# Minimum replacements to make adjacent characters unequal in a ternary string

Given a string of ‘0’, ‘1’ and ‘2’. The task is to find the minimum number of replacements such that the adjacent characters are not equal.

Examples:

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

Input: s = “0120102”
Output: 0

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The following problem can be solved using greedy method. We can greedily compare every adjacent pair. If the adjacent pairs which is character at ith and i-1th are same, then replace ith th charcater with a character which is not equal to the character at i-1th and i+1th index. In case of the last adjacent pair, just replace it with the character which is not equal to the character at i-1th index.

Below is the implementation of the above approach:

## 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
# 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 `

## PHP

 ` `

Output:

```2
```

Time Complexity : O(n)

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