Given an array of n elements, find the longest subsequence of k (k<=n) elements which contain only elements greater than or equal to k.
For example:
Array: [3, 1, 5, 1, 6]
Output: 3 (the subsequence 3, 5, 6 meets the requirements
because 3,5,6 are all greater than 3.)
How can I do this in O(n)?
I tried my best, but all I can come up with is O(n logn).
Related
I am going through this program mentioned here.
Given an array arr[] of N integers. The task is to count the total number of subarrays of the given array such that the difference between the consecutive elements in the subarrays is one. That is, for any index i in the subarrays, arr[i+1] – arr[i] = 1.
Examples:
Input : arr[] = {1, 2, 3}
Output : 3
The subarrays are {1, 2}. {2, 3} and {1, 2, 3}
Input : arr[] = {1, 2, 3, 5, 6, 7}
Output : 6
Efficient Approach: An efficient approach is to observe that in an array of length say K, the total number of subarrays of size greater than 1 = (K)*(K-1)/2.
So, the idea is to traverse the array by using two pointers to calculate subarrays with consecutive elements in a window of maximum length and then calculate all subarrays in that window using the above formula.
Below is the step-by-step algorithm:
Take two pointers to say fast and slow, for maintaining a window of consecutive elements.
Start traversing the array.
If elements differ by 1 increment only the fast pointer.
Else, calculate the length of the current window between the indexes fast and slow.
My question is about the statement An efficient approach is to observe that in an array of length say K, the total number of subarrays of size greater than 1 = (K)*(K-1)/2 How this formula (K)*(K-1)/2 is derived?
The number of subarrays of size 1 is K.
The number of subarrays of size 2 is K-1.
( We need to select subarrays of size 2, hence we can have pair with the indices (0,1), (1,2), .......(K-1,K). In total we can have K-1 such pairs)
The number of subarrays of size 3 is K-2.
...
...
The number of subarrays of size K is 1
So, the number of subarrays of size greater than 1 is
= K-1 + K-2 + K-3+ ...... 1
= (K + K-1 + K-2 + K-3+ ...... 1) - K //adding and removing K
= (K(K+1)/2) - K
= K(K-1)/2
we are given an array of size n (n is even), we have to divide it into two equal-sized subarrays array1 and array2, sized n/2 each such that product of all the numbers of array1 equals to the product of all the numbers in array2.
Given array:
arr = [2, 4, 5, 12, 15, 18]
solution:
array2 = [4, 5, 18]
array1 = [2, 12, 15]
Explanation:
product of all elements in array1 is 360
product of all elements in array2 is 360.
This problem might be soved using dynamic programming. You need to get subset of size n/2 with product equal to p = sqrt(overall_product). Note that there is no solution when overall_product is not exact square.
Recursion might look like
solution(p, n/2, arr) = choose valid solution from
solution(p / arr[i], n/2-1, arr without arr[i])
return true for arguments (1,0,...)
use memoization or table to solve problem with merely large n values.
I have an unsorted array of size n and I need to find k-1 divisors so every subset is of the same size (like after the array is sorted).
I have seen this question with k-1=3. I guess I need the median of medians and this is will take o(n). But I think we should do it k times so o(nk).
I would like to understand why it would take o(n logk).
For example: I have an unsorted array with integers and I want find the k'th divisors which is the k-1 integers that split the array into k (same sized) subarrays according to their values.
If I have [1, 13, 6, 7, 81, 9, 10, 11] the 3=k dividers is [7 ,11] spliting to [1 6, 9 10 13 81] where every subset is big as 2 and equal.
You can use a divide-and-conquer approach. First, find the (k-1)/2th divider using the median-of-medians algorithm. Next, use the selected element to partition the list into two sub-lists. Repeat the algorithm on each sub-list to find the remaining dividers.
The maximum recursion depth is O(log k) and the total cost across all sub-lists at each level is O(n), so this is an O(n log k) algorithm.
Given an array A of size N, we construct a list containing all possible subarrays of A in descending order.
Two subarrays B and C are compare by padding zeroes until both are of size N. Then, we compare the two subarrays element by element and return as soon as a point of difference is observed.
We are given multiple queries where given x we have to find the maximum element in the xth subarray sorted according to the order given above.
For example, if the array A is [3, 1, 2, 4]; then the sorted subarrays will be:
[4]
[3, 1, 2, 4]
[3, 1, 2]
[3, 1]
[3]
[2, 4]
[2]
[1, 2, 4]
[1, 2]
[1]
A query where x = 3 corresponds to finding the maximum element in the subarray [3, 1, 2]; so here the answer would be 3.
Since the number of queries are large (of the order of 10^5) and the number of elements in the array can also be large (of the order of 10^5), we would need to do some preprocessing to answer each query in O(1) or O(log N) or O(sqrt N) time. I can't seem to figure out how to do this. I have solved it for when the array contains unique elements, however how could we do this for when the array contains repetitions? Is there any data structure which could help in storing the required information?
Build suffix array in back order for this array (consider it like string)
For every entry store it's length and cumulative count (sum of lengths from the beginning of suffix array)
For query find needed index by binary search for cumulative counts, and get needed prefix of found suffix
For your examples suffixes with cumul.counts are
4 (0)
3124 (1)
34 (5)
124 (7)
query 3 finds entry 3124 (1<=3<5), and gets 3-1=2-nd (by length) prefix = 312
Does anyone know an Algorithm that sorts k-approximately an array?
We were asked to find and Algorithm for k-approximate sorting, and it should run in O(n log(n/k)). but I can't seem to find any.
K-approx. sorting means that an array and any 1 <= i <= n-k such that sum a[j] <= sum a[j] i<=j<= i+k-1 i+1<=j<= i+k
I know I'm very late to the question ... But under the assumption that k is some approximation value between 0 and 1 (when 0 is completely unsorted and 1 is perfectly sorted) surely the answer to this is quicksort (or mergesort).
Consider the following array:
[4, 6, 9, 1, 10, 8, 2, 7, 5, 3]
Let's say this array is 'unsorted' - now apply one iteration of quicksort to this array with the (length[array]/2)th element as a pivot: length[array]/2 = 5. So the 5th element is our pivot (i.e. 8):
[4, 6, 2, 1, 3, 9, 7, 10, 8]
Now this is array is not sorted - but it is more sorted than one iteration ago, i.e. its approximately sorted but for a low approximation, i.e. a low value of k. Repeat this step again on the two halves of the array and it becomes more sorted. As k increases towards 1 - i.e. perfectly sorted - the complexity becomes O(N log(N/1)) = O(N log(N)).