feat(problems): add LeetCode array problem solutions

work/Easy/Array/1413.minimum-value-to-get-positive-step-by-step-sum.js

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+/*
+ * 1413. Minimum Value to Get Positive Step by Step Sum
+ * Difficulty: Easy
+ * https://leetcode.com/problems/minimum-value-to-get-positive-step-by-step-sum/
+ * 
+ * ──────────────────────────────────────────────────
+ * 
+ * Given an array of integers nums, you start with an initial positive
+ * value startValue.
+ * 
+ * In each iteration, you calculate the step by step sum of startValue
+ * plus elements in nums (from left to right).
+ * 
+ * Return the minimum positive value of startValue such that the step by
+ * step sum is never less than 1.
+ * 
+ *  
+ * 
+ * Example 1:
+ * 
+ * Input: nums = [-3,2,-3,4,2]
+ * Output: 5
+ * Explanation: If you choose startValue = 4, in the third iteration
+ * your step by step sum is less than 1.
+ * step by step sum
+ * startValue = 4 | startValue = 5 | nums
+ *   (4 -3 ) = 1  | (5 -3 ) = 2    |  -3
+ *   (1 +2 ) = 3  | (2 +2 ) = 4    |   2
+ *   (3 -3 ) = 0  | (4 -3 ) = 1    |  -3
+ *   (0 +4 ) = 4  | (1 +4 ) = 5    |   4
+ *   (4 +2 ) = 6  | (5 +2 ) = 7    |   2
+ * 
+ * Example 2:
+ * 
+ * Input: nums = [1,2]
+ * Output: 1
+ * Explanation: Minimum start value should be positive. 
+ * 
+ * Example 3:
+ * 
+ * Input: nums = [1,-2,-3]
+ * Output: 5
+ * 
+ *  
+ * 
+ * Constraints:
+ * 
+ * 	• 1 <= nums.length <= 100
+ * 
+ * 	• -100 <= nums[i] <= 100
+*/
+
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var minStartValue = function(nums) {
+  let prefix = [nums[0]];
+  // Make prefix sum start at 1 after initializing seed value
+  for (let i = 1; i < nums.length; i++){
+    prefix.push(prefix[i - 1] + nums[i]);
+  }
+
+  let min = Math.min(...prefix);
+  return Math.max(1, 1 - min);
+};

work/Easy/Array/1480.running-sum-of-1d-array.js

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+/*
+ * 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;
+};