Add solution #1419

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,294 LeetCode solutions in JavaScript
+# 1,295 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1085,6 +1085,7 @@
 1416|[Restore The Array](./solutions/1416-restore-the-array.js)|Hard|
 1417|[Reformat The String](./solutions/1417-reformat-the-string.js)|Easy|
 1418|[Display Table of Food Orders in a Restaurant](./solutions/1418-display-table-of-food-orders-in-a-restaurant.js)|Medium|
+1419|[Minimum Number of Frogs Croaking](./solutions/1419-minimum-number-of-frogs-croaking.js)|Medium|
 1422|[Maximum Score After Splitting a String](./solutions/1422-maximum-score-after-splitting-a-string.js)|Easy|
 1431|[Kids With the Greatest Number of Candies](./solutions/1431-kids-with-the-greatest-number-of-candies.js)|Easy|
 1436|[Destination City](./solutions/1436-destination-city.js)|Easy|

solutions/1419-minimum-number-of-frogs-croaking.js

@@ -0,0 +1,44 @@
+/**
+ * 1419. Minimum Number of Frogs Croaking
+ * https://leetcode.com/problems/minimum-number-of-frogs-croaking/
+ * Difficulty: Medium
+ *
+ * You are given the string croakOfFrogs, which represents a combination of the string "croak"
+ * from different frogs, that is, multiple frogs can croak at the same time, so multiple "croak"
+ * are mixed.
+ *
+ * Return the minimum number of different frogs to finish all the croaks in the given string.
+ *
+ * A valid "croak" means a frog is printing five letters 'c', 'r', 'o', 'a', and 'k' sequentially.
+ * The frogs have to print all five letters to finish a croak. If the given string is not a
+ * combination of a valid "croak" return -1.
+ */
+
+/**
+ * @param {string} croakOfFrogs
+ * @return {number}
+ */
+var minNumberOfFrogs = function(croakOfFrogs) {
+  let activeFrogs = 0;
+  let maxFrogs = 0;
+  const counts = { c: 0, r: 0, o: 0, a: 0, k: 0 };
+  const order = 'croak';
+
+  for (const char of croakOfFrogs) {
+    if (!order.includes(char)) return -1;
+    counts[char]++;
+
+    if (char === 'c') {
+      activeFrogs++;
+      maxFrogs = Math.max(maxFrogs, activeFrogs);
+    } else if (char === 'k') {
+      activeFrogs--;
+    }
+
+    for (let i = 1; i < order.length; i++) {
+      if (counts[order[i]] > counts[order[i - 1]]) return -1;
+    }
+  }
+
+  return activeFrogs === 0 && counts.c === counts.k ? maxFrogs : -1;
+};