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
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 character 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 <bits/stdc++.h> 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; } |
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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 |
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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 |
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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 |
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PHP
<?php // PHP 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 $n = strlen($s); $cnt = 0; $str = "012"; // Iterate in the string for ($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 != ($n - 1)) { // Check for character which is // not same in i+1 and i-1 for ($it = 0 ; $it < strlen($str); $it++) { if ($str[$it] != $s[$i + 1] && $str[$it] != $s[$i - 1]) { $s[$i] = $str[$it]; break; } } } else // Last pair { // Check for character which is // not same in i-1 index for ($it = 0 ; $it< strlen($str); $it++) { if ($str[$it] != $s[$i - 1]) { $s[$i] = $str[$it]; break; } } } } } return $cnt; } // Driver Code $s = "201220211"; echo countMinimalReplacements($s); // This code is contributed by Ryuga ?> |
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Output:
2
Time Complexity : O(n)
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