Add solution #1345

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,244 LeetCode solutions in JavaScript
+# 1,245 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1021,6 +1021,7 @@
 1342|[Number of Steps to Reduce a Number to Zero](./solutions/1342-number-of-steps-to-reduce-a-number-to-zero.js)|Easy|
 1343|[Number of Sub-arrays of Size K and Average Greater than or Equal to Threshold](./solutions/1343-number-of-sub-arrays-of-size-k-and-average-greater-than-or-equal-to-threshold.js)|Medium|
 1344|[Angle Between Hands of a Clock](./solutions/1344-angle-between-hands-of-a-clock.js)|Medium|
+1345|[Jump Game IV](./solutions/1345-jump-game-iv.js)|Hard|
 1351|[Count Negative Numbers in a Sorted Matrix](./solutions/1351-count-negative-numbers-in-a-sorted-matrix.js)|Easy|
 1352|[Product of the Last K Numbers](./solutions/1352-product-of-the-last-k-numbers.js)|Medium|
 1356|[Sort Integers by The Number of 1 Bits](./solutions/1356-sort-integers-by-the-number-of-1-bits.js)|Easy|

solutions/1345-jump-game-iv.js

@@ -0,0 +1,69 @@
+/**
+ * 1345. Jump Game IV
+ * https://leetcode.com/problems/jump-game-iv/
+ * Difficulty: Hard
+ *
+ * Given an array of integers arr, you are initially positioned at the first index of the array.
+ *
+ * In one step you can jump from index i to index:
+ * - i + 1 where: i + 1 < arr.length.
+ * - i - 1 where: i - 1 >= 0.
+ * - j where: arr[i] == arr[j] and i != j.
+ *
+ * Return the minimum number of steps to reach the last index of the array.
+ *
+ * Notice that you can not jump outside of the array at any time.
+ */
+
+/**
+ * @param {number[]} arr
+ * @return {number}
+ */
+var minJumps = function(arr) {
+  const n = arr.length;
+  if (n <= 1) return 0;
+
+  const valueToIndices = new Map();
+  for (let i = 0; i < n; i++) {
+    if (!valueToIndices.has(arr[i])) {
+      valueToIndices.set(arr[i], []);
+    }
+    valueToIndices.get(arr[i]).push(i);
+  }
+
+  const visited = new Set([0]);
+  let queue = [0];
+  let result = 0;
+
+  while (queue.length) {
+    const nextQueue = [];
+
+    for (const current of queue) {
+      if (current === n - 1) return result;
+
+      if (current + 1 < n && !visited.has(current + 1)) {
+        visited.add(current + 1);
+        nextQueue.push(current + 1);
+      }
+
+      if (current - 1 >= 0 && !visited.has(current - 1)) {
+        visited.add(current - 1);
+        nextQueue.push(current - 1);
+      }
+
+      for (const next of valueToIndices.get(arr[current]) || []) {
+        if (!visited.has(next)) {
+          visited.add(next);
+          nextQueue.push(next);
+        }
+      }
+
+      valueToIndices.delete(arr[current]);
+    }
+
+    queue = nextQueue;
+    result++;
+  }
+
+  return result;
+};