for every number in the list. Then look for the first 0-bit in that bitmask. That turns it into an O(n) operation in terms of time but pretty damned expensive in terms of memory requirements :-)
I doubt you could improve on O(n) since I can't see a way of doing it that doesn't involve looking at each number at least once.
The algorithm for that one would be along the lines of:
def smallest_not_in_list(list):
bitmask = mask_make(2^64) // might take a while :-)
mask_clear_all (bitmask)
for i = 1 to list.last:
mask_set (bitmask, list[i])
for i = 0 to 2^64 - 1:
if mask_is_clear (bitmask, i):
return i
assert ("No gaps")
Sort the list, look at the first and second elements, and start going up until there is a gap.
We could use a hash table to hold the numbers. Once all numbers are done, run a counter from 0 till we find the lowest. A reasonably good hash will hash and store in constant time, and retrieves in constant time.
for every i in X // One scan Θ(1)
hashtable.put(i, i); // O(1)
low = 0;
while (hashtable.get(i) <> null) // at most n+1 times
low++;
print low;
The worst case if there are n elements in the array, and are {0, 1, ... n-1}, in which case, the answer will be obtained at n, still keeping it O(n).
You can do it in O(n) time and O(1) additional space, although the hidden factor is quite large. This isn't a practical way to solve the problem, but it might be interesting nonetheless.
For every unsigned 64-bit integer (in ascending order) iterate over the list until you find the target integer or you reach the end of the list. If you reach the end of the list, the target integer is the smallest integer not in the list. If you reach the end of the 64-bit integers, every 64-bit integer is in the list.
Here it is as a Python function:
def smallest_missing_uint64(source_list):
the_answer = None
target = 0L
while target < 2L**64:
target_found = False
for item in source_list:
if item == target:
target_found = True
if not target_found and the_answer is None:
the_answer = target
target += 1L
return the_answer
This function is deliberately inefficient to keep it O(n). Note especially that the function keeps checking target integers even after the answer has been found. If the function returned as soon as the answer was found, the number of times the outer loop ran would be bound by the size of the answer, which is bound by n. That change would make the run time O(n^2), even though it would be a lot faster.
Thanks to egon, swilden, and Stephen C for my inspiration. First, we know the bounds of the goal value because it cannot be greater than the size of the list. Also, a 1GB list could contain at most 134217728 (128 * 2^20) 64-bit integers.
Hashing part
I propose using hashing to dramatically reduce our search space. First, square root the size of the list. For a 1GB list, that's N=11,586. Set up an integer array of size N. Iterate through the list, and take the square root* of each number you find as your hash. In your hash table, increment the counter for that hash. Next, iterate through your hash table. The first bucket you find that is not equal to it's max size defines your new search space.
Bitmap part
Now set up a regular bit map equal to the size of your new search space, and again iterate through the source list, filling out the bitmap as you find each number in your search space. When you're done, the first unset bit in your bitmap will give you your answer.
This will be completed in O(n) time and O(sqrt(n)) space.
(*You could use use something like bit shifting to do this a lot more efficiently, and just vary the number and size of buckets accordingly.)
Well if there is only one missing number in a list of numbers, the easiest way to find the missing number is to sum the series and subtract each value in the list. The final value is the missing number.
int i = 0;
while ( i < Array.Length)
{
if (Array[i] == i + 1)
{
i++;
}
if (i < Array.Length)
{
if (Array[i] <= Array.Length)
{//SWap
int temp = Array[i];
int AnoTemp = Array[temp - 1];
Array[temp - 1] = temp;
Array[i] = AnoTemp;
}
else
i++;
}
}
for (int j = 0; j < Array.Length; j++)
{
if (Array[j] > Array.Length)
{
Console.WriteLine(j + 1);
j = Array.Length;
}
else
if (j == Array.Length - 1)
Console.WriteLine("Not Found !!");
}
}
Here's my answer written in Java:
Basic Idea:
1- Loop through the array throwing away duplicate positive, zeros, and negative numbers while summing up the rest, getting the maximum positive number as well, and keep the unique positive numbers in a Map.
2- Compute the sum as max * (max+1)/2.
3- Find the difference between the sums calculated at steps 1 & 2
4- Loop again from 1 to the minimum of [sums difference, max] and return the first number that is not in the map populated in step 1.
public static int solution(int[] A) {
if (A == null || A.length == 0) {
throw new IllegalArgumentException();
}
int sum = 0;
Map<Integer, Boolean> uniqueNumbers = new HashMap<Integer, Boolean>();
int max = A[0];
for (int i = 0; i < A.length; i++) {
if(A[i] < 0) {
continue;
}
if(uniqueNumbers.get(A[i]) != null) {
continue;
}
if (A[i] > max) {
max = A[i];
}
uniqueNumbers.put(A[i], true);
sum += A[i];
}
int completeSum = (max * (max + 1)) / 2;
for(int j = 1; j <= Math.min((completeSum - sum), max); j++) {
if(uniqueNumbers.get(j) == null) { //O(1)
return j;
}
}
//All negative case
if(uniqueNumbers.isEmpty()) {
return 1;
}
return 0;
}
As Stephen C smartly pointed out, the answer must be a number smaller than the length of the array. I would then find the answer by binary search. This optimizes the worst case (so the interviewer can't catch you in a 'what if' pathological scenario). In an interview, do point out you are doing this to optimize for the worst case.
The way to use binary search is to subtract the number you are looking for from each element of the array, and check for negative results.
I like the "guess zero" apprach. If the numbers were random, zero is highly probable. If the "examiner" set a non-random list, then add one and guess again:
LowNum=0
i=0
do forever {
if i == N then leave /* Processed entire array */
if array[i] == LowNum {
LowNum++
i=0
}
else {
i++
}
}
display LowNum
The worst case is n*N with n=N, but in practice n is highly likely to be a small number (eg. 1)
I am not sure if I got the question. But if for list 1,2,3,5,6 and the missing number is 4, then the missing number can be found in O(n) by:
(n+2)(n+1)/2-(n+1)n/2
EDIT: sorry, I guess I was thinking too fast last night. Anyway, The second part should actually be replaced by sum(list), which is where O(n) comes. The formula reveals the idea behind it: for n sequential integers, the sum should be (n+1)*n/2. If there is a missing number, the sum would be equal to the sum of (n+1) sequential integers minus the missing number.
Thanks for pointing out the fact that I was putting some middle pieces in my mind.
Well done Ants Aasma! I thought about the answer for about 15 minutes and independently came up with an answer in a similar vein of thinking to yours:
#define SWAP(x,y) { numerictype_t tmp = x; x = y; y = tmp; }
int minNonNegativeNotInArr (numerictype_t * a, size_t n) {
int m = n;
for (int i = 0; i < m;) {
if (a[i] >= m || a[i] < i || a[i] == a[a[i]]) {
m--;
SWAP (a[i], a[m]);
continue;
}
if (a[i] > i) {
SWAP (a[i], a[a[i]]);
continue;
}
i++;
}
return m;
}
m represents "the current maximum possible output given what I know about the first i inputs and assuming nothing else about the values until the entry at m-1".
This value of m will be returned only if (a[i], ..., a[m-1]) is a permutation of the values (i, ..., m-1). Thus if a[i] >= m or if a[i] < i or if a[i] == a[a[i]] we know that m is the wrong output and must be at least one element lower. So decrementing m and swapping a[i] with the a[m] we can recurse.
If this is not true but a[i] > i then knowing that a[i] != a[a[i]] we know that swapping a[i] with a[a[i]] will increase the number of elements in their own place.
Otherwise a[i] must be equal to i in which case we can increment i knowing that all the values of up to and including this index are equal to their index.
The proof that this cannot enter an infinite loop is left as an exercise to the reader. :)
The Dafny fragment from Ants' answer shows why the in-place algorithm may fail. The requires pre-condition describes that the values of each item must not go beyond the bounds of the array.
method AntsAasma(A: array<int>) returns (M: int)
requires A != null && forall N :: 0 <= N < A.Length ==> 0 <= A[N] < A.Length;
modifies A;
{
// Pass 1, move every value to the position of its value
var N := A.Length;
var cursor := 0;
while (cursor < N)
{
var target := A[cursor];
while (0 <= target < N && target != A[target])
{
var new_target := A[target];
A[target] := target;
target := new_target;
}
cursor := cursor + 1;
}
// Pass 2, find first location where the index doesn't match the value
cursor := 0;
while (cursor < N)
{
if (A[cursor] != cursor)
{
return cursor;
}
cursor := cursor + 1;
}
return N;
}
Paste the code into the validator with and without the forall ... clause to see the verification error. The second error is a result of the verifier not being able to establish a termination condition for the Pass 1 loop. Proving this is left to someone who understands the tool better.
Here's an answer in Java that does not modify the input and uses O(N) time and N bits plus a small constant overhead of memory (where N is the size of the list):
int smallestMissingValue(List<Integer> values) {
BitSet bitset = new BitSet(values.size() + 1);
for (int i : values) {
if (i >= 0 && i <= values.size()) {
bitset.set(i);
}
}
return bitset.nextClearBit(0);
}
def solution(A):
index = 0
target = []
A = [x for x in A if x >=0]
if len(A) ==0:
return 1
maxi = max(A)
if maxi <= len(A):
maxi = len(A)
target = ['X' for x in range(maxi+1)]
for number in A:
target[number]= number
count = 1
while count < maxi+1:
if target[count] == 'X':
return count
count +=1
return target[count-1] + 1
Got 100% for the above solution.
1)Filter negative and Zero
2)Sort/distinct
3)Visit array
Complexity: O(N) or O(N * log(N))
using Java8
public int solution(int[] A) {
int result = 1;
boolean found = false;
A = Arrays.stream(A).filter(x -> x > 0).sorted().distinct().toArray();
//System.out.println(Arrays.toString(A));
for (int i = 0; i < A.length; i++) {
result = i + 1;
if (result != A[i]) {
found = true;
break;
}
}
if (!found && result == A.length) {
//result is larger than max element in array
result++;
}
return result;
}
An unordered_set can be used to store all the positive numbers, and then we can iterate from 1 to length of unordered_set, and see the first number that does not occur.
int firstMissingPositive(vector<int>& nums) {
unordered_set<int> fre;
// storing each positive number in a hash.
for(int i = 0; i < nums.size(); i +=1)
{
if(nums[i] > 0)
fre.insert(nums[i]);
}
int i = 1;
// Iterating from 1 to size of the set and checking
// for the occurrence of 'i'
for(auto it = fre.begin(); it != fre.end(); ++it)
{
if(fre.find(i) == fre.end())
return i;
i +=1;
}
return i;
}
Solution through basic javascript
var a = [1, 3, 6, 4, 1, 2];
function findSmallest(a) {
var m = 0;
for(i=1;i<=a.length;i++) {
j=0;m=1;
while(j < a.length) {
if(i === a[j]) {
m++;
}
j++;
}
if(m === 1) {
return i;
}
}
}
console.log(findSmallest(a))
Hope this helps for someone.
With python it is not the most efficient, but correct
#!/usr/bin/env python3
# -*- coding: UTF-8 -*-
import datetime
# write your code in Python 3.6
def solution(A):
MIN = 0
MAX = 1000000
possible_results = range(MIN, MAX)
for i in possible_results:
next_value = (i + 1)
if next_value not in A:
return next_value
return 1
test_case_0 = [2, 2, 2]
test_case_1 = [1, 3, 44, 55, 6, 0, 3, 8]
test_case_2 = [-1, -22]
test_case_3 = [x for x in range(-10000, 10000)]
test_case_4 = [x for x in range(0, 100)] + [x for x in range(102, 200)]
test_case_5 = [4, 5, 6]
print("---")
a = datetime.datetime.now()
print(solution(test_case_0))
print(solution(test_case_1))
print(solution(test_case_2))
print(solution(test_case_3))
print(solution(test_case_4))
print(solution(test_case_5))
def solution(A):
A.sort()
j = 1
for i, elem in enumerate(A):
if j < elem:
break
elif j == elem:
j += 1
continue
else:
continue
return j
this can help:
0- A is [5, 3, 2, 7];
1- Define B With Length = A.Length; (O(1))
2- initialize B Cells With 1; (O(n))
3- For Each Item In A:
if (B.Length <= item) then B[Item] = -1 (O(n))
4- The answer is smallest index in B such that B[index] != -1 (O(n))