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]…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.