要删除的最小元素数，以使成对的连续元素相同

给定一个字符串str 。任务是计算要删除的最小元素数，以使成对的连续元素相同
例子：

Input : str = “11344” 
Output: 1 
Remove the digit 3 from 3rd place so that the string becomes 1144. Thus pairwise two consecutive elements are same. Hence answer is 1.
Input : str = “55553” 
Output : 1 
Remove the digit 3 from the 5th place so that the string becomes 5555. Thus pairwise two consecutive elements are same. Hence answer is 1. 
 

方法：检查当前两个连续的元素是否相同。如果是，则将索引增加2并继续检查，直到遍历所有元素。否则将索引增加1并以1计数。
下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
 
// Function to count the minimum number of elements
// to remove from a number so that pairwise two
// consecutive digits are same.
int countConsecutive(string s)
{
 
    // initialize counting variable
    int count = 0;
 
    for (int i = 0; i < s.size(); i++) {
 
        // check if two consecutive digits are same
        if (s[i] == s[i + 1])
            i++;
        else
            count++;
    }
 
    return count;
}
 
// Driver code
int main()
{
    string str = "44522255";
    cout << countConsecutive(str);
    return 0;
}

Java

// Java implementation of the above approach
 
class GFG {
 
// Function to count the minimum number of elements
// to remove from a number so that pairwise two
// consecutive digits are same.
    static int countConsecutive(String s) {
 
        // initialize counting variable
        int count = 0;
 
        for (int i = 0; i < s.length(); i++) {
 
            // check if two consecutive digits are same
            if (s.charAt(i) == s.charAt(i + 1)) {
                i++;
            } else {
                count++;
            }
        }
 
        return count;
    }
 
// Driver code
    public static void main(String args[]) {
        String str = "44522255";
        System.out.println(countConsecutive(str));
 
    }
}
 
// This code is contributed by PrinciRaj19992

Python3

# Python 3 implementation of the above approach
 
# Function to count the minimum number of
# elements to remove from a number so that
# pairwise two consecutive digits are same.
def countConsecutive(s):
 
    # initialize counting variable
    count = -1
 
    for i in range(len(s)-1):
 
        # check if two consecutive
        # digits are same
        if(i <= len(s)):
            if (s[i] is s[i + 1]):
                i += 1
            else:
                count += 1
    return count
 
# Driver code
if __name__ == '__main__':
    str = "44522255"
    print(countConsecutive(str))
     
# This code is contributed by PrinciRaj1992

C#

// C# implementation of above approach
using System;
public class GFG {
 
    // Function to count the minimum number of elements
    // to remove from a number so that pairwise two
    // consecutive digits are same.
    static int countConsecutive(String s) {
 
        // initialize counting variable
        int count = 0;
 
        for (int i = 0; i < s.Length; i++) {
 
            // check if two consecutive digits are same
            if (s[i] == s[i+1]) {
                i++;
            } else {
                count++;
            }
        }
 
        return count;
    }
 
// Driver code
    public static void Main() {
        String str = "44522255";
        Console.WriteLine(countConsecutive(str));
 
    }
}
 
// This code is contributed by 29AjayKumar

Javascript

// Javascript implementation of the above approach
 
// Function to count the minimum number of elements
// to remove from a number so that pairwise two
// consecutive digits are same.
function countConsecutive(s)
{
 
    // initialize counting variable
    let count = 0;
 
    for (let i = 0; i < s.length; i++) {
 
        // check if two consecutive digits are same
        if (s[i] == s[i + 1])
            i++;
        else
            count++;
    }
 
    return count;
}
 
// Driver code
    let str = "44522255";
    document.write(countConsecutive(str));
 
 
 
 
// This code is contributed by Surbhi Tyagi.

输出：

2

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