Minimum replacements such that the difference between the index of the same characters is divisible by 3

• Last Updated : 15 Sep, 2022

Given a string of ‘0’, ‘1’ and ‘2’. The task is to find the minimum replacements in the string such that the differences between the indexes of the same characters is divisible by 3.

Examples:

Input: s = “2101200”
Output:
1201201 or 2102102 can be the resultant string
which has 3 replacements.

Input: s = “012”
Output:

Approach: There can be 6 different strings such that the difference between the index of similar characters is divisible by 3. Hence generate all 6 different strings, and compare the replacements done with the original string. Store the string which has a minimal number of replacements. The different strings can be generated using next_permutation in C++.

Below is the implementation of the above approach:

C++

 `// C++ program to find minimum replacements``// such that the difference between the``// index of the same characters``// is divisible by 3``#include ``using` `namespace` `std;` `// Function to count the number of``// minimal replacements``int` `countMinimalReplacements(string s)``{` `    ``int` `n = s.length();` `    ``int` `mini = INT_MAX;``    ``string dup = ``"012"``;` `    ``// Generate all permutations``    ``do` `{` `        ``// Count the replacements``        ``int` `dif = 0;``        ``for` `(``int` `i = 0; i < n; i++)``            ``if` `(s[i] != dup[i % 3])``                ``dif++;` `        ``mini = min(mini, dif);``    ``} ``while` `(next_permutation(dup.begin(), dup.end()));` `    ``// Return the replacements``    ``return` `mini;``}` `// Driver Code``int` `main()``{``    ``string s = ``"2101200"``;``    ``cout << countMinimalReplacements(s);``    ``return` `0;``}`

Java

 `// Java program to find minimum replacements``// such that the difference between the``// index of the same characters``// is divisible by 3``class` `GFG {` `    ``// Function to count the number of``    ``// minimal replacements``    ``static` `int` `countMinimalReplacements(String s)``    ``{` `        ``int` `n = s.length();` `        ``int` `mini = Integer.MAX_VALUE;``        ``char``[] dup = ``"012"``.toCharArray();` `        ``// Generate all permutations``        ``do` `{` `            ``// Count the replacements``            ``int` `dif = ``0``;``            ``for` `(``int` `i = ``0``; i < n; i++) {``                ``if` `(s.charAt(i) != dup[i % ``3``]) {``                    ``dif++;``                ``}``            ``}` `            ``mini = Math.min(mini, dif);``        ``} ``while` `(next_permutation(dup));` `        ``// Return the replacements``        ``return` `mini;``    ``}` `    ``static` `boolean` `next_permutation(``char``[] p)``    ``{``        ``for` `(``int` `a = p.length - ``2``; a >= ``0``; --a) {``            ``if` `(p[a] < p[a + ``1``]) {``                ``for` `(``int` `b = p.length - ``1``;; --b) {``                    ``if` `(p[b] > p[a]) {``                        ``char` `t = p[a];``                        ``p[a] = p[b];``                        ``p[b] = t;``                        ``for` `(++a, b = p.length - ``1``; a < b; ++a, --b) {``                            ``t = p[a];``                            ``p[a] = p[b];``                            ``p[b] = t;``                        ``}``                        ``return` `true``;``                    ``}``                ``}``            ``}``        ``}``        ``return` `false``;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String args[])``    ``{``        ``String s = ``"2101200"``;``        ``System.out.println(countMinimalReplacements(s));``    ``}``}` `/* This code contributed by PrinciRaj1992 */`

C#

 `// C# program to find minimum replacements``// such that the difference between the``// index of the same characters``// is divisible by 3``using` `System;` `class` `GFG {` `    ``// Function to count the number of``    ``// minimal replacements``    ``static` `int` `countMinimalReplacements(String s)``    ``{` `        ``int` `n = s.Length;` `        ``int` `mini = ``int``.MaxValue;``        ``char``[] dup = ``"012"``.ToCharArray();` `        ``// Generate all permutations``        ``do` `{` `            ``// Count the replacements``            ``int` `dif = 0;``            ``for` `(``int` `i = 0; i < n; i++) {``                ``if` `(s[i] != dup[i % 3]) {``                    ``dif++;``                ``}``            ``}` `            ``mini = Math.Min(mini, dif);``        ``} ``while` `(next_permutation(dup));` `        ``// Return the replacements``        ``return` `mini;``    ``}` `    ``static` `bool` `next_permutation(``char``[] p)``    ``{``        ``for` `(``int` `a = p.Length - 2; a >= 0; --a) {``            ``if` `(p[a] < p[a + 1]) {``                ``for` `(``int` `b = p.Length - 1;; --b) {``                    ``if` `(p[b] > p[a]) {``                        ``char` `t = p[a];``                        ``p[a] = p[b];``                        ``p[b] = t;``                        ``for` `(++a, b = p.Length - 1; a < b; ++a, --b) {``                            ``t = p[a];``                            ``p[a] = p[b];``                            ``p[b] = t;``                        ``}``                        ``return` `true``;``                    ``}``                ``}``            ``}``        ``}``        ``return` `false``;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``String s = ``"2101200"``;``        ``Console.WriteLine(countMinimalReplacements(s));``    ``}``}` `// This code has been contributed by 29AjayKumar`

Javascript

 ``

Output

`3`

Complexity Analysis:

• Time Complexity: O(3*N), as we are a loop to traverse N times. Where n is the length of the string.
• Auxiliary Space: O(1), as we are not using any extra space.

My Personal Notes arrow_drop_up