Separate Black and White Balls

class Solution {
public:
    long long minimumSwapsToSort(string balls) {
        long long swapCount = 0;
        int countOfOnes = 0;
    
        for (int index = 0; index < balls.size(); index++) {
            if (balls[index] == '0') {
                swapCount += countOfOnes;
            } else {
                countOfOnes++;
            }
        }
    
        return swapCount;
    }
};

This solution addresses the problem of sorting a string containing black and white balls represented by the characters '0' and '1'. The approach taken works on counting the number of '1's (white balls) encountered as one iterates through the string while accumulating a swap count whenever a '0' (black ball) is encountered after any '1'.

Key steps in the solution:

• Initialize a swapCount as 0 to store the resultant minimum number of swaps needed.
• Initialize countOfOnes as 0 to keep track of the number of white balls ('1's) encountered.
• Loop through each character in the string:
  • If a black ball ('0') is discovered, add the current countOfOnes to swapCount.
  • If a white ball ('1') is found, increment the countOfOnes.
• Return swapCount at the end.

This approach ensures that every black ball found after encountering white balls will account for swaps needed to move all those white balls past each black ball, effectively sorting the string. The algorithm is efficient, operating in linear time relative to the size of the string.

class Solution {
    
    public long calculateSwaps(String sequence) {
        long swaps = 0;
        int blacks = 0;
    
        for (int index = 0; index < sequence.length(); index++) {
            if (sequence.charAt(index) == '0') {
                swaps += (long) blacks;
            } else {
                blacks++;
            }
        }
    
        return swaps;
    }
}

The provided Java solution focuses on calculating the minimum number of swaps needed to separate black and white balls represented by '0's and '1's in a given sequence. The solution leverages a single traversal of the sequence string, utilizing a variable to track the number of swaps and the count of black balls encountered.

Explanation of the Code:

• A class named Solution contains a method called calculateSwaps.
• The method calculateSwaps accepts a string sequence as a parameter, which represents a sequence of black and white balls.
• Two main variables are used:
  • swaps to store the cumulative number of swaps required.
  • blacks to count the number of black balls ('1') encountered as the sequence is processed.
• As the program iterates through each character in the sequence:
  • If a white ball (character '0') is found, the number of swaps is incremented by the number of black balls found up to that point (blacks).
  • If a black ball (character '1') is found, the blacks counter is incremented.
• Finally, the variable swaps is returned as the result, indicating the total number of swaps needed.

Key Points:

• The solution efficiently processes the sequence in O(n) time complexity where n is the length of the sequence string.
• The method only uses O(1) additional space for its counters, making it space-efficient as well.

class Solution:
    def minSwapsRequired(self, t: str) -> int:
        swap_count = 0
        count_black = 0
    
        for ball in t:
            if ball == "0":
                swap_count += count_black
            else:
                count_black += 1
    
        return swap_count

Separating black and white balls efficiently involves counting and minimizing swaps, a classic sorting problem. In Python, the solution provided uses a straightforward method to solve this problem by iteratively counting swaps needed to group all black (represented as "1") balls and white (represented as "0") balls together.

The approach works as follows:

• Initialize swap_count to track the number of swaps required and count_black to count the black balls encountered so far.
• Traverse through each ball in the string t.
• For each white ball ("0") encountered, increase swap_count by the number of black balls count_black found before it. This step ensures that each white ball will count the swaps needed to move past all preceding black balls to be grouped correctly.
• If a black ball ("1") is encountered, simply increment the count_black.
• Finally, return swap_count which will be the minimum number of swaps required to group all black balls to the left of all white balls.

This elegant solution ensures that you can determine the minimum swaps required with a single pass through the string, providing an efficient O(n) complexity where n is the number of balls. This avoids the need for a more cumbersome and less efficient sorting algorithm. Ensure the input string t consists only of "0"s and "1"s to avoid errors or unexpected behavior.