Minimum operations required to convert all characters of a String to a given Character

Given a string str, a character ch and an integer K, the task is to find the minimum number of operations required to convert all the characters of string str to ch. Each operation involves converting K closest characters from each side of the index i, i.e., characters in the indices [i – K, i + K] can be converted to ch
Note: Each index can be part of a single operation only.
Examples:

Input: str = “abcsdx”, K = 2, ch = ‘#’ 
Output:
Explanation: 
Operation 1: Select i = 1, therefore, str = “abcsdx” modifies to “###sdx”. 
Operation 2: Select i = 6, therefore, str = “###sdx” modifies to “######”.

Input: str = “Hellomypkfsg”, k = 3, ch = ‘$’ 
Output:
Explanation: 
Operation 1: Select i = 2, therefore, str = “Hellomypkfsg” modifies to “$$$$$mypkfsg”. 
Operation 2: Select i = 9, therefore, str = “$$$$$mypkfsg” modifies to “$$$$$$$$$$$$”.

Approach: 
Follow the steps below to solve the problem:

  1. For any index i, the maximum number of characters than can be converted are 2 * K + 1. Therefore, if the total number of characters in the string does not exceed 2 * K, a single operation is required only to convert the entire string into ch.
  2. Otherwise, the number of operations required will be ceil(n / (2*k+1)).
  3. Iterate over the string, starting from the leftmost possible index that can be used for the operation and print every (2*k+1)th index after it .

Below is the implementation of the above approach:



C++

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// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the minimum
// number of operations required
void countOperations(int n, int k)
{
    // Maximum number of characters that
    // can be changed in one operation
    int div = 2 * k + 1;
  
    // If length of the string less than
    // maximum number of characters that
    // can be changed in an operation
    if (n / 2 <= k) {
        cout << 1 << "\n";
  
        // Set the last index as the
        // index for the operation
        if (n > k)
            cout << k + 1;
  
        else
            cout << n;
    }
  
    // Otherwise
    else {
  
        // If size of the string is
        // equal to the maximum number
        // of characters in an operation
        if (n % div == 0) {
  
            // Find the number of
            // operations required
            int oprn = n / div;
  
            cout << oprn << "\n";
  
            // Find the starting postion
            int pos = k + 1;
  
            cout << pos << " ";
            for (int i = 1; i <= oprn; i++) {
                // Print i-th index
                cout << pos << " ";
  
                // Shift to next index
                pos += div;
            }
        }
  
        // Otherwise
        else {
  
            // Find the number of
            // operations required
            int oprn = n / div + 1;
            cout << oprn << "\n";
  
            int pos = n % div;
  
            // If n % div exceeds k
            if (n % div > k)
                pos -= k;
  
            for (int i = 1; i <= oprn; i++) {
  
                // Print i-th index
                cout << pos << " ";
  
                // Shift to next index
                pos += div;
            }
        }
    }
}
  
// Driver Code
int main()
{
    string str = "edfreqwsazxet";
    char ch = '$';
    int n = str.size();
    int k = 4;
    countOperations(n, k);
  
    return 0;
}

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Java

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// Java program to implement
// the above approach
class GFG{
  
// Function to find the minimum
// number of operations required
static void countOperations(int n, int k)
{
      
    // Maximum number of characters that
    // can be changed in one operation
    int div = 2 * k + 1;
  
    // If length of the String less than
    // maximum number of characters that
    // can be changed in an operation
    if (n / 2 <= k)
    {
        System.out.print(1 + "\n");
  
        // Set the last index as the
        // index for the operation
        if (n > k)
            System.out.print(k + 1);
  
        else
            System.out.print(n);
    }
  
    // Otherwise
    else
    {
          
        // If size of the String is
        // equal to the maximum number
        // of characters in an operation
        if (n % div == 0)
        {
              
            // Find the number of
            // operations required
            int oprn = n / div;
  
            System.out.print(oprn + "\n");
  
            // Find the starting postion
            int pos = k + 1;
  
            System.out.print(pos + " ");
            for(int i = 1; i <= oprn; i++)
            {
                  
                // Print i-th index
                System.out.print(pos + " ");
  
                // Shift to next index
                pos += div;
            }
        }
  
        // Otherwise
        else 
        {
              
            // Find the number of
            // operations required
            int oprn = n / div + 1;
            System.out.print(oprn + "\n");
  
            int pos = n % div;
  
            // If n % div exceeds k
            if (n % div > k)
                pos -= k;
  
            for(int i = 1; i <= oprn; i++)
            {
                  
                // Print i-th index
                System.out.print(pos + " ");
  
                // Shift to next index
                pos += div;
            }
        }
    }
}
  
// Driver Code
public static void main(String[] args)
{
    String str = "edfreqwsazxet";
    char ch = '$';
    int n = str.length();
    int k = 4;
      
    countOperations(n, k);
}
}
  
// This code is contributed by amal kumar choubey

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Python3

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# Python3 program to implement
# the above approach
  
# Function to find the minimum
# number of operations required
def countOperations(n, k):
      
    # Maximum number of characters that
    # can be changed in one operation
    div = 2 * k + 1
  
    # If length of the less than
    # maximum number of characters that
    # can be changed in an operation
    if (n // 2 <= k):
        print(1)
  
        # Set the last index as the
        # index for the operation
        if (n > k):
            print(k + 1)
        else:
            print(n)
  
    # Otherwise
    else:
  
        # If size of the is
        # equal to the maximum number
        # of characters in an operation
        if (n % div == 0):
  
            # Find the number of
            # operations required
            oprn = n // div
  
            print(oprn)
  
            # Find the starting postion
            pos = k + 1
  
            print(pos, end = " ")
            for i in range(1, oprn + 1):
                  
                # Print i-th index
                print(pos, end = " ")
  
                # Shift to next index
                pos += div
  
        # Otherwise
        else:
  
            # Find the number of
            # operations required
            oprn = n // div + 1
            print(oprn)
  
            pos = n % div
  
            # If n % div exceeds k
            if (n % div > k):
                pos -= k
  
            for i in range(1, oprn + 1):
  
                # Print i-th index
                print(pos, end = " ")
  
                # Shift to next index
                pos += div
  
# Driver Code
if __name__ == '__main__':
      
    str = "edfreqwsazxet"
    ch = '$'
    n = len(str)
    k = 4
      
    countOperations(n, k)
  
# This code is contributed by mohit kumar 29

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C#

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// C# program to implement
// the above approach
using System;
class GFG{
  
// Function to find the minimum
// number of operations required
static void countOperations(int n, int k)
{
      
    // Maximum number of characters that
    // can be changed in one operation
    int div = 2 * k + 1;
  
    // If length of the String less than
    // maximum number of characters that
    // can be changed in an operation
    if (n / 2 <= k)
    {
        Console.Write(1 + "\n");
  
        // Set the last index as the
        // index for the operation
        if (n > k)
            Console.Write(k + 1);
  
        else
            Console.Write(n);
    }
  
    // Otherwise
    else
    {
          
        // If size of the String is
        // equal to the maximum number
        // of characters in an operation
        if (n % div == 0)
        {
              
            // Find the number of
            // operations required
            int oprn = n / div;
  
            Console.Write(oprn + "\n");
  
            // Find the starting postion
            int pos = k + 1;
  
            Console.Write(pos + " ");
            for(int i = 1; i <= oprn; i++)
            {
                  
                // Print i-th index
                Console.Write(pos + " ");
  
                // Shift to next index
                pos += div;
            }
        }
  
        // Otherwise
        else 
        {
              
            // Find the number of
            // operations required
            int oprn = n / div + 1;
            Console.Write(oprn + "\n");
  
            int pos = n % div;
  
            // If n % div exceeds k
            if (n % div > k)
                pos -= k;
  
            for(int i = 1; i <= oprn; i++)
            {
                  
                // Print i-th index
                Console.Write(pos + " ");
  
                // Shift to next index
                pos += div;
            }
        }
    }
}
  
// Driver Code
public static void Main(String[] args)
{
    String str = "edfreqwsazxet";
    int n = str.Length;
    int k = 4;
      
    countOperations(n, k);
}
}
   
// This code is contributed by Rohit_ranjan

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Output: 

2
4 13

Time Complexity: O(N) 
Auxiliary Space: O(1)
 

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