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- I verified this with certain small values though for n=1024 and k=16 I got different answer. Following is the java implementation of the problem. It has three independent part: Divide S in K parts. The parts can have value 0 and order matters. i.e. 10 = 0+5+5+0 and 10=5+5+0+0 are different; Divide S in K parts.
- A rather straightforward approach: Iterate over the maximum possible subarray sum, say [math]i[/math]. Now, you can define [math]dp(j, k, l)[/math] as the minimum possible subarray sum if you need to partition subarray [math]A[0, j][/math] in [mat...
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Partitioning An Array Into K Subarrays To Minimize The Maximum Difference Return the count of distinct elements in all windows (or in all sub-arrays) of size k. NET array or ListGatech cs 2110 reddit
- class Solution { public int subarraysWithKDistinct(int[] A, int K) { int count = 0; int[] table = new int[A.length+1]; for(int i = 0; i < A.length; ++i) { Arrays.fill(table, 0); int size = 0; for(int j = i; j < A.length; ++j) { table[A[j]]++; if(table[A[j]] == 1) { size++; } if(size == K) { count++; } } } return count; } }
- Given an array of integers and an integer k, you need to find the total number of continuous subarrays whose sum equals to k. Example 1: Input:nums = [1,1,1], k = 2 Output: 2 Constraints: The length of the array is in range [1, 20,000]. The range of numbers in the array is [-1000, 1000] and the range of the integer k is [-1e7, 1e7].
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