LeetCode #3621 — HARD

Number of Integers With Popcount-Depth Equal to K I

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

You are given two integers n and k.

For any positive integer x, define the following sequence:

p₀ = x
p_i+1 = popcount(p_i) for all i >= 0, where popcount(y) is the number of set bits (1's) in the binary representation of y.

This sequence will eventually reach the value 1.

The popcount-depth of x is defined as the smallest integer d >= 0 such that p_d = 1.

For example, if x = 7 (binary representation "111"). Then, the sequence is: 7 → 3 → 2 → 1, so the popcount-depth of 7 is 3.

Your task is to determine the number of integers in the range [1, n] whose popcount-depth is exactly equal to k.

Return the number of such integers.

Example 1:

Input: n = 4, k = 1

Output: 2

Explanation:

The following integers in the range [1, 4] have popcount-depth exactly equal to 1:

x	Binary	Sequence
2	`"10"`	`2 → 1`
4	`"100"`	`4 → 1`

Thus, the answer is 2.

Example 2:

Input: n = 7, k = 2

Output: 3

Explanation:

The following integers in the range [1, 7] have popcount-depth exactly equal to 2:

x	Binary	Sequence
3	`"11"`	`3 → 2 → 1`
5	`"101"`	`5 → 2 → 1`
6	`"110"`	`6 → 2 → 1`

Thus, the answer is 3.

Constraints:

1 <= n <= 10¹⁵
0 <= k <= 5

Step 01

Brute Force Baseline

Problem summary: You are given two integers n and k. For any positive integer x, define the following sequence: p0 = x pi+1 = popcount(pi) for all i >= 0, where popcount(y) is the number of set bits (1's) in the binary representation of y. This sequence will eventually reach the value 1. The popcount-depth of x is defined as the smallest integer d >= 0 such that pd = 1. For example, if x = 7 (binary representation "111"). Then, the sequence is: 7 → 3 → 2 → 1, so the popcount-depth of 7 is 3. Your task is to determine the number of integers in the range [1, n] whose popcount-depth is exactly equal to k. Return the number of such integers.

Baseline thinking

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

Pattern signal: Math · Dynamic Programming · Bit Manipulation

Example 1

4
1

Example 2

7
2

Core Insight

What unlocks the optimal approach

Use digit dynamic programming on the binary representation of <code>n</code>: let <code>dp[pos][ones][tight]</code> = number of ways to choose bits from the most significant down to position <code>pos</code> with exactly <code>ones</code> ones so far, where <code>tight</code> indicates whether you're still matching the prefix of <code>n</code>.
Precompute <code>depth[j]</code> for all <code>j</code> from <code>0</code> to <code>64</code> by repeatedly applying <code>popcount(j)</code> until you reach <code>1</code>.
After your DP, let <code>dp_final[j]</code> be the count of numbers <= <code>n</code> that have exactly <code>j</code> ones; the answer is the sum of all <code>dp_final[j]</code> for which <code>depth[j] == k</code>.

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

Largest constraint values

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 #3621: Number of Integers With Popcount-Depth Equal to K I
// Auto-generated Java example from go.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (go):
// // Accepted solution for LeetCode #3621: Number of Integers With Popcount-Depth Equal to K I
// package main
// 
// import (
// 	"math/bits"
// 	"strconv"
// )
// 
// // https://space.bilibili.com/206214
// func popcountDepth(n int64, k int) (ans int64) {
// 	if k == 0 {
// 		return 1
// 	}
// 
// 	// 注：也可以不转成字符串，下面 dfs 用位运算取出 n 的第 i 位
// 	// 但转成字符串的通用性更好
// 	s := strconv.FormatInt(n, 2)
// 	m := len(s)
// 	if k == 1 {
// 		return int64(m - 1)
// 	}
// 
// 	memo := make([][]int64, m)
// 	for i := range memo {
// 		memo[i] = make([]int64, m+1)
// 		for j := range memo[i] {
// 			memo[i][j] = -1
// 		}
// 	}
// 
// 	var dfs func(int, int, bool) int64
// 	dfs = func(i, left1 int, isLimit bool) (res int64) {
// 		if i == m {
// 			if left1 == 0 {
// 				return 1
// 			}
// 			return
// 		}
// 		if !isLimit {
// 			p := &memo[i][left1]
// 			if *p >= 0 {
// 				return *p
// 			}
// 			defer func() { *p = res }()
// 		}
// 		up := 1
// 		if isLimit {
// 			up = int(s[i] - '0')
// 		}
// 		for d := 0; d <= min(up, left1); d++ {
// 			res += dfs(i+1, left1-d, isLimit && d == up)
// 		}
// 		return
// 	}
// 
// 	f := make([]int, m+1)
// 	for i := 1; i <= m; i++ {
// 		f[i] = f[bits.OnesCount(uint(i))] + 1
// 		if f[i] == k {
// 			// 计算有多少个二进制数恰好有 i 个 1
// 			ans += dfs(0, i, true)
// 		}
// 	}
// 	return
// }

// Accepted solution for LeetCode #3621: Number of Integers With Popcount-Depth Equal to K I
package main

import (
	"math/bits"
	"strconv"
)

// https://space.bilibili.com/206214
func popcountDepth(n int64, k int) (ans int64) {
	if k == 0 {
		return 1
	}

	// 注：也可以不转成字符串，下面 dfs 用位运算取出 n 的第 i 位
	// 但转成字符串的通用性更好
	s := strconv.FormatInt(n, 2)
	m := len(s)
	if k == 1 {
		return int64(m - 1)
	}

	memo := make([][]int64, m)
	for i := range memo {
		memo[i] = make([]int64, m+1)
		for j := range memo[i] {
			memo[i][j] = -1
		}
	}

	var dfs func(int, int, bool) int64
	dfs = func(i, left1 int, isLimit bool) (res int64) {
		if i == m {
			if left1 == 0 {
				return 1
			}
			return
		}
		if !isLimit {
			p := &memo[i][left1]
			if *p >= 0 {
				return *p
			}
			defer func() { *p = res }()
		}
		up := 1
		if isLimit {
			up = int(s[i] - '0')
		}
		for d := 0; d <= min(up, left1); d++ {
			res += dfs(i+1, left1-d, isLimit && d == up)
		}
		return
	}

	f := make([]int, m+1)
	for i := 1; i <= m; i++ {
		f[i] = f[bits.OnesCount(uint(i))] + 1
		if f[i] == k {
			// 计算有多少个二进制数恰好有 i 个 1
			ans += dfs(0, i, true)
		}
	}
	return
}

# Accepted solution for LeetCode #3621: Number of Integers With Popcount-Depth Equal to K I
#
# @lc app=leetcode id=3621 lang=python3
#
# [3621] Number of Integers With Popcount-Depth Equal to K I
#

# @lc code=start
import math

class Solution:
    def popcountDepth(self, n: int, k: int) -> int:
        if k == 0:
            # The only number with depth 0 is 1.
            # Since range is [1, n] and n >= 1, the answer is 1.
            return 1

        # Precompute depths for small numbers.
        # Max n is 10^15, which is less than 2^50.
        # The maximum popcount for a number <= 10^15 is roughly 50.
        # We can compute depths up to 60 to be safe.
        
        depth_map = {}
        depth_map[1] = 0
        
        # Function to get depth for small numbers recursively
        def get_depth(val):
            if val in depth_map:
                return depth_map[val]
            # Recursive step: depth(x) = 1 + depth(popcount(x))
            # popcount(val) will be strictly less than val for val > 1
            d = 1 + get_depth(val.bit_count())
            depth_map[val] = d
            return d

        # Fill depth map for possible popcounts (1 to 60)
        for i in range(1, 61):
            get_depth(i)
            
        # We are looking for x in [1, n] such that depth(x) == k.
        # This is equivalent to depth(popcount(x)) == k - 1.
        # Let c = popcount(x). Then we need depth(c) == k - 1.
        # Since x <= n < 2^60, c will be between 1 and 60.
        
        target_popcounts = []
        for c in range(1, 61):
            if depth_map[c] == k - 1:
                target_popcounts.append(c)
        
        if not target_popcounts:
            return 0
            
        # Function to count integers in [1, num] with exactly `cnt` set bits.
        # Using combinatorics logic (Digit DP style without memoization table).
        def count_with_bits(num, cnt):
            if cnt < 0:
                return 0
            if num == 0:
                return 0
            
            s = bin(num)[2:]
            length = len(s)
            res = 0
            current_bits = 0
            
            for i in range(length):
                if s[i] == '1':
                    # If we put '0' at this position:
                    # Remaining positions: length - 1 - i
                    # Bits needed: cnt - current_bits
                    remaining_len = length - 1 - i
                    needed = cnt - current_bits
                    if 0 <= needed <= remaining_len:
                        res += math.comb(remaining_len, needed)
                    
                    # Now assume we put '1' at this position and continue
                    current_bits += 1
            
            # Check the number itself
            if current_bits == cnt:
                res += 1
            
            return res

        total_count = 0
        for c in target_popcounts:
            total_count += count_with_bits(n, c)
            
        return total_count

# @lc code=end

// Accepted solution for LeetCode #3621: Number of Integers With Popcount-Depth Equal to K I
// Rust example auto-generated from go reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (go):
// // Accepted solution for LeetCode #3621: Number of Integers With Popcount-Depth Equal to K I
// package main
// 
// import (
// 	"math/bits"
// 	"strconv"
// )
// 
// // https://space.bilibili.com/206214
// func popcountDepth(n int64, k int) (ans int64) {
// 	if k == 0 {
// 		return 1
// 	}
// 
// 	// 注：也可以不转成字符串，下面 dfs 用位运算取出 n 的第 i 位
// 	// 但转成字符串的通用性更好
// 	s := strconv.FormatInt(n, 2)
// 	m := len(s)
// 	if k == 1 {
// 		return int64(m - 1)
// 	}
// 
// 	memo := make([][]int64, m)
// 	for i := range memo {
// 		memo[i] = make([]int64, m+1)
// 		for j := range memo[i] {
// 			memo[i][j] = -1
// 		}
// 	}
// 
// 	var dfs func(int, int, bool) int64
// 	dfs = func(i, left1 int, isLimit bool) (res int64) {
// 		if i == m {
// 			if left1 == 0 {
// 				return 1
// 			}
// 			return
// 		}
// 		if !isLimit {
// 			p := &memo[i][left1]
// 			if *p >= 0 {
// 				return *p
// 			}
// 			defer func() { *p = res }()
// 		}
// 		up := 1
// 		if isLimit {
// 			up = int(s[i] - '0')
// 		}
// 		for d := 0; d <= min(up, left1); d++ {
// 			res += dfs(i+1, left1-d, isLimit && d == up)
// 		}
// 		return
// 	}
// 
// 	f := make([]int, m+1)
// 	for i := 1; i <= m; i++ {
// 		f[i] = f[bits.OnesCount(uint(i))] + 1
// 		if f[i] == k {
// 			// 计算有多少个二进制数恰好有 i 个 1
// 			ans += dfs(0, i, true)
// 		}
// 	}
// 	return
// }

// Accepted solution for LeetCode #3621: Number of Integers With Popcount-Depth Equal to K I
// Auto-generated TypeScript example from go.
function exampleSolution(): void {
}
// Reference (go):
// // Accepted solution for LeetCode #3621: Number of Integers With Popcount-Depth Equal to K I
// package main
// 
// import (
// 	"math/bits"
// 	"strconv"
// )
// 
// // https://space.bilibili.com/206214
// func popcountDepth(n int64, k int) (ans int64) {
// 	if k == 0 {
// 		return 1
// 	}
// 
// 	// 注：也可以不转成字符串，下面 dfs 用位运算取出 n 的第 i 位
// 	// 但转成字符串的通用性更好
// 	s := strconv.FormatInt(n, 2)
// 	m := len(s)
// 	if k == 1 {
// 		return int64(m - 1)
// 	}
// 
// 	memo := make([][]int64, m)
// 	for i := range memo {
// 		memo[i] = make([]int64, m+1)
// 		for j := range memo[i] {
// 			memo[i][j] = -1
// 		}
// 	}
// 
// 	var dfs func(int, int, bool) int64
// 	dfs = func(i, left1 int, isLimit bool) (res int64) {
// 		if i == m {
// 			if left1 == 0 {
// 				return 1
// 			}
// 			return
// 		}
// 		if !isLimit {
// 			p := &memo[i][left1]
// 			if *p >= 0 {
// 				return *p
// 			}
// 			defer func() { *p = res }()
// 		}
// 		up := 1
// 		if isLimit {
// 			up = int(s[i] - '0')
// 		}
// 		for d := 0; d <= min(up, left1); d++ {
// 			res += dfs(i+1, left1-d, isLimit && d == up)
// 		}
// 		return
// 	}
// 
// 	f := make([]int, m+1)
// 	for i := 1; i <= m; i++ {
// 		f[i] = f[bits.OnesCount(uint(i))] + 1
// 		if f[i] == k {
// 			// 计算有多少个二进制数恰好有 i 个 1
// 			ans += dfs(0, i, true)
// 		}
// 	}
// 	return
// }

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 × m)

Space

O(n × m)

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.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

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.