# Leetcode 1480: Running Sum of 1d Array Solution

Here, we will see how to solve Running Sum of 1d Array Solution of leet code 1480 problem.

You are given an arrayÂ `nums`. We define a running sum of an array asÂ `runningSum[i] = sum(nums[0]â€¦nums[i])`.

You have to return the running sum ofÂ `nums`.

Example 1:

```Input: nums = [1,2,3,4,5]
Output: [1,3,6,10,15]
Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4, 1+2+3+4+5].```

Example 2:

```Input: nums = [1,1,1,1,1,1]
Output: [1,2,3,4,5,6]
Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 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]```

## Running Sum of 1d Array Solution code in C++ and Go lang:

Here, we will be solving problem in multiple ways with code.

C++ code 1:

```class Solution {
public:
vector<int> runningSum(vector<int>& nums) {
vector<int> v(nums.size());
v[0] = nums[0];

for(int i = 1; i < nums.size(); i++) {
v[i] = nums[i] + v[i-1];
}
return v;
}
};```

C++ code 2:

```class Solution {
public:
vector<int> runningSum(vector<int>& nums) {
int sum=0;

for(int i=0;i<nums.size();i++){
sum += nums[i];
nums[i] = sum;
}
return nums;
}
};```

Go code 1:

```func runningSum(nums []int) []int {
for i, _ := range nums {
if i == 0 {
continue
}
nums[i] += nums[i-1]
}
return nums
}```

Go code 2:

```func runningSum(nums []int) []int {
size := len(nums)

res := make([]int, size)

res[0] = nums[0]

for i := 1; i < size; i++ {
res[i] = nums[i] + res[i-1]
}
return res
}```

Output:

```Input: nums = [1,2,3,4,5]
Output: [1,3,6,10,15]```

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