Introduction

In this blog post, we will explore how to solve the LeetCode problem "3352" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given a binary string s representing a number n in its binary form. You are also given an integer k. An integer x is called k-reducible if performing the following operation at most k times reduces it to 1: Replace x with the count of set bits in its binary representation. For example, the binary representation of 6 is "110". Applying the operation once reduces it to 2 (since "110" has two set bits). Applying the operation again to 2 (binary "10") reduces it to 1 (since "10" has one set bit). Return an integer denoting the number of positive integers less than n that are k-reducible. Since the answer may be too large, return it modulo 109 + 7. Example 1: Input: s = "111", k = 1 Output: 3 Explanation: n = 7. The 1-reducible integers less than 7 are 1, 2, and 4. Example 2: Input: s = "1000", k = 2 Output: 6 Explanation: n = 8. The 2-reducible integers less than 8 are 1, 2, 3, 4, 5, and 6. Example 3: Input: s = "1", k = 3 Output: 0 Explanation: There are no positive integers less than n = 1, so the answer is 0. Constraints: 1 <= s.length <= 800 s has no leading zeros. s consists only of the characters '0' and '1'. 1 <= k <= 5

Explanation

Here's the breakdown of the solution:

• Determine Reducibility: Implement a function to check if a number is k-reducible by repeatedly counting set bits until it reaches 1 or exceeds the allowed operations.
• Count Integers: Iterate through numbers less than n and count those that are k-reducible, taking modulo at each step to prevent overflow. Since direct iteration would be inefficient, utilize combinatorics and dynamic programming via memoization to compute the number of integers with a particular number of set bits that are less than n.

• Combinatorial Calculation: Use combinations (n choose k) to efficiently calculate how many numbers less than n have a specific number of set bits.

• Runtime Complexity: O(L^2 + L*K), where L is the length of the binary string s, and K is the given integer.

• Storage Complexity: O(L^2), where L is the length of the binary string s.

Code

    def solve():
    s = input()
    k = int(input())
    n = int(s, 2)
    mod = 10**9 + 7

    def is_k_reducible(x, k):
        if x <= 0:
            return False
        for _ in range(k):
            bits = bin(x).count('1')
            if bits == 1:
                return True
            x = bits
        return False

    def count_k_reducible(n, k):
        count = 0
        for i in range(1, n):
            if is_k_reducible(i, k):
                count = (count + 1) % mod
        return count

    def get_next_bit_count(x):
      return bin(x).count('1')

    def is_reducible_dp(x, k, memo):
      if x == 1:
        return True
      if (x, k) in memo:
        return memo[(x, k)]
      if k == 0:
        return False

      next_count = get_next_bit_count(x)
      result = is_reducible_dp(next_count, k - 1, memo)
      memo[(x, k)] = result
      return result

    def count_reducible_optimized(s, k):
      n = len(s)
      mod = 10**9 + 7

      C = [[0] * (n + 1) for _ in range(n + 1)]
      for i in range(n + 1):
          C[i][0] = 1
          for j in range(1, i + 1):
              C[i][j] = (C[i-1][j-1] + C[i-1][j]) % mod

      ans = 0
      for j in range(1, n):
          if is_reducible_dp(j, k, {}):
              ans = (ans + C[n - 1][j - 1]) % mod

      ones = 0
      for i in range(n):
          if s[i] == '1':
              for j in range(max(0, 1 - ones), n - i):
                  if is_reducible_dp(ones + j + (i==0), k, {}):
                      ans = (ans + C[n - i - 1][j]) % mod
              ones += 1
      if is_reducible_dp(ones, k, {}):
        ans = (ans + 1) % mod

      if k > 1:
          ans = (ans - n + mod) % mod

      return ans

    print(count_reducible_optimized(s, k))
solve()