/*
 * 1480. Running Sum of 1d Array
 * Difficulty: Easy
 * https://leetcode.com/problems/running-sum-of-1d-array/
 * 
 * ──────────────────────────────────────────────────
 * 
 * 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
*/

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var runningSum = function(nums) {
  let prefix = [nums[0]];
  for (let i = 1; i < nums.length; i++){
    prefix.push(prefix[i - 1] + nums[i]);
  }
  return prefix;
};