最小替换，以使三进制字符串的相邻字符不相等

给定字符串“ 0”，“ 1”和“ 2”。任务是找到最少的替换数，以使相邻字符不相等。
例子：

Input: s = “201220211” 
Output: 2 
Resultant string after changes is 201210210 
Input: s = “0120102” 
Output: 0

方法：可以使用贪婪法解决以下问题。我们可以贪婪地比较每对相邻的货币对。如果相邻对，其是字符在^第i^个和^第i-1相同，则更换^第i个字符与未^第i-1和i + 1^个索引处等于所述字符的字符。如果是最后一对，则将字符替换为不等于^第i-1^个索引处的字符。
下面是上述方法的实现：

C++

// C++ program to count the minimal
// replacements such that adjacent characters
// are unequal
#include 
using namespace std;
 
// Function to count the number of
// minimal replacements
int countMinimalReplacements(string s)
{
    // Find the length of the string
    int n = s.length();
 
    int cnt = 0;
 
    // Iterate in the string
    for (int i = 1; i < n; i++) {
 
        // Check if adjacent is similar
        if (s[i] == s[i - 1]) {
            cnt += 1;
 
            // If not the last pair
            if (i != (n - 1)) {
 
                // Check for character which is
                // not same  in i+1 and i-1
                for (auto it : "012") {
                    if (it != s[i + 1] &&
                        it != s[i - 1]) {
                        s[i] = it;
                        break;
                    }
                }
            }
 
            else // Last pair
            {
                // Check for character which is
                // not same in i-1 index
                for (auto it : "012") {
                    if (it != s[i - 1]) {
                        s[i] = it;
                        break;
                    }
                }
            }
        }
    }
 
    return cnt;
}
 
// Driver Code
int main()
{
    string s = "201220211";
    cout << countMinimalReplacements(s);
    return 0;
}

Java

// Java program to count the minimal
// replacements such that adjacent
// characters are unequal
class GFG
{
 
    static final int MAX = 26;
     
    // Function to count the number of
    // minimal replacements
    static int countMinimalReplacements(char[] s)
    {
        // Find the length of the String
        int n = s.length;
 
        int cnt = 0;
 
        // Iterate in the String
        for (int i = 1; i < n; i++)
        {
 
            // Check if adjacent is similar
            if (s[i] == s[i - 1])
            {
                cnt += 1;
 
                // If not the last pair
                if (i != (n - 1))
                {
 
                    // Check for character which is
                    // not same in i+1 and i-1
                    for (char it : "012".toCharArray())
                    {
                        if (it != s[i + 1]
                                && it != s[i - 1])
                        {
                            s[i] = it;
                            break;
                        }
                    }
                }
                else // Last pair
                {
                    // Check for character which is
                    // not same in i-1 index
                    for (char it : "012".toCharArray())
                    {
                        if (it != s[i - 1])
                        {
                            s[i] = it;
                            break;
                        }
                    }
                }
            }
        }
        return cnt;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        String s = "201220211";
        System.out.println(countMinimalReplacements(s.toCharArray()));
    }
}
 
// This code is contributed by 29AjayKumar

Python3

# Python 3 program to count the minimal
# replacements such that adjacent
# characters are unequal
 
# Function to count the number of
# minimal replacements
def countMinimalReplacements(s):
     
    # Find the length of the string
    n = len(s)
 
    cnt = 0
 
    # Iterate in the string
    for i in range(1, n):
         
        # Check if adjacent is similar
        if (s[i] == s[i - 1]):
            cnt += 1;
 
            # If not the last pair
            if (i != (n - 1)):
                 
                # Check for character which is
                # not same in i+1 and i-1
                s = list(s)
                for j in "012":
                    if (j != s[i + 1] and
                        j != s[i - 1]):
                        s[i] = j
                        break
 
                s = ''.join(s)
                 
            # Last pair
            else:
                 
                # Check for character which is
                # not same in i-1 index
                s = list(s)
 
                for k in "012":
                    if (k != s[i - 1]):
                        s[i] = k
                        break
                s = ''.join(s)
         
    return cnt
 
# Driver Code
if __name__ == '__main__':
    s = "201220211"
    print(countMinimalReplacements(s))
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# program to count the minimal
// replacements such that adjacent
// characters are unequal
using System;
 
class GFG
{
    static readonly int MAX = 26;
     
    // Function to count the number of
    // minimal replacements
    static int countMinimalReplacements(char[] s)
    {
        // Find the length of the String
        int n = s.Length;
 
        int cnt = 0;
 
        // Iterate in the String
        for (int i = 1; i < n; i++)
        {
 
            // Check if adjacent is similar
            if (s[i] == s[i - 1])
            {
                cnt += 1;
 
                // If not the last pair
                if (i != (n - 1))
                {
 
                    // Check for character which is
                    // not same in i+1 and i-1
                    foreach (char it in "012".ToCharArray())
                    {
                        if (it != s[i + 1] &&
                            it != s[i - 1])
                        {
                            s[i] = it;
                            break;
                        }
                    }
                }
                else // Last pair
                {
                    // Check for character which is
                    // not same in i-1 index
                    foreach (char it in "012".ToCharArray())
                    {
                        if (it != s[i - 1])
                        {
                            s[i] = it;
                            break;
                        }
                    }
                }
            }
        }
        return cnt;
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        String s = "201220211";
        Console.WriteLine(countMinimalReplacements(s.ToCharArray()));
    }
}
 
// This code is contributed by Rajput-Ji

Javascript

      // JavaScript program to count the minimal
      // replacements such that adjacent
      // characters are unequal
 
      // Function to count the number of
      // minimal replacements
      function countMinimalReplacements(s) {
        // Find the length of the string
        var n = s.length;
 
        var cnt = 0;
        var str = "012";
 
        // Iterate in the string
        for (var i = 1; i < n; i++) {
          // Check if adjacent is similar
          if (s[i] === s[i - 1]) {
            cnt += 1;
 
            // If not the last pair
            if (i !== n - 1) {
              // Check for character which is
              // not same in i+1 and i-1
              for (var it = 0; it < str.length; it++) {
                if (str[it] !== s[i + 1] && str[it] !== s[i - 1]) {
                  s[i] = str[it];
                  break;
                }
              }
            } // Last pair
            else {
              // Check for character which is
              // not same in i-1 index
              for (var it = 0; it < str.length; it++) {
                if (str[it] !== s[i - 1]) {
                  s[i] = str[it];
                  break;
                }
              }
            }
          }
        }
 
        return cnt;
      }
 
      // Driver Code
      var s = "201220211";
      document.write(countMinimalReplacements(s));
    

输出：

2

时间复杂度： O(n)

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