algoadvance

You are given three binary strings s1, s2, and target, each of length n. You need to determine whether it is possible to make all characters in each position of the three strings equal to the corresponding character in the target string using the following operation:

1. Choose a non-negative integer i (0 <= i < n) and select a pair of indices (j, k) with 0 <= j < k < 3. Then you can set s[j][i] = s[k][i].

Return true if it’s possible to make the three strings equal to the target string, and false otherwise.

Clarifying Questions

1. Can s1, s2, and target have different lengths?
   • No, all three strings are guaranteed to have the same length n.
2. Are there any constraints on n (length of the strings)?
   • Typical constraints for this problem have not been provided. For edge cases, consider very large or very small values of n.
3. Is the input constrained to be binary strings (containing only ‘0’ and ‘1’)?
   • Yes, all provided strings contain only ‘0’ and ‘1’.
4. Is the operation allowed to be performed any number of times?
   • Yes, the operation can be applied any non-negative number of times.

Strategy

Given the functionality of the operation allowed, each position in the strings can be independently analyzed to see if they can be adjusted to match the corresponding position in the target string.

For each position i:

• Check the characters s1[i], s2[i], and target[i].
• Based on the possible combinations of values in s1[i] and s2[i], check if it’s feasible to adjust both to match target[i].

Code

def make_three_strings_equal(s1, s2, target):
    n = len(s1)
    
    for i in range(n):
        if target[i] == '0':
            # To make all zeros, we need both to have at least one zero at position i
            if not (s1[i] == '0' or s2[i] == '0'):
                return False
        elif target[i] == '1':
            # To make all ones, we need both to have at least one one at position i
            if not (s1[i] == '1' or s2[i] == '1'):
                return False
    
    return True

Time Complexity

• Time Complexity: The algorithm iterates through each character of the string exactly once. Therefore, the time complexity is O(n), where n is the length of the strings.
• Space Complexity: The space complexity is O(1) since we are only using a constant amount of space for variables.

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