How do you solve Number of Integers With Popcount-Depth Equal to K I on LeetCode?

Number of Integers With Popcount-Depth Equal to K I (LeetCode #3621) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Popcount-depth reduces the problem to counting set bits.

What is the time complexity of Number of Integers With Popcount-Depth Equal to K I?

The optimal solution for Number of Integers With Popcount-Depth Equal to K I runs in O(n) time and uses O(n) space. Dynamic programming allows us to efficiently count valid integers without direct iteration.

What is the brute force solution for Number of Integers With Popcount-Depth Equal to K I?

The brute force approach for Number of Integers With Popcount-Depth Equal to K I is Brute Force, with O(n²) time complexity and O(1) space complexity. Check each integer from 1 to n and compute its popcount-depth by repeatedly applying popcount until reaching 1. Count how many have a depth equal to k. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Number of Integers With Popcount-Depth Equal to K I?

Explain your thought process clearly. Discuss edge cases and constraints. Practice similar problems to build confidence.

What are common mistakes when solving Number of Integers With Popcount-Depth Equal to K I?

Overlooking the need to precompute depths. Not handling the binary representation correctly.

#3621

Number of Integers With Popcount-Depth Equal to K I

Hard

MathDynamic ProgrammingBit ManipulationCombinatoricsDynamic ProgrammingBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

Use digit dynamic programming to count valid integers based on their binary representation and precomputed depths.

⚙️

Algorithm

3 steps

1Step 1: Precompute popcount-depth for all integers from 0 to 64.
2Step 2: Use dynamic programming to count integers with exactly k ones in their binary representation while ensuring they do not exceed n.
3Step 3: Return the count of integers with the desired popcount-depth.

solution.py1 lines

1def precompute_depths(): depths = [0] * 65; for i in range(65): x = i; while x > 1: x = bin(x).count('1'); depths[i] += 1; return depths; def count_with_depth(n, k): depths = precompute_depths(); dp = [[0] * (k + 1) for _ in range(65)]; dp[0][0] = 1; for i in range(64): for j in range(k + 1): dp[i + 1][j] = dp[i][j]; if j > 0: dp[i + 1][j] += dp[i][j - 1]; return dp[64][k];

ℹ

Complexity note: Dynamic programming allows us to efficiently count valid integers without direct iteration.

1Popcount-depth reduces the problem to counting set bits.
2Dynamic programming can optimize counting based on binary representation.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.