LeetCode #1387 — MEDIUM

Sort Integers by The Power Value

Move from brute-force thinking to an efficient approach using dynamic programming strategy.

The Problem

Problem Statement

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 k^th 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.

Example 1:

Input: lo = 12, hi = 15, k = 2
Output: 13
Explanation: The power of 12 is 9 (12 --> 6 --> 3 --> 10 --> 5 --> 16 --> 8 --> 4 --> 2 --> 1)
The power of 13 is 9
The power of 14 is 17
The power of 15 is 17
The interval sorted by the power value [12,13,14,15]. For k = 2 answer is the second element which is 13.
Notice that 12 and 13 have the same power value and we sorted them in ascending order. Same for 14 and 15.

Example 2:

Input: lo = 7, hi = 11, k = 4
Output: 7
Explanation: The power array corresponding to the interval [7, 8, 9, 10, 11] is [16, 3, 19, 6, 14].
The interval sorted by power is [8, 10, 11, 7, 9].
The fourth number in the sorted array is 7.

Constraints:

1 <= lo <= hi <= 1000
1 <= k <= hi - lo + 1

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Dynamic Programming

Example 1

12
15
2

Example 2

7
11
4

Core Insight

What unlocks the optimal approach

Use dynamic programming to get the power of each integer of the intervals.
Sort all the integers of the interval by the power value and return the k-th in the sorted list.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1387: Sort Integers by The Power Value
class Solution {
    public int getKth(int lo, int hi, int k) {
        Integer[] nums = new Integer[hi - lo + 1];
        for (int i = lo; i <= hi; ++i) {
            nums[i - lo] = i;
        }
        Arrays.sort(nums, (a, b) -> {
            int fa = f(a), fb = f(b);
            return fa == fb ? a - b : fa - fb;
        });
        return nums[k - 1];
    }

    private int f(int x) {
        int ans = 0;
        for (; x != 1; ++ans) {
            if (x % 2 == 0) {
                x /= 2;
            } else {
                x = x * 3 + 1;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1387: Sort Integers by The Power Value
func getKth(lo int, hi int, k int) int {
	f := func(x int) (ans int) {
		for ; x != 1; ans++ {
			if x%2 == 0 {
				x /= 2
			} else {
				x = 3*x + 1
			}
		}
		return
	}
	nums := make([]int, hi-lo+1)
	for i := range nums {
		nums[i] = lo + i
	}
	sort.Slice(nums, func(i, j int) bool {
		fx, fy := f(nums[i]), f(nums[j])
		if fx != fy {
			return fx < fy
		}
		return nums[i] < nums[j]
	})
	return nums[k-1]
}

# Accepted solution for LeetCode #1387: Sort Integers by The Power Value
@cache
def f(x: int) -> int:
    ans = 0
    while x != 1:
        if x % 2 == 0:
            x //= 2
        else:
            x = 3 * x + 1
        ans += 1
    return ans


class Solution:
    def getKth(self, lo: int, hi: int, k: int) -> int:
        return sorted(range(lo, hi + 1), key=f)[k - 1]

// Accepted solution for LeetCode #1387: Sort Integers by The Power Value
impl Solution {
    pub fn get_kth(lo: i32, hi: i32, k: i32) -> i32 {
        let f = |mut x: i32| -> i32 {
            let mut ans = 0;
            while x != 1 {
                if x % 2 == 0 {
                    x /= 2;
                } else {
                    x = 3 * x + 1;
                }
                ans += 1;
            }
            ans
        };

        let mut nums: Vec<i32> = (lo..=hi).collect();
        nums.sort_by(|&x, &y| f(x).cmp(&f(y)));
        nums[(k - 1) as usize]
    }
}

// Accepted solution for LeetCode #1387: Sort Integers by The Power Value
function getKth(lo: number, hi: number, k: number): number {
    const f = (x: number): number => {
        let ans = 0;
        for (; x !== 1; ++ans) {
            if (x % 2 === 0) {
                x >>= 1;
            } else {
                x = x * 3 + 1;
            }
        }
        return ans;
    };
    const nums = new Array(hi - lo + 1).fill(0).map((_, i) => i + lo);
    nums.sort((a, b) => {
        const fa = f(a),
            fb = f(b);
        return fa === fb ? a - b : fa - fb;
    });
    return nums[k - 1];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n × M)

Space

O(n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.