696. Count Binary Substrings

/**
 * Problem: LeetCode 696 - Count Binary Substrings
 * Difficulty: Easy
 * Language: Java
 * Author: CodeByTushu
 */
class Solution {
    public int countBinarySubstrings(String s) {
        int prevRun = 0;
        int currRun = 1;
        int count = 0;

        for (int i = 1; i < s.length(); i++) {
            if (s.charAt(i) == s.charAt(i - 1)) {
                currRun++;
            } else {
                count += Math.min(prevRun, currRun);
                prevRun = currRun;
                currRun = 1;
            }
        }
        count += Math.min(prevRun, currRun);

        return count;
    }
}

"""
Problem: LeetCode 696 - Count Binary Substrings
Difficulty: Easy
Language: Python
Author: CodeByTushu
"""
class Solution:
    def countBinarySubstrings(self, s: str) -> int:
        prev_run = 0
        curr_run = 1
        count = 0

        for i in range(1, len(s)):
            if s[i] == s[i - 1]:
                curr_run += 1
            else:
                count += min(prev_run, curr_run)
                prev_run = curr_run
                curr_run = 1

        count += min(prev_run, curr_run)
        return count

/**
 * Problem: LeetCode 696 - Count Binary Substrings
 * Difficulty: Easy
 * Language: C++
 * Author: CodeByTushu
 */
#include <string>
#include <algorithm>

using namespace std;

class Solution {
public:
    int countBinarySubstrings(string s) {
        int prevRun = 0;
        int currRun = 1;
        int count = 0;

        for (int i = 1; i < (int)s.length(); i++) {
            if (s[i] == s[i - 1]) {
                currRun++;
            } else {
                count += min(prevRun, currRun);
                prevRun = currRun;
                currRun = 1;
            }
        }
        count += min(prevRun, currRun);

        return count;
    }
};

/**
 * Problem: LeetCode 696 - Count Binary Substrings
 * Difficulty: Easy
 * Language: JavaScript
 * Author: CodeByTushu
 */
/**
 * @param {string} s
 * @return {number}
 */
var countBinarySubstrings = function(s) {
    let prevRun = 0;
    let currRun = 1;
    let count = 0;

    for (let i = 1; i < s.length; i++) {
        if (s[i] === s[i - 1]) {
            currRun++;
        } else {
            count += Math.min(prevRun, currRun);
            prevRun = currRun;
            currRun = 1;
        }
    }
    count += Math.min(prevRun, currRun);

    return count;
};