1480. Running Sum of 1d Array

Description

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

Solutions

Solution 1: Prefix Sum

We directly traverse the array. For the current element $nums[i]$, we add it with the prefix sum $nums[i-1]$ to get the prefix sum $nums[i]$ of the current element.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

Python Code
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class Solution:
    def runningSum(self, nums: List[int]) -> List[int]:
        return list(accumulate(nums))

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

C++ Code
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class Solution {
public:
    vector<int> runningSum(vector<int>& nums) {
        for (int i = 1; i < nums.size(); ++i) nums[i] += nums[i - 1];
        return nums;
    }
};

Go Code
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func runningSum(nums []int) []int {
	for i := 1; i < len(nums); i++ {
		nums[i] += nums[i-1]
	}
	return nums
}

TypeScript Code
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function runningSum(nums: number[]): number[] {
    for (let i = 1; i < nums.length; ++i) {
        nums[i] += nums[i - 1];
    }
    return nums;
}

C# Code
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public class Solution {
    public int[] RunningSum(int[] nums) {
        for (int i = 1; i < nums.Length; ++i) {
            nums[i] += nums[i - 1];
        }
        return nums;
    }
}

PHP Code
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class Solution {
    /**
     * @param Integer[] $nums
     * @return Integer[]
     */
    function runningSum($nums) {
        for ($i = 1; $i < count($nums); $i++) {
            $nums[$i] += $nums[$i - 1];
        }
        return $nums;
    }
}