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

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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;
    }
}


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