Approach: Prefix sum

def runningSum(nums):
    s = 0; out = []
    for x in nums:
        s += x; out.append(s)
    return out

Complexity: O(n) time, O(1) space.

Full Solution — all languages

class Solution:
    def runningSum(self, nums):
        s = 0
        out = []
        for x in nums:
            s += x
            out.append(s)
        return out

import java.util.*;
class Solution {
    public int[] runningSum(int[] nums) {
        int[] out = new int[nums.length];
        int s = 0;
        for (int i = 0; i < nums.length; i++) {
            s += nums[i];
            out[i] = s;
        }
        return out;
    }
}

#include <bits/stdc++.h> using namespace std;
class Solution {
    public: vector<int> runningSum(vector<int>& nums) {
        vector<int> out(nums.size());
        int s = 0;
        for (size_t i = 0; i < nums.size(); i++) {
            s += nums[i];
            out[i] = s;
        }
        return out;
    }
};

/**
 * @param {number[]} nums
 * @return {number[]}
 */ var runningSum = function(nums) {
    const out = [];
    let s = 0;
    for (const x of nums) {
        s += x;
        out.push(s);
    }
    return out;
};

Running Sum of 1d Array

Example 1

Example 2

Constraints