Add solution #1387

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,271 LeetCode solutions in JavaScript
+# 1,272 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1057,6 +1057,7 @@
 1383|[Maximum Performance of a Team](./solutions/1383-maximum-performance-of-a-team.js)|Hard|
 1385|[Find the Distance Value Between Two Arrays](./solutions/1385-find-the-distance-value-between-two-arrays.js)|Easy|
 1386|[Cinema Seat Allocation](./solutions/1386-cinema-seat-allocation.js)|Medium|
+1387|[Sort Integers by The Power Value](./solutions/1387-sort-integers-by-the-power-value.js)|Medium|
 1389|[Create Target Array in the Given Order](./solutions/1389-create-target-array-in-the-given-order.js)|Easy|
 1400|[Construct K Palindrome Strings](./solutions/1400-construct-k-palindrome-strings.js)|Medium|
 1402|[Reducing Dishes](./solutions/1402-reducing-dishes.js)|Hard|

solutions/1387-sort-integers-by-the-power-value.js

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+/**
+ * 1387. Sort Integers by The Power Value
+ * https://leetcode.com/problems/sort-integers-by-the-power-value/
+ * Difficulty: Medium
+ *
+ * The power of an integer x is defined as the number of steps needed to transform x into
+ * 1 using the following steps:
+ * - if x is even then x = x / 2
+ * - if x is odd then x = 3 * x + 1
+ *
+ * For example, the power of x = 3 is 7 because 3 needs 7 steps to become 1
+ * (3 --> 10 --> 5 --> 16 --> 8 --> 4 --> 2 --> 1).
+ *
+ * Given three integers lo, hi and k. The task is to sort all integers in the interval
+ * [lo, hi] by the power value in ascending order, if two or more integers have the same
+ * power value sort them by ascending order.
+ *
+ * Return the kth integer in the range [lo, hi] sorted by the power value.
+ *
+ * Notice that for any integer x (lo <= x <= hi) it is guaranteed that x will transform
+ * into 1 using these steps and that the power of x is will fit in a 32-bit signed integer.
+ */
+
+/**
+ * @param {number} lo
+ * @param {number} hi
+ * @param {number} k
+ * @return {number}
+ */
+var getKth = function(lo, hi, k) {
+  const powerValues = new Map();
+  const numbers = Array.from({ length: hi - lo + 1 }, (_, i) => lo + i);
+
+  numbers.sort((a, b) => {
+    const powerA = getPower(a);
+    const powerB = getPower(b);
+    return powerA === powerB ? a - b : powerA - powerB;
+  });
+
+  return numbers[k - 1];
+
+  function calculatePower(num) {
+    let steps = 0;
+    while (num !== 1) {
+      if (num % 2 === 0) {
+        num = num / 2;
+      } else {
+        num = 3 * num + 1;
+      }
+      steps++;
+    }
+    return steps;
+  }
+
+  function getPower(num) {
+    if (!powerValues.has(num)) {
+      powerValues.set(num, calculatePower(num));
+    }
+    return powerValues.get(num);
+  }
+};