Minimize the number of replacements to get a string with same number of ‘a’, ‘b’ and ‘c’ in it

Given a string consisting of only three possible characters ‘a’, ‘b’ or ‘c’. The task is to replace characters of the given string with ‘a’, ‘b’ or ‘c’ only such that there is the equal number of characters of ‘a’, ‘b’ and ‘c’ in the string. The task is to minimize the number of replacements and print the lexicographically smallest string possible of all such strings with minimal replacements. If it is not possible to obtain such a string, print -1. Examples:

Input : s = "bcabba"
Output : bcabca
Number of replacements done is 1 and this is
the lexicographically smallest possible 

Input : "aaaaaa"
Output : aabbcc

Input : "aaaaa"
Output : -1

Approach:

• Count the number of ‘a’, ‘b’ and ‘c’ in the string.
• If the count of them is equal then the same string will be the answer.
• If length of the string is not a multiple of 3, then it is not possible.
• First, reduce the number of exceeding a’s in the string.
  1. Replace ‘c’ by ‘a’ if there are extra ‘c’ using a sliding window technique from left.
  2. Replace ‘b’ by ‘a’ if there are extra ‘b’ using a sliding window technique from left in case of no ‘c’ at an index.
• Secondly, reduce the number of exceeding b’s in the string by replacing ‘c’ from front using sliding window.
• Thirdly, reduce the number of exceeding c’s by reducing the number of extra ‘a’ from the back using sliding window.
• Fourthly, reduce the number of exceeding b’s in the string by reducing the number of extra ‘a’ from the back.
• Fifthly, reduce the number of exceeding c’s if any by reducing the number of extra ‘b’ from the back.

We keep on replacing from back in order to get the lexicographically smallest string. Below is the implementation of the above approach: 

C++

// CPP program to Minimize the number of
// replacements to get a string with same
// number of ‘a’, ‘b’ and ‘c’ in it

#include <bits/stdc++.h>
using namespace std;

// Function to count numbers
string lexoSmallest(string s, int n)
{
    // Count the number of 'a', 'b' and
    // 'c' in string
    int ca = 0, cb = 0, cc = 0;

    for (int i = 0; i < n; i++) {
        if (s[i] == 'a')
            ca++;
        else if (s[i] == 'b')
            cb++;
        else
            cc++;
    }

    // If equal previously
    if (ca == cb && cb == cc) {
        return s;
    }

    int cnt = n / 3;

    // If not a multiple of 3
    if (cnt * 3 != n) {
        return "-1";
    }

    int i = 0;

    // Increase the number of a's by
    // removing extra 'b' and ;c;
    while (ca < cnt && i < n) {

        // Check if it is 'b' and it more
        // than n/3
        if (s[i] == 'b' && cb > cnt) {
            cb--;
            s[i] = 'a';
            ca++;
        }

        // Check if it is 'c' and it
        // more than n/3
        else if (s[i] == 'c' && cc > cnt) {
            cc--;
            s[i] = 'a';
            ca++;
        }

        i++;
    }

    i = 0;

    // Increase the number of b's by
    // removing extra 'c'
    while (cb < cnt && i < n) {

        // Check if it is 'c' and it more
        // than n/3
        if (s[i] == 'c' && cc > cnt) {
            cc--;
            s[i] = '1';

            cb++;
        }
        i++;
    }

    i = n - 1;

    // Increase the number of c's from back
    while (cc < cnt && i >= 0) {

        // Check if it is 'a' and it more
        // than n/3
        if (s[i] == 'a' && ca > cnt) {
            ca--;
            s[i] = 'c';
            cc++;
        }

        i--;
    }

    i = n - 1;

    // Increase the number of b's from back
    while (cb < cnt && i >= 0) {

        // Check if it is 'a' and it more
        // than n/3
        if (s[i] == 'a' && ca > cnt) {
            ca--;
            s[i] = 'b';
            cb++;
        }

        i--;
    }

    i = n - 1;

    // Increase the number of c's from back
    while (cc < cnt && i >= 0) {

        // Check if it is 'b' and it more
        // than n/3
        if (s[i] == 'b' && cb > cnt) {
            cb--;
            s[i] = 'c';
            cc++;
        }

        i--;
    }

    return s;
}

// Driver Code
int main()
{
    string s = "aaaaaa";
    int n = s.size();

    cout << lexoSmallest(s, n);

    return 0;
}

Java

// Java program to minimize the number of
// replacements to get a string with same
// number of ‘a’, ‘b’ and ‘c’ in it
import java.util.*;

public class GFG {

  // Function to count numbers
  static String lexoSmallest(String str, int n)
  {

    // Count the number of 'a', 'b' and
    // 'c' in string
    int ca = 0, cb = 0, cc = 0;
    StringBuilder s = new StringBuilder(str);

    for (int j = 0; j < n; j++) {
      if (s.charAt(j) == 'a')
        ca++;
      else if (s.charAt(j) == 'b')
        cb++;
      else
        cc++;
    }

    // If equal previously
    if (ca == cb && cb == cc) {
      return s.toString();
    }

    int cnt = n / 3;

    // If not a multiple of 3
    if (cnt * 3 != n) {
      return "-1";
    }

    int i = 0;

    // Increase the number of a's by
    // removing extra 'b' and ;c;
    while (ca < cnt && i < n) {

      // Check if it is 'b' and it more
      // than n/3
      if (s.charAt(i) == 'b' && cb > cnt) {
        cb--;
        s.setCharAt(i, 'a');
        ca++;
      }

      // Check if it is 'c' and it
      // more than n/3
      else if (s.charAt(i) == 'c' && cc > cnt) {
        cc--;
        s.setCharAt(i, 'a');
        ca++;
      }
      i++;
    }

    i = 0;

    // Increase the number of b's by
    // removing extra 'c'
    while (cb < cnt && i < n) {

      // Check if it is 'c' and it more
      // than n/3
      if (s.charAt(i) == 'c' && cc > cnt) {
        cc--;
        s.setCharAt(i, '1');

        cb++;
      }
      i++;
    }

    i = n - 1;

    // Increase the number of c's from back
    while (cc < cnt && i >= 0) {

      // Check if it is 'a' and it more
      // than n/3
      if (s.charAt(i) == 'a' && ca > cnt) {
        ca--;
        s.setCharAt(i, 'c');
        cc++;
      }
      i--;
    }

    i = n - 1;

    // Increase the number of b's from back
    while (cb < cnt && i >= 0) {

      // Check if it is 'a' and it more
      // than n/3
      if (s.charAt(i) == 'a' && ca > cnt) {
        ca--;
        s.setCharAt(i, 'b');
        cb++;
      }
      i--;
    }

    i = n - 1;

    // Increase the number of c's from back
    while (cc < cnt && i >= 0) {

      // Check if it is 'b' and it more
      // than n/3
      if (s.charAt(i) == 'b' && cb > cnt) {
        cb--;
        s.setCharAt(i, 'c');
        cc++;
      }
      i--;
    }
    return s.toString();
  }

  // Driver Code
  public static void main(String[] args)
  {
    String s = "aaaaaa";
    int n = s.length();

    System.out.println(lexoSmallest(s, n));
  }
}

// This code is contributed by Samim Hossain Mondal.

Python3

# Python3 program to Minimize the number of
# replacements to get a string with same
# number of 'a', 'b' and 'c' in it

# Function to count numbers
def lexoSmallest(s, n):

    # Count the number of 'a', 'b' and
    # 'c' in string
    ca = 0
    cb = 0
    cc = 0
    for i in range(n):
        if (s[i] == 'a'):
            ca += 1
        elif (s[i] == 'b'):
            cb += 1
        else:
            cc += 1

    # If equal previously
    if (ca == cb and cb == cc):
        return s

    cnt = n // 3

    # If not a multiple of 3
    if (cnt * 3 != n):
        return "-1"

    i = 0

    # Increase the number of a's by
    # removing extra 'b' and c
    while (ca < cnt and i < n):

        # Check if it is 'b' and it more
        # than n/3
        if (s[i] == 'b' and cb > cnt) :
            cb -= 1
            s[i] = 'a'
            ca += 1

        # Check if it is 'c' and it
        # more than n/3
        elif (s[i] == 'c' and cc > cnt):
            cc -= 1
            s[i] = 'a'
            ca += 1

        i += 1
    i = 0

    # Increase the number of b's by
    # removing extra 'c'
    while (cb < cnt and i < n):

        # Check if it is 'c' and it more
        # than n/3
        if (s[i] == 'c' and cc > cnt):
            cc -= 1
            s[i] = '1'
            cb += 1

        i += 1

    i = n - 1

    # Increase the number of c's from back
    while (cc < cnt and i >= 0):

        # Check if it is 'a' and it more
        # than n/3
        if (s[i] == 'a' and ca > cnt):
            ca -= 1
            s[i] = 'c'
            cc += 1

        i -= 1

    i = n - 1

    # Increase the number of b's from back
    while (cb < cnt and i >= 0):

        # Check if it is 'a' and it more
        # than n/3
        if (s[i] == 'a' and ca > cnt):
            ca -= 1
            s[i] = 'b'
            cb += 1

        i -= 1

    i = n - 1

    # Increase the number of c's from back
    while (cc < cnt and i >= 0):

        # Check if it is 'b' and it more
        # than n/3
        if (s[i] == 'b' and cb > cnt):
            cb -= 1
            s[i] = 'c'
            cc += 1

        i -= 1
    return s

# Driver Code

s = "aaaaaa"
n = len(s)

print(*lexoSmallest(list(s), n),sep="")

# This code is contributed by shivanisinghss2110

C#

// C# program to minimize the number of
// replacements to get a string with same
// number of ‘a’, ‘b’ and ‘c’ in it
using System;
using System.Text;

class GFG{

// Function to count numbers
static string lexoSmallest(string str, int n)
{

    // Count the number of 'a', 'b' and
    // 'c' in string
    int ca = 0, cb = 0, cc = 0;
    StringBuilder s = new StringBuilder(str);

    for(int j = 0; j < n; j++)
    {
        if (s[j] == 'a')
            ca++;
        else if (s[j] == 'b')
            cb++;
        else
            cc++;
    }

    // If equal previously
    if (ca == cb && cb == cc)
    {
        return s.ToString();
    }

    int cnt = n / 3;

    // If not a multiple of 3
    if (cnt * 3 != n)
    {
        return "-1";
    }

    int i = 0;

    // Increase the number of a's by
    // removing extra 'b' and ;c;
    while (ca < cnt && i < n)
    {

        // Check if it is 'b' and it more
        // than n/3
        if (s[i] == 'b' && cb > cnt)
        {
            cb--;
            s[i] = 'a';
            ca++;
        }

        // Check if it is 'c' and it
        // more than n/3
        else if (s[i] == 'c' && cc > cnt)
        {
            cc--;
            s[i] = 'a';
            ca++;
        }
        i++;
    }

    i = 0;

    // Increase the number of b's by
    // removing extra 'c'
    while (cb < cnt && i < n)
    {

        // Check if it is 'c' and it more
        // than n/3
        if (s[i] == 'c' && cc > cnt)
        {
            cc--;
            s[i] = '1';

            cb++;
        }
        i++;
    }

    i = n - 1;

    // Increase the number of c's from back
    while (cc < cnt && i >= 0)
    {

        // Check if it is 'a' and it more
        // than n/3
        if (s[i] == 'a' && ca > cnt)
        {
            ca--;
            s[i] = 'c';
            cc++;
        }
        i--;
    }

    i = n - 1;

    // Increase the number of b's from back
    while (cb < cnt && i >= 0)
    {

        // Check if it is 'a' and it more
        // than n/3
        if (s[i] == 'a' && ca > cnt)
        {
            ca--;
            s[i] = 'b';
            cb++;
        }
        i--;
    }

    i = n - 1;

    // Increase the number of c's from back
    while (cc < cnt && i >= 0)
    {

        // Check if it is 'b' and it more
        // than n/3
        if (s[i] == 'b' && cb > cnt)
        {
            cb--;
            s[i] = 'c';
            cc++;
        }
        i--;
    }
    return s.ToString();
}

// Driver Code
public static void Main(string[] args)
{
    string s = "aaaaaa";
    int n = s.Length;

    Console.Write(lexoSmallest(s, n));
}
}

// This code is contributed by rutvik_56

PHP

<?php
// PHP program to Minimize the number of
// replacements to get a string with same
// number of ‘a’, ‘b’ and ‘c’ in it

// Function to count numbers
function lexoSmallest($s, $n)
{
    // Count the number of 'a', 'b'
    // and 'c' in string
    $ca = 0;
    $cb = 0;
    $cc = 0;

    for ($i = 0; $i < $n; $i++)
    {
        if ($s[$i] == 'a')
            $ca++;
        else if ($s[$i] == 'b')
            $cb++;
        else
            $cc++;
    }

    // If equal previously
    if ($ca == $cb && $cb == $cc)
    {
        return $s;
    }

    $cnt = floor($n / 3);

    // If not a multiple of 3
    if ($cnt * 3 != $n)
    {
        return "-1";
    }

    $i = 0;

    // Increase the number of a's by
    // removing extra 'b' and ;c;
    while ($ca < $cnt && $i < $n)
    {

        // Check if it is 'b' and it is
        // more than n/3
        if ($s[$i] == 'b' && $cb > $cnt)
        {
            $cb--;
            $s[$i] = 'a';
            $ca++;
        }

        // Check if it is 'c' and it
        // more than n/3
        else if ($s[$i] == 'c' && $cc > $cnt)
        {
            $cc--;
            $s[$i] = 'a';
            $ca++;
        }

        $i++;
    }

    $i = 0;

    // Increase the number of b's by
    // removing extra 'c'
    while ($cb < $cnt && $i < $n)
    {

        // Check if it is 'c' and it more
        // than n/3
        if ($s[$i] == 'c' && $cc > $cnt)
        {
            $cc--;
            $s[$i] = '1';

            $cb++;
        }
        $i++;
    }

    $i = $n - 1;

    // Increase the number of c's from back
    while ($cc < $cnt && $i >= 0)
    {

        // Check if it is 'a' and it is
        // more than n/3
        if ($s[$i] == 'a' && $ca > $cnt)
        {
            $ca--;
            $s[$i] = 'c';
            $cc++;
        }

        $i--;
    }

    $i = $n - 1;

    // Increase the number of b's from back
    while ($cb < $cnt && $i >= 0)
    {

        // Check if it is 'a' and it is
        // more than n/3
        if ($s[$i] == 'a' && $ca > $cnt)
        {
            $ca--;
            $s[$i] = 'b';
            $cb++;
        }

        $i--;
    }

    $i = $n - 1;

    // Increase the number of c's from back
    while ($cc < $cnt && $i >= 0)
    {

        // Check if it is 'b' and it more
        // than n/3
        if ($s[$i] == 'b' && $cb > $cnt)
        {
            $cb--;
            $s[$i] = 'c';
            $cc++;
        }

        $i--;
    }

    return $s;
}

// Driver Code
$s = "aaaaaa";
$n = strlen($s);

echo lexoSmallest($s, $n);

// This code is contributed by Ryuga.
?>

Javascript

// Javascript program to Minimize the number of
// replacements to get a string with same
// number of ‘a’, ‘b’ and ‘c’ in it

// Utility code for replacing a character
// at a specified index
function setCharAt(str,index,chr) {
    if(index > str.length-1) {
        return str;
    }
    return str.substring(0,index) + chr + str.substring(index+1);
}

// Function to count numbers
function lexoSmallest(s, n)
{
    // Count the number of 'a', 'b' and
    // 'c' in string
    let ca = 0, cb = 0, cc = 0;

    for (let i = 0; i < n; i++) {
        if (s[i] == 'a')
            ca++;
        else if (s[i] == 'b')
            cb++;
        else
            cc++;
    }

    // If equal previously
    if (ca == cb && cb == cc) {
        return s;
    }

    let cnt = n / 3;

    // If not a multiple of 3
    if (cnt * 3 != n) {
        return "-1";
    }

    let i = 0;

    // Increase the number of a's by
    // removing extra 'b' and ;c;
    while (ca < cnt && i < n) {

        // Check if it is 'b' and it more
        // than n/3
        if (s[i] == 'b' && cb > cnt) {
            cb--;
            s = setCharAt(s, i, 'a');
            ca++;
        }

        // Check if it is 'c' and it
        // more than n/3
        else if (s[i] == 'c' && cc > cnt) {
            cc--;
            s = setCharAt(s, i, 'a');
            ca++;
        }

        i++;
    }

    i = 0;

    // Increase the number of b's by
    // removing extra 'c'
    while (cb < cnt && i < n) {

        // Check if it is 'c' and it more
        // than n/3
        if (s[i] == 'c' && cc > cnt) {
            cc--;
            s = setCharAt(s, i, '1');

            cb++;
        }
        i++;
    }

    i = n - 1;

    // Increase the number of c's from back
    while (cc < cnt && i >= 0) {

        // Check if it is 'a' and it more
        // than n/3
        if (s[i] == 'a' && ca > cnt) {
            ca--;
            s = setCharAt(s, i, 'c');
            cc++;
        }

        i--;
    }

    i = n - 1;

    // Increase the number of b's from back
    while (cb < cnt && i >= 0) {

        // Check if it is 'a' and it more
        // than n/3
        if (s[i] == 'a' && ca > cnt) {
            ca--;
            s = setCharAt(s, i, 'b');
            cb++;
        }

        i--;
    }

    i = n - 1;

    // Increase the number of c's from back
    while (cc < cnt && i >= 0) {

        // Check if it is 'b' and it more
        // than n/3
        if (s[i] == 'b' && cb > cnt) {
            cb--;
            s = setCharAt(s, i, 'c');
            cc++;
        }

        i--;
    }

    return s;
}

// Driver Code
let s = "aaaaaa";
let n = s.length;

console.log(lexoSmallest(s, n));

// This code is contributed by Samim Hossain Mondal.

Output:

aabbcc

Time Complexity: O(N*6)
Auxiliary Space: O(1), no extra space is required, so it is a constant.