Leetcode 1758: Minimum Changes To Make Alternating Binary String

Input: You are given a string `s` which only contains the characters '0' and '1'.

Example: Input: s = '1001'

Constraints:

• 1 <= s.length <= 10^4

• s[i] is either '0' or '1'.

Output: Return the minimum number of operations needed to make `s` alternating.

Example: Output: 1

Constraints:

• The number of operations should be minimized.

Goal: The goal is to minimize the number of changes required to convert the given string into an alternating string.

Steps:

• 1. Traverse the string `s` and check each adjacent pair of characters.

• 2. If two adjacent characters are the same, a change is needed.

• 3. Calculate the number of changes required for both possible alternating patterns: '010101...' and '101010...'.

• 4. Return the minimum of these two values.

Goal: Make sure the solution works for large inputs and performs within time limits.

Steps:

• The solution should handle strings of length up to 10^4 efficiently.

Assumptions:

• The string `s` is not empty and contains only '0' or '1'.

• Input: Input: s = '1001'

• Explanation: To make the string alternating, change the second '0' to '1' or the last '1' to '0'. The minimum number of operations needed is 1.

• Input: Input: s = '1101'

• Explanation: Here, the first two '1's are adjacent and need to be altered to make the string alternating. Two changes are required.

Approach: The approach is to calculate the number of operations required for both alternating patterns: '010101...' and '101010...'. For each pattern, count how many characters need to be changed and then choose the minimum number of changes.

Observations:

• The problem is about comparing the string to two alternating patterns and finding the minimum number of changes needed.

• Since there are only two possible alternating patterns, we can compare the string to each pattern and count the differences.

Steps:

• 1. Initialize two counters, one for each pattern ('010101...' and '101010...').

• 2. Traverse the string and compare each character with the corresponding character in both alternating patterns.

• 3. Count the differences for both patterns.

• 4. Return the minimum count as the result.

Empty Inputs:

• If the string is empty, return 0 (no changes needed).

Large Inputs:

• Ensure the solution can handle strings of length up to 10^4 efficiently.

Special Values:

• If the string is already alternating, the result is 0.

Constraints:

• The solution should be optimized to handle the maximum string length.

int minOperations(string s) {
    int zero = 0, one = 0, zf = false, of = false;
    for(int i = 0; i < s.size(); i++) {
        if(zf   && s[i] == '0') zero++;
        if(!zf  && s[i] == '1') zero++;
        if(!of   && s[i] == '0') one++;
        if(of  && s[i] == '1') one++;
        zf = !zf, of = !of;
    }
    return min(zero, one);
}

1 : Function Declaration

int minOperations(string s) {

Declares the function to calculate the minimum operations for an alternating binary string.

2 : Variable Initialization

    int zero = 0, one = 0, zf = false, of = false;

Initializes counters and flags for tracking mismatches with alternating patterns.

3 : Loop

    for(int i = 0; i < s.size(); i++) {

Iterates over the characters of the input string to count mismatches.

4 : Condition Check

        if(zf   && s[i] == '0') zero++;

Checks if the current character matches the '0' expectation for the first pattern.

5 : Condition Check

        if(!zf  && s[i] == '1') zero++;

Checks if the current character matches the '1' expectation for the first pattern.

6 : Condition Check

        if(!of   && s[i] == '0') one++;

Checks if the current character matches the '0' expectation for the second pattern.

7 : Condition Check

        if(of  && s[i] == '1') one++;

Checks if the current character matches the '1' expectation for the second pattern.

8 : Flag Update

        zf = !zf, of = !of;

Flips the flags to alternate the expected pattern for the next iteration.

9 : Return Statement

    return min(zero, one);

Returns the minimum of the two mismatch counts as the result.

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

Average Case: O(n), as each character is checked against both alternating patterns.

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

Description: The time complexity is linear, as the string is traversed once.

Best Case: O(1), as no additional space is needed regardless of the input.

Worst Case: O(1), since only a few variables are used for comparison.

Description: The space complexity is constant, as only a few counters are used.

Leetcode 1758: Minimum Changes To Make Alternating Binary String

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