#include<stdio.h>// Function to print subarraysvoidprint(intarr1[],inti,intj){// Print subarray elements within range [i..j]printf("[%d..%d] -- { ",i,j);for(intk=i;k<=j;k++){printf("%d ",arr1[k]);}printf("}\n");}// Function to find subarrays with a given sumvoid...
1publicclassSubarrayWithGivenSum {2publicstaticint[] findSubarrayWithGivenSum(int[] arr,intsum) {3int[] range =newint[2];4range[0] = -1; range[1] = -1;5if(arr ==null|| arr.length == 0 || sum < 0) {6returnrange;7}8intstart = 0;9intcurrSum = arr[0];10for(intend = ...
Given anarrayof integers and a target sum, find a continuous subarray in the array that adds up to the given sum. The problem can be divided into two main variants: Subarray with Positive Numbers: The array contains only positive numbers. Subarray with Mixed Numbers: The array contains both ...
You are given an integer arraynums. The range of a subarray ofnumsis the difference between the largest and smallest element in the subarray. Returnthe sum of all subarray ranges ofnums. A subarray is a contiguous non-empty sequence of elements within an array. Example 1: Input: nums = [...
Given an array nums and a target value k, find the maximum length of a subarray that sums to k. If there isn't one, return 0 instead. Note:The sum of the entire nums array is guaranteed to fit within the 32-bit signed integer range. Example 1: Input: nums = [1, -1, 5, -2...
[leetcode] 1508. Range Sum of Sorted Subarray Sums Description Given the array nums consisting of n positive integers. You computed the sum of all non-empty continous subarrays from the array and then sort them in non-decreasing order, creating a new array of n * (n + 1) / 2 numbers...
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[leetcode] 209. Minimum Size Subarray Sum Description Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn’t one, return 0 instead.
1508. Range Sum of Sorted Subarray Sums 难度:m BF: class Solution: def rangeSum(self, nums: List[int], n: int, left: int, right: int) -> int: helper = [] for i in range(n): helper.append(nums[i]) for j in range(i+1,n): ...