### LeetCode - Running Sum of 1d Array Solution - The Coding Shala

Home >> LeetCode >> running sum of 1d array

In this post, we will learn how to solve LeetCode's Running Sum of 1d Array problem and its solution in Java.

## Running Sum of 1d Array

Given an array nums. We define a running sum of an array as runningSum[i] = sum(nums[0]…nums[i]). Return the running sum of nums.

Example 1:

Input: nums = [1,2,3,4]

Output: [1,3,6,10]

Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].

## Running Sum of 1d Array Java Solution

**Approach 1:**

Creating a new Array. consider given array is A and we will create a new Array B then:

B[i] = A[0] + A[1] + .... + A[i] or or we can write as B[i] = A[i] + B[i-1].

Time: O(n).

Space: O(n).

Java Program:

class Solution { public int[] runningSum(int[] nums) { int len = nums.length; int[] ans = new int[len]; ans[0] = nums[0]; for(int i=1;i<len;i++){ ans[i] = ans[i-1] + nums[i]; } return ans; } }

**Approach 2:**

We can modify the same array.

Time: O(n).

Space: O(1).

Java Program:

class Solution { public int[] runningSum(int[] nums) { for(int i=1; i<nums.length; i++) { nums[i] += nums[i-1]; } return nums; } }

**Other Posts You May Like**- LeetCode - Next greater element
- LeetCode - Climbing stairs
- LeetCode - swap nodes
- LeetCode - degree of an array
- LeetCode - contains duplicate

## Comments

## Post a Comment