# Running Sum of 1d Array – Leetcode Solution

In this post, we are going to solve the 1480. Running Sum of 1d Array problem of Leetcode. This problem 1480. Running Sum of 1d Array is a Leetcode easy level problem. Let’s see the code, 1480. Running Sum of 1d Array – Leetcode Solution.

## Problem

Given an array `nums`. We define a running sum of an array as `runningSum[i] = sum(nums[0]â€¦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`

Now, let’s see the code of 1480. Running Sum of 1d Array – Leetcode Solution.

# Running Sum of 1d Array – Leetcode Solution

### 1480. Running Sum of 1d Array – Solution in Java

```class Solution {
public int[] runningSum(int[] nums) {
int[] ans = new int[nums.length];

ans[0] = nums[0];
for(int i=1;i<nums.length;i++){
ans[i] = nums[i]+ans[i-1];
}
return ans;
}
}```

### 1480. Running Sum of 1d Array – Solution in C++

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

### 1480. Running Sum of 1d Array– Solution in Python

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

Note: This problem 1480. Running Sum of 1d Array is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.