Add solution #1658

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,455 LeetCode solutions in JavaScript
+# 1,456 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1278,6 +1278,7 @@
 1655|[Distribute Repeating Integers](./solutions/1655-distribute-repeating-integers.js)|Hard|
 1656|[Design an Ordered Stream](./solutions/1656-design-an-ordered-stream.js)|Easy|
 1657|[Determine if Two Strings Are Close](./solutions/1657-determine-if-two-strings-are-close.js)|Medium|
+1658|[Minimum Operations to Reduce X to Zero](./solutions/1658-minimum-operations-to-reduce-x-to-zero.js)|Medium|
 1668|[Maximum Repeating Substring](./solutions/1668-maximum-repeating-substring.js)|Easy|
 1669|[Merge In Between Linked Lists](./solutions/1669-merge-in-between-linked-lists.js)|Medium|
 1672|[Richest Customer Wealth](./solutions/1672-richest-customer-wealth.js)|Easy|

solutions/1658-minimum-operations-to-reduce-x-to-zero.js

@@ -0,0 +1,42 @@
+/**
+ * 1658. Minimum Operations to Reduce X to Zero
+ * https://leetcode.com/problems/minimum-operations-to-reduce-x-to-zero/
+ * Difficulty: Medium
+ *
+ * You are given an integer array nums and an integer x. In one operation, you can either remove
+ * the leftmost or the rightmost element from the array nums and subtract its value from x. Note
+ * that this modifies the array for future operations.
+ *
+ * Return the minimum number of operations to reduce x to exactly 0 if it is possible, otherwise,
+ * return -1.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} x
+ * @return {number}
+ */
+var minOperations = function(nums, x) {
+  const targetSum = nums.reduce((sum, num) => sum + num, 0) - x;
+  if (targetSum < 0) return -1;
+  if (targetSum === 0) return nums.length;
+
+  let currentSum = 0;
+  let maxLength = -1;
+  let left = 0;
+
+  for (let right = 0; right < nums.length; right++) {
+    currentSum += nums[right];
+
+    while (currentSum > targetSum && left <= right) {
+      currentSum -= nums[left];
+      left++;
+    }
+
+    if (currentSum === targetSum) {
+      maxLength = Math.max(maxLength, right - left + 1);
+    }
+  }
+
+  return maxLength === -1 ? -1 : nums.length - maxLength;
+};