Add solution #1737

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,510 LeetCode solutions in JavaScript
+# 1,511 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1338,6 +1338,7 @@
 1734|[Decode XORed Permutation](./solutions/1734-decode-xored-permutation.js)|Medium|
 1735|[Count Ways to Make Array With Product](./solutions/1735-count-ways-to-make-array-with-product.js)|Hard|
 1736|[Latest Time by Replacing Hidden Digits](./solutions/1736-latest-time-by-replacing-hidden-digits.js)|Easy|
+1737|[Change Minimum Characters to Satisfy One of Three Conditions](./solutions/1737-change-minimum-characters-to-satisfy-one-of-three-conditions.js)|Medium|
 1748|[Sum of Unique Elements](./solutions/1748-sum-of-unique-elements.js)|Easy|
 1749|[Maximum Absolute Sum of Any Subarray](./solutions/1749-maximum-absolute-sum-of-any-subarray.js)|Medium|
 1752|[Check if Array Is Sorted and Rotated](./solutions/1752-check-if-array-is-sorted-and-rotated.js)|Easy|

solutions/1737-change-minimum-characters-to-satisfy-one-of-three-conditions.js

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+/**
+ * 1737. Change Minimum Characters to Satisfy One of Three Conditions
+ * https://leetcode.com/problems/change-minimum-characters-to-satisfy-one-of-three-conditions/
+ * Difficulty: Medium
+ *
+ * You are given two strings a and b that consist of lowercase letters. In one operation, you can
+ * change any character in a or b to any lowercase letter.
+ *
+ * Your goal is to satisfy one of the following three conditions:
+ * - Every letter in a is strictly less than every letter in b in the alphabet.
+ * - Every letter in b is strictly less than every letter in a in the alphabet.
+ * - Both a and b consist of only one distinct letter.
+ *
+ * Return the minimum number of operations needed to achieve your goal.
+ */
+
+/**
+ * @param {string} a
+ * @param {string} b
+ * @return {number}
+ */
+var minCharacters = function(a, b) {
+  const aFreq = getFrequency(a);
+  const bFreq = getFrequency(b);
+
+  const condition1 = operationsForCondition1(aFreq, bFreq);
+  const condition2 = operationsForCondition1(bFreq, aFreq);
+  const condition3 = operationsForCondition3(aFreq, bFreq);
+
+  return Math.min(condition1, condition2, condition3);
+
+  function getFrequency(str) {
+    const freq = Array(26).fill(0);
+    for (const char of str) {
+      freq[char.charCodeAt(0) - 97]++;
+    }
+    return freq;
+  }
+
+  function operationsForCondition1(aFreq, bFreq) {
+    let minOps = Infinity;
+    for (let i = 0; i < 25; i++) {
+      let ops = 0;
+      for (let j = 0; j <= i; j++) ops += aFreq[j];
+      for (let j = i + 1; j < 26; j++) ops += bFreq[j];
+      minOps = Math.min(minOps, ops);
+    }
+    return minOps;
+  }
+
+  function operationsForCondition3(aFreq, bFreq) {
+    let minOps = Infinity;
+    for (let i = 0; i < 26; i++) {
+      const ops = a.length + b.length - aFreq[i] - bFreq[i];
+      minOps = Math.min(minOps, ops);
+    }
+    return minOps;
+  }
+};