Running Sum of 1d Array LeetCode Solution

Problem – Running Sum of 1d Array LeetCode Solution

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].

Example 2:

Input: nums = [1,1,1,1,1]
Output: [1,2,3,4,5]
Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1].

Example 3:

Input: nums = [3,1,2,10,1]
Output: [3,4,6,16,17]

Constraints:

  • 1 <= nums.length <= 1000
  • -10^6 <= nums[i] <= 10^6

Running Sum of 1d Array LeetCode Solution in Java

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

Running Sum of 1d Array LeetCode Solution in C++

vector<int> runningSum(vector<int>& nums) {
        int i = 1;
        while (i<nums.size()){
            nums[i]+=nums[i-1];
            i++;
        }
        return nums;
    }

Running Sum of 1d Array LeetCode Solution in Python

 def runningSum(self, nums):
        i = 1
        while i<len(nums):
            nums[i]+=nums[i-1]
            i+=1
        return nums
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