连续字符的最小翻转以使字符串中的所有字符相同

给定一个仅由 1 和 0 组成的字符串。我们可以一次性更改此字符串的任何连续序列。找到这个最小翻转次数，使字符串只包含相同的字符。
例子：

Input : 00011110001110
Output : 2
We need to convert 1's sequence
so string consist of all 0's.

Input : 010101100011
Output : 4

我们需要找到字符串中的最小翻转，以便所有字符都相等。我们只需要找到仅由 0 或 1 组成的序列数。然后所需的翻转次数将是这个数字的一半，因为我们可以更改全 0 或全 1。

C++

// CPP program to find min flips in binary
// string to make all characters equal
#include 
using namespace std;
 
// To find min number of flips in binary string
int findFlips(char str[], int n)
{
    char last = ' '; int res = 0;
 
    for (int i = 0; i < n; i++) {
 
        // If last character is not equal
        // to str[i] increase res
        if (last != str[i])
            res++;
        last = str[i];
    }
 
    // To return min flips
    return res / 2;
}
 
// Driver program to check findFlips()
int main()
{
    char str[] = "00011110001110";
    int n = strlen(str);
 
    cout << findFlips(str, n);
 
    return 0;
}

Java

// Java program to find min flips in binary
// string to make all characters equal
public class minFlips {
 
    // To find min number of flips in binary string
    static int findFlips(String str, int n)
    {
        char last = ' '; int res = 0;
 
        for (int i = 0; i < n; i++) {
 
            // If last character is not equal
            // to str[i] increase res
            if (last != str.charAt(i))
                res++;
            last = str.charAt(i);
        }
 
        // To return min flips
        return res / 2;
    }
 
    // Driver program to check findFlips()
    public static void main(String[] args)
    {
        String str = "00011110001110";
        int n = str.length();
 
        System.out.println(findFlips(str, n));
    }
}

Python 3

# Python 3 program to find min flips in
# binary string to make all characters equal
 
# To find min number of flips in
# binary string
def findFlips(str, n):
 
    last = ' '
    res = 0
 
    for i in range( n) :
 
        # If last character is not equal
        # to str[i] increase res
        if (last != str[i]):
            res += 1
        last = str[i]
 
    # To return min flips
    return res // 2
 
# Driver Code
if __name__ == "__main__":
     
    str = "00011110001110"
    n = len(str)
 
    print(findFlips(str, n))
 
# This code is contributed by ita_c

C#

// C# program to find min flips in
// binary string to make all
// characters equal
using System;
 
public class GFG {
 
    // To find min number of flips
    // in binary string
    static int findFlips(String str, int n)
    {
        char last = ' '; int res = 0;
 
        for (int i = 0; i < n; i++) {
 
            // If last character is not
            // equal to str[i] increase
            // res
            if (last != str[i])
                res++;
            last = str[i];
        }
 
        // To return min flips
        return res / 2;
    }
 
    // Driver program to check findFlips()
    public static void Main()
    {
        String str = "00011110001110";
        int n = str.Length;
 
        Console.Write(findFlips(str, n));
    }
}
 
// This code is contributed by nitin mittal

PHP

Javascript

时间复杂度： O(n)
辅助空间： O(1)

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