algoadvance

You are given an array of binary strings strs and two integers m and n.

• strs[i] represents a binary string consisting only of ‘0’s and ‘1’s.
• m represents the maximum number of ‘0’s you can have.
• n represents the maximum number of ‘1’s you can have.

Your task is to find the maximum number of strings you can form from the given array strs without exceeding the given values of m and n.

Clarifying Questions

1. What should be done if strs is empty?
   • Return 0 as no strings can be formed.
2. Are there any constraints on the length of the string strs or values of m and n?
   • The constraints are as follows:
     • 1 <= strs.length <= 600
     • 1 <= strs[i].length <= 100
     • strs[i] consists only of ‘0’s and ‘1’s.
     • 0 <= m, n <= 100.
3. Is there any specific output format required?
   • The output should be a single integer representing the maximum number of strings that can be formed.

Strategy

This problem requires a dynamic programming (DP) approach because it involves making decisions at each step to maximize the outcome while keeping track of resource constraints (m and n). We can use a 2D DP array where dp[i][j] represents the maximum number of strings that can be formed with at most i ‘0’s and j ‘1’s.

Steps:

1. Initialize a DP Array:
   • Create a 2D DP array of size (m+1) x (n+1) initialized to 0.
2. Count ‘0’s and ‘1’s for Each String:
   • For each string in strs, count the number of ‘0’s and ‘1’s.
3. Update the DP Array:
   • For each string (represented by its ‘0’ and ‘1’ count), update the DP array backward to ensure that each string is only considered once. Iterate backward to avoid overwriting values that haven’t been used yet.
4. Extract the Result:
   • The desired result will be the maximum value in the DP array after processing all strings.

Code

def findMaxForm(strs, m, n):
    # Initialize the DP table
    dp = [[0] * (n + 1) for _ in range(m + 1)]

    # Process each string in strs
    for s in strs:
        zeros = s.count('0')
        ones = s.count('1')

        # Update the DP table backwards
        for i in range(m, zeros - 1, -1):
            for j in range(n, ones - 1, -1):
                dp[i][j] = max(dp[i][j], dp[i - zeros][j - ones] + 1)

    return dp[m][n]

# Example Usage
strs = ["10", "0001", "111001", "1", "0"]
m = 5
n = 3
print(findMaxForm(strs, m, n))  # Output will be 4

Time Complexity

• Initialization: O(m * n)
• Processing Strings:
  • For each string, count ‘0’s and ‘1’s in O(L) where L is the length of the string. This contributes O(S * L).
  • Updating the DP table in the worst case involves O(m * n), but this is nested inside the string processing loop.

Total Time Complexity: O(S * L + S * m * n), where S is the number of strings and L is the maximum length of a string in strs.

This ensures the solution is efficient given the constraints.

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