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| 1 | +package com.fishercoder.solutions; |
| 2 | + |
| 3 | +import java.util.ArrayList; |
| 4 | +import java.util.Collections; |
| 5 | +import java.util.List; |
| 6 | + |
| 7 | +/** |
| 8 | + * 1387. Sort Integers by The Power Value |
| 9 | + * |
| 10 | + * The power of an integer x is defined as the number of steps needed to transform x into 1 using the following steps: |
| 11 | + * if x is even then x = x / 2 |
| 12 | + * if x is odd then x = 3 * x + 1 |
| 13 | + * |
| 14 | + * 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). |
| 15 | + * 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. |
| 16 | + * Return the k-th integer in the range [lo, hi] sorted by the power value. |
| 17 | + * 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 32 bit signed integer. |
| 18 | + * |
| 19 | + * Example 1: |
| 20 | + * Input: lo = 12, hi = 15, k = 2 |
| 21 | + * Output: 13 |
| 22 | + * Explanation: The power of 12 is 9 (12 --> 6 --> 3 --> 10 --> 5 --> 16 --> 8 --> 4 --> 2 --> 1) |
| 23 | + * The power of 13 is 9 |
| 24 | + * The power of 14 is 17 |
| 25 | + * The power of 15 is 17 |
| 26 | + * The interval sorted by the power value [12,13,14,15]. For k = 2 answer is the second element which is 13. |
| 27 | + * Notice that 12 and 13 have the same power value and we sorted them in ascending order. Same for 14 and 15. |
| 28 | + * |
| 29 | + * Example 2: |
| 30 | + * Input: lo = 1, hi = 1, k = 1 |
| 31 | + * Output: 1 |
| 32 | + * |
| 33 | + * Example 3: |
| 34 | + * Input: lo = 7, hi = 11, k = 4 |
| 35 | + * Output: 7 |
| 36 | + * Explanation: The power array corresponding to the interval [7, 8, 9, 10, 11] is [16, 3, 19, 6, 14]. |
| 37 | + * The interval sorted by power is [8, 10, 11, 7, 9]. |
| 38 | + * The fourth number in the sorted array is 7. |
| 39 | + * |
| 40 | + * Example 4: |
| 41 | + * Input: lo = 10, hi = 20, k = 5 |
| 42 | + * Output: 13 |
| 43 | + * |
| 44 | + * Example 5: |
| 45 | + * Input: lo = 1, hi = 1000, k = 777 |
| 46 | + * Output: 570 |
| 47 | + * |
| 48 | + * Constraints: |
| 49 | + * 1 <= lo <= hi <= 1000 |
| 50 | + * 1 <= k <= hi - lo + 1 |
| 51 | + * */ |
| 52 | +public class _1387 { |
| 53 | + public static class Solution1 { |
| 54 | + public int getKth(int lo, int hi, int k) { |
| 55 | + List<int[]> power = new ArrayList<>(); |
| 56 | + for (int i = lo; i <= hi; i++) { |
| 57 | + power.add(new int[]{getSteps(i), i}); |
| 58 | + } |
| 59 | + Collections.sort(power, (a, b) -> a[0] != b[0] ? a[0] - b[0] : a[1] - b[1]); |
| 60 | + return power.get(k - 1)[1]; |
| 61 | + } |
| 62 | + |
| 63 | + private int getSteps(int number) { |
| 64 | + int steps = 0; |
| 65 | + while (number != 1) { |
| 66 | + if (number % 2 == 0) { |
| 67 | + number /= 2; |
| 68 | + } else { |
| 69 | + number = 3 * number + 1; |
| 70 | + } |
| 71 | + steps++; |
| 72 | + } |
| 73 | + return steps; |
| 74 | + } |
| 75 | + } |
| 76 | +} |
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