numsand an integerthe number of non-emptysubarrayswith a sumgoal. Asubarrayis a contiguous part of the array. Example 1: Input:nums = [1,0,1,0,1], goal = 2Output:4Explanation:The 4 subarrays are bolded and unde
public int numSubarraysWithSum(int[]A,intS) { return atMost(A,S) - atMost(A,S- 1); } private int atMost(int[]A,intS) { if (S< 0) return 0; int res = 0, n =A.length; for (inti= 0,j= 0;j<n; ++j) {S-=A[j]; while (S< 0)S+=A[i++]; res += j - i...
public int numSubarraysWithSum(int[] A, int S) { if (A.length == 0) return 0; int result = 0; int[] sumOne = new int[A.length]; int st = 0; int ed = 0; sumOne[0] = A[0] == 1 ? 1 : 0; for (int i = 1; i < A.length; i++) { sumOne[i] = A[i] =...
public:intnumSubarraysWithSum(vector<int>& A,intS) {intres =0,sum=0, left =0, n = A.size();for(inti =0; i < n; ++i) {sum+= A[i];while(left < i &∑> S)sum-= A[left++];if(sum< S)continue;if(sum== S) ++res;for(intj = left; j < i && A[j] ==0; ++j)...
D[0] exists in our deque, it means that before B[i], we didn’t find a subarray whose sum at least K. B[i] is the first prefix sum that valid this condition. In other words, A[D[0]] ~ A[i-1] is the shortest subarray starting at A[D[0]] with sum at least K. ...
第一种是prefix sum + 双指针。 例如Minimum Size Subarray Sum - LeetCode: Given an array of n positive integers and a positive integers, find the minimal length of a contiguous subarray of which the sum ≥s. If there isn't one, return 0 instead. 用O(N)时间得到一个prefix sum array,可...
Explanation: There are 7 subarrays with a sum divisible by K = 5:[4, 5, 0, -2, -3, 1], [5], [5, 0], [5, 0, -2, -3], [0], [0, -2, -3], [-2, -3]Note:数组最长30000。如果O(n^2) 那时间委实长久。1 <= A.length <= 30000-10000 <= A[i] <= 100002 <=...
给定一个正整数数组 nums和一个整数 k,返回 nums 中「好子数组」 的数目。 如果nums 的某个子数组中不同整数的个数恰好为 k,则称 nums 的这个连续、不一定不同的子数组为 「好子数组 」。 例如,[1,2,3,1,2] 中有3 个不同的整数:1,2,以及 3。 子数组 是数组的 连续 部分。 示例1: 输入:nums...
这题跟560. subarray sum equals k几乎一毛一样。 我想到了用map和sum - k 来实现ONE PASS,但是不知道怎么维护一个类似560题里面那个count的值了。正确答案是用Math.max每次对比当前maxLen 和 i - map.get(sum-k)。 思维难度还是有的。 这题leetcode收费了。我贴一个别人的答案。注意这里的put始终是put...
The question asks for three non-overlapping intervals with maximum sum of all 3 intervals. If the middle interval is [i, i+k-1], where k <= i <= n-2k, the left interval has to be in subrange [0, i-1], and the right interval is from subrange [i+k, n-1]. ...