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# 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…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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