Add solution #1590

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,388 LeetCode solutions in JavaScript
+# 1,389 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1213,6 +1213,7 @@
 1585|[Check If String Is Transformable With Substring Sort Operations](./solutions/1585-check-if-string-is-transformable-with-substring-sort-operations.js)|Hard|
 1588|[Sum of All Odd Length Subarrays](./solutions/1588-sum-of-all-odd-length-subarrays.js)|Easy|
 1589|[Maximum Sum Obtained of Any Permutation](./solutions/1589-maximum-sum-obtained-of-any-permutation.js)|Medium|
+1590|[Make Sum Divisible by P](./solutions/1590-make-sum-divisible-by-p.js)|Medium|
 1598|[Crawler Log Folder](./solutions/1598-crawler-log-folder.js)|Easy|
 1657|[Determine if Two Strings Are Close](./solutions/1657-determine-if-two-strings-are-close.js)|Medium|
 1668|[Maximum Repeating Substring](./solutions/1668-maximum-repeating-substring.js)|Easy|

solutions/1590-make-sum-divisible-by-p.js

@@ -0,0 +1,41 @@
+/**
+ * 1590. Make Sum Divisible by P
+ * https://leetcode.com/problems/make-sum-divisible-by-p/
+ * Difficulty: Medium
+ *
+ * Given an array of positive integers nums, remove the smallest subarray (possibly empty) such that
+ * the sum of the remaining elements is divisible by p. It is not allowed to remove the whole array.
+ *
+ * Return the length of the smallest subarray that you need to remove, or -1 if it's impossible.
+ *
+ * A subarray is defined as a contiguous block of elements in the array.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} p
+ * @return {number}
+ */
+var minSubarray = function(nums, p) {
+  const totalSum = nums.reduce((sum, num) => sum + num, 0);
+  const remainder = totalSum % p;
+
+  if (remainder === 0) return 0;
+
+  const prefixSums = new Map([[0, -1]]);
+  let currentSum = 0;
+  let minLength = nums.length;
+
+  for (let i = 0; i < nums.length; i++) {
+    currentSum = (currentSum + nums[i]) % p;
+    const target = (currentSum - remainder + p) % p;
+
+    if (prefixSums.has(target)) {
+      minLength = Math.min(minLength, i - prefixSums.get(target));
+    }
+
+    prefixSums.set(currentSum, i);
+  }
+
+  return minLength < nums.length ? minLength : -1;
+};