💯✅ LeetCode 1480. Running Sum of 1d Array | Go

Link 👉🏻 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
• -106 <= nums[i] <= 106

Intuition

We utilize the concept of Prefix Sum to solve this problem. We need a variable to store the running sum of the array. We iterate over the input array nums and calculate the running sum. We store the running sum in a new array and return it.

Approach

1. Initialize a variable tmp to store the sum of the array.
2. Create a new array sum with the same length as the input array.
3. Iterate over the input array nums and calculate the running sum.
4. Store the running sum in the sum array.
5. Return the sum array.

Complexity

• Time Complexity:
• Space Complexity:

Code

func runningSum(nums []int) []int {
    tmp := 0
	sum := make([]int, len(nums))

	for i := 0; i < len(nums); i++ {
		tmp += nums[i]
		sum[i] = tmp
	}

	return sum
}