Leetcode 2546: Apply Bitwise Operations to Make Strings Equal

Input: You are given two binary strings `s` and `target` with the same length `n`. Each string consists of only '0's and '1's.

Example: s = '1101', target = '1011'

Constraints:

• 2 <= n <= 10^5

• s and target consist of only '0' and '1'.

Output: Return `true` if it is possible to transform `s` into `target` using the defined operation; otherwise return `false`.

Example: true

Constraints:

Goal: The task is to determine whether it's possible to make `s` equal to `target` by repeatedly applying the specified operation.

Steps:

• 1. Check if both strings `s` and `target` contain at least one '1'.

• 2. If both strings contain '1' or both do not contain '1', return true.

• 3. Otherwise, return false.

Goal: The solution needs to efficiently handle the constraints, especially for large values of `n`.

Steps:

• 2 <= n <= 10^5

• s and target consist of only '0' and '1'.

Assumptions:

• The input strings `s` and `target` will always have the same length.

• Only binary strings (composed of '0' and '1') are considered.

• Input: s = '1101', target = '1011'

• Explanation: In this case, the strings `s` and `target` can be transformed into one another using a series of operations, so the result is `true`.

• Input: s = '1000', target = '1111'

• Explanation: It is not possible to transform `s` into `target` because no sequence of operations will change the '0' in `s` to '1' independently.

Approach: The solution can be determined by checking whether both `s` and `target` contain at least one '1'. If both do or both do not, it is possible to transform `s` into `target`.

Observations:

• The key observation is that the operation preserves the presence of '1's in both strings.

• If both `s` and `target` contain at least one '1', the transformation is possible.

• We need to check if both strings have the same pattern in terms of the presence of '1's.

Steps:

• 1. Check whether both `s` and `target` contain any '1'.

• 2. If both contain '1' or neither contain '1', return `true`.

• 3. If one contains '1' and the other does not, return `false`.

Empty Inputs:

• Handle cases where `s` and `target` are minimal in size, i.e., 2 characters long.

Large Inputs:

• Ensure the solution can handle the upper constraint where `n` can be as large as 10^5.

Special Values:

• Consider cases where the strings `s` and `target` are identical or completely different.

Constraints:

• The solution needs to handle edge cases like strings with no '1' or where `s` and `target` differ significantly.

bool makeStringsEqual(string s, string target) {
    
    return (s.find('1') != string::npos) == (target.find('1') != string::npos);
}

1 : Function Definition

bool makeStringsEqual(string s, string target) {

The function 'makeStringsEqual' is defined to take two strings, 's' and 'target', and return a boolean value indicating whether both strings have the same presence of the character '1'.

2 : Return Statement

    return (s.find('1') != string::npos) == (target.find('1') != string::npos);

The function uses the 'find' method to check if the character '1' exists in both strings. If both strings either contain or do not contain the character '1', the function returns 'true'; otherwise, it returns 'false'.

Best Case: O(n), where n is the length of the string.

Average Case: O(n), since checking if '1' exists in both strings is linear.

Worst Case: O(n), since checking for '1' in both strings is a linear operation.

Description: The time complexity is O(n) since we need to scan both strings to check for '1'.

Best Case: O(1), as no additional space is required in the solution.

Worst Case: O(1), since we do not need any extra space apart from the input.

Description: The space complexity is O(1), as the solution operates in constant space.

Leetcode 2546: Apply Bitwise Operations to Make Strings Equal

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