How do you solve Count K-Reducible Numbers Less Than N on LeetCode?

Count K-Reducible Numbers Less Than N (LeetCode #3352) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Understanding how to count set bits is crucial for this problem.

What is the time complexity of Count K-Reducible Numbers Less Than N?

The optimal solution for Count K-Reducible Numbers Less Than N runs in O(n log n) time and uses O(n) space. This complexity arises because we are iterating through all numbers up to n and counting set bits requires log n operations.

What is the brute force solution for Count K-Reducible Numbers Less Than N?

The brute force approach for Count K-Reducible Numbers Less Than N is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking each number less than n to see if it can be reduced to 1 in k or fewer operations. This is straightforward but inefficient for large inputs. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Count K-Reducible Numbers Less Than N?

Count K-Reducible Numbers Less Than N uses the following concepts: Math, String, Dynamic Programming, Combinatorics. The recommended patterns to study are: Dynamic Programming, Bit Manipulation.

What are the interview tips for Count K-Reducible Numbers Less Than N?

Always clarify the constraints and edge cases before diving into coding. Explain your thought process as you code; it helps the interviewer follow your logic. Practice writing clean and efficient code, as readability is key in interviews.

What are common mistakes when solving Count K-Reducible Numbers Less Than N?

Not considering edge cases where n is very small. Failing to correctly implement the bit counting logic.

#3352

Count K-Reducible Numbers Less Than N

Hard

MathStringDynamic ProgrammingCombinatoricsDynamic ProgrammingBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(n)

The optimal solution leverages dynamic programming and precomputation to efficiently determine the number of k-reducible numbers without iterating through each number individually.

⚙️

Algorithm

3 steps

1Step 1: Precompute the number of operations needed to reduce each number to 1 for all numbers up to n.
2Step 2: Use a dynamic programming approach to count how many numbers can be reduced to 1 in k or fewer operations.
3Step 3: Return the count of k-reducible numbers less than n.

solution.py16 lines

1# Full working Python code
2MOD = 10**9 + 7
3
4def countKReducible(s, k):
5    n = int(s, 2)
6    dp = [0] * (n + 1)
7    for i in range(1, n + 1):
8        x = i
9        operations = 0
10        while x > 1:
11            x = bin(x).count('1')
12            operations += 1
13            if operations > k:
14                break
15        dp[i] = operations
16    return sum(1 for i in range(1, n) if dp[i] <= k) % MOD

ℹ

Complexity note: This complexity arises because we are iterating through all numbers up to n and counting set bits requires log n operations.

1Understanding how to count set bits is crucial for this problem.
2Dynamic programming can significantly reduce the time complexity by avoiding repeated calculations.

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