分析:矩阵每一行,计算prefix sum,然后对于任意两列,计算行的累加和,问题就转换为了Subarray Sum Equals K time O(n3n3) space O(n) classSolution {public:intnumSubmatrixSumTarget(vector<vector<int>>& matrix,inttarget) {intm = matrix.size(), n = matrix[0].size();for(inti =0; i < m; ++i...
无独有偶,leetcode最近的contest中出了一题非常类似的问题,本质上依然是最长target sum的子数组,当时的想法也是陷入了范式中,错误的认为子数组的问题都可以使用dp或者滑动数组进行求解,导致走入了死胡同。 本题也是一条变种题,可见如何直接出subarray of target sum 是没有多大的价值的,这个题目直接问的话,就非常简...
本题也是一条变种题,可见如何直接出subarray of target sum 是没有多大的价值的,这个题目直接问的话,就非常简单直白,很容易就会想到使用hashmap去求解,但是如果进行一下包装,就未必能想到了。 总之,碰到subarray的问题,不仅要能联想到dp/sliding window,还要有意识去想想其他的解法。 publicintlongestWPI(int[]hours...
Given a binary array arr[] and an integer target, return the number of non-empty subarrays with a sum equal to the target. Note: A subarray is the contiguous part of the array. Examples: Input: arr[] = [1, 0, 1, 0, 1], target = 2 Output: 4 Explanation: There are four subar...
leetcode 209 长度最小的子数组 minimum-size-subarray-sum【ct】 思路:设置start和end都为数组第0个元素 , sum为数组的第一个元素,如果和>=target,就start++ ,sum-=nums[start] 否则end-- ,sum+=nums[end] 每次循环中去计算minLength,并返回 etc 数组 leetcode 560. Subarray Sum Equals K 动态规划DP...
Given a binary array, find the maximum length of a contiguous subarray with equal number of 0 and 1. 这里我们所关心的状态其实是0和1出现个数的差异。如果A[0 .. i] 中0和1出现个数差异与A[0 .. j]中0和1出现个数差异相等的话,那么A[i+1 .. j]中0和1的个数应当相等(assume i < j)...
{ start, end }; } } return new int[] { -1, -1 }; // Return -1 if no subarray is found } public static void main(String[] args) { int[] arr = { 1, 2, 3, 7, 5 }; int targetSum = 12; int[] result = findSubarrayWithGivenSum(arr, targetSum); if (result[0] != ...
2.Smallest Subarray with a given sum(easy) 2.1 问题描述 Given an array of positive integers nums and a positive integer target, return the minimal length of a 「contiguous subarray」 [numsl, numsl+1, ..., numsr-1, numsr] of which the sum is greater than or equal to target. If th...
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How to find total number of subarrays with sum atmost k? Constrains : -1e4 <= a[i] <= 1e4 1 <= n <= 1e5 +3 shivam565 3 years ago 11 anujdhillxn 3 years ago,#| +8 Convert the statement to it's equivalent in terms of prefix sums i.eprer−prel<=kprer−prel<=k. ...