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...
leetcode 930. Binary Subarrays With Sum This remains me of some 'subarray count' type problems….. classSolution{publicintnumSubarraysWithSum(int[] A,intS){int[] ps =newint[A.length +1]; ps[0] =1;intsum=0;intret=0;for(intv: A) { sum += v;if(sum - S >=0) { ret +=...
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] =...
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...
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. ...
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 <=...
【LeetCode】3186. Maximum Total Damage With Spell Casting 17:36 【LeetCode】3193. Count the Number of Inversions 46:21 【LeetCode】3097. Shortest Subarray With OR at Least K II 09:21 【LeetCode】 3097. Shortest Subarray With OR at Least K II 09:21 【LeetCode】3161. Block Placem...
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...
1186 Maximum Subarray Sum with One Deletion 删除一次得到子数组最大和 Description: Given an array of integers, return the maximum sum for a non-empty subarray (contiguous elements) with at most one element deletion. In other words, you want to choose a subarray and optionally delete one element...
https://leetcode.com/problems/subarray-sums-divisible-by-k/ 解题方法: IfSum[0 to i] % K == Sum[0 to j]andi < j, thenSum[i + 1 to j] % K == 0, so each time we find out an existing mod result, it means we find out a sub-array which sum of this sub-array is divisib...