Add solution #322

Author: JoshCrozier

README.md

@@ -215,6 +215,7 @@
 290|[Word Pattern](./0290-word-pattern.js)|Easy|
 295|[Find Median from Data Stream](./0295-find-median-from-data-stream.js)|Hard|
 316|[Remove Duplicate Letters](./0316-remove-duplicate-letters.js)|Medium|
+322|[Coin Change](./0322-coin-change.js)|Medium|
 326|[Power of Three](./0326-power-of-three.js)|Easy|
 328|[Odd Even Linked List](./0328-odd-even-linked-list.js)|Medium|
 334|[Increasing Triplet Subsequence](./0334-increasing-triplet-subsequence.js)|Medium|

solutions/0322-coin-change.js

@@ -0,0 +1,33 @@
+/**
+ * 322. Coin Change
+ * https://leetcode.com/problems/coin-change/
+ * Difficulty: Medium
+ *
+ * You are given an integer array coins representing coins of different denominations and an
+ * integer amount representing a total amount of money.
+ *
+ * Return the fewest number of coins that you need to make up that amount. If that amount of
+ * money cannot be made up by any combination of the coins, return -1.
+ *
+ * You may assume that you have an infinite number of each kind of coin.
+ */
+
+/**
+ * @param {number[]} coins
+ * @param {number} amount
+ * @return {number}
+ */
+var coinChange = function(coins, amount) {
+  const counts = new Array(amount + 1).fill(amount + 1);
+  counts[0] = 0;
+
+  for (let i = 1; i <= amount; i++) {
+    for (let j = 0; j < coins.length; j++) {
+      if (i - coins[j] >= 0) {
+        counts[i] = Math.min(counts[i], 1 + counts[i - coins[j]]);
+      }
+    }
+  }
+
+  return counts[amount] !== amount + 1 ? counts[amount] : -1;
+};