Replace even-indexed characters of minimum number of substrings to convert a string to another

Given two strings, str1 and str2 of length N, the task is to convert the string str1 to string str2 by selecting a substring and replacing all characters present at even indices of the substring by any possible characters, even number of times.

Examples:

Input: str1 = “abcdef”, str2 = “ffffff” 
Output: 2 
Explanation:
Selecting the substring {str1[0], …, str[4]} and replacing all the even indices of the substring by ‘f’ modifies str1 to “fbfdff”. 
Selecting the substring {str1[1], …, str[3]} and replacing all the even indices of the substring by ‘f’ modifies str1 to “ffffff”, which is the same as str2. 
Therefore, the required output is 2.

Input: str1 = “rtrtyy”, str2 = “wtwtzy” 
Output: 1 
Explanation:
Selecting the substring {str1[0], …, str[4]} and replacing str1[0] by ‘w’, str1[[2] by ‘w’ and str[4] by ‘t’ modifies str1 to “wtwtzy”, which is same as str2. Therefore, the required output is 1.

Approach: The problem can be solved using Greedy technique. Follow the steps below to solve the problem:

• Initialize a variable, say cntOp, to store the minimum count of given operations required to convert the string str1 to str2.
• Iterate over the characters of the string. For every i^th index, check if str1[i] and str2[i] are same or not. If found to be different, then find the longest substring possible that contains different characters at even indices in both the strings. Replace even indexed characters of that substring of str1 and increment the value of cntOp by 1.
• Finally, print the value of cntOp.

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach

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

// Function to count the minimum number of
// substrings of str1 such that replacing
// even-indexed characters of those substrings
// converts the string str1 to str2
int minOperationsReq(string str1, string str2)
{
    // Stores length of str1
    int N = str1.length();

    // Stores minimum count of operations
    // to convert str1 to str2
    int cntOp = 0;

    // Traverse both the given string
    for (int i = 0; i < N; i++) {

        // If current character in
        // both the strings are equal
        if (str1[i] == str2[i]) {
            continue;
        }

        // Stores current index
        // of the string
        int ptr = i;

        // If current character in both
        // the strings are not equal
        while (ptr < N && str1[ptr] != str2[ptr]) {

            // Replace str1[ptr]
            // by str2[ptr]
            str1[ptr] = str2[ptr];

            // Update ptr
            ptr += 2;
        }

        // Update cntOp
        cntOp++;
    }

    return cntOp;
}

// Driver Code
int main()
{
    string str1 = "abcdef";
    string str2 = "ffffff";

    cout << minOperationsReq(str1, str2);
    return 0;
}

Java

// Java program to implement
// the above approach

import java.io.*;

class GFG {

    // Function to count the minimum number of
    // substrings of str1 such that replacing
    // even-indexed characters of those substrings
    // converts the string str1 to str2
    static int min_Operations(String str1,
                              String str2)
    {
        // Stores length of str1
        int N = str1.length();

        // Convert str1 to character array
        char[] str = str1.toCharArray();

        // Stores minimum count of operations
        // to convert str1 to str2
        int cntOp = 0;

        // Traverse both the given string
        for (int i = 0; i < N; i++) {

            // If current character in both
            // the strings are equal
            if (str[i] == str2.charAt(i)) {
                continue;
            }

            // Stores current index
            // of the string
            int ptr = i;

            // If current character in both the
            // string are not equal
            while (ptr < N && str[ptr] != str2.charAt(ptr)) {

                // Replace str1[ptr]
                // by str2[ptr]
                str[ptr] = str2.charAt(ptr);

                // Update ptr
                ptr += 2;
            }

            // Update cntOp
            cntOp++;
        }

        return cntOp;
    }

    // Driver Code
    public static void main(String[] args)
    {
        String str1 = "abcdef";
        String str2 = "ffffff";
        System.out.println(
            min_Operations(str1, str2));
    }
}

Python3

# Python3 program to implement
# the above approach

# Function to count the minimum number of
# substrings of str1 such that replacing
# even-indexed characters of those substrings
# converts the str1 to str2
def minOperationsReq(str11, str22):

    str1 = list(str11)
    str2 = list(str22)

    # Stores length of str1
    N = len(str1)

    # Stores minimum count of operations
    # to convert str1 to str2
    cntOp = 0

    # Traverse both the given string
    for i in range(N):

        # If current character in
        # both the strings are equal
        if (str1[i] == str2[i]):
            continue

        # Stores current index
        # of the string
        ptr = i

        # If current character in both
        # the strings are not equal
        while (ptr < N and str1[ptr] != str2[ptr]):

            # Replace str1[ptr]
            # by str2[ptr]
            str1[ptr] = str2[ptr]

            # Update ptr
            ptr += 2

        # Update cntOp
        cntOp += 1

    return cntOp

# Driver Code
str1 = "abcdef"
str2 = "ffffff"

print(minOperationsReq(str1, str2))

# This code is contributed by code_hunt

C#

// C# program to implement
// the above approach
using System;

class GFG
{

    // Function to count the minimum number of
    // substrings of str1 such that replacing
    // even-indexed characters of those substrings
    // converts the string str1 to str2
    static int min_Operations(String str1,
                              String str2)
    {

        // Stores length of str1
        int N = str1.Length;

        // Convert str1 to character array
        char[] str = str1.ToCharArray();

        // Stores minimum count of operations
        // to convert str1 to str2
        int cntOp = 0;

        // Traverse both the given string
        for (int i = 0; i < N; i++)
        {

            // If current character in both
            // the strings are equal
            if (str[i] == str2[i])
            {
                continue;
            }

            // Stores current index
            // of the string
            int ptr = i;

            // If current character in both the
            // string are not equal
            while (ptr < N && str[ptr] != str2[ptr])
            {

                // Replace str1[ptr]
                // by str2[ptr]
                str[ptr] = str2[ptr];

                // Update ptr
                ptr += 2;
            }

            // Update cntOp
            cntOp++;
        }

        return cntOp;
    }

    // Driver Code
    public static void Main(String[] args)
    {
        String str1 = "abcdef";
        String str2 = "ffffff";
        Console.WriteLine(
            min_Operations(str1, str2));
    }
}

// This code contributed by gauravrajput1

Javascript

<script>

// Javascript program to implement
// the above approach

// Function to count the minimum number of
    // substrings of str1 such that replacing
    // even-indexed characters of those substrings
    // converts the string str1 to str2
    function min_Operations( str1,  str2)
    {
        // Stores length of str1
    var N = str1.length;

    // Stores minimum count of operations
    // to convert str1 to str2
    var cntOp = 0;

    // Traverse both the given string
    for (var i = 0; i < N; i++) {

        // If current character in
        // both the strings are equal
        if (str1.charCodeAt(i)== str2.charCodeAt(i))
        {

            continue;
        }

        // Stores current index
        // of the string
        var ptr = i;

        // If current character in both
        // the strings are not equal
        while (ptr < N && str1[ptr] != str2[ptr]) {

            // Replace str1[ptr]
            // by str2[ptr]

            str1 = str1.substring(0, ptr) + str2[ptr]
            + str1.substring(ptr + 1);
            // Update ptr
            ptr += 2;
        }

        // Update cntOp
        cntOp++;
    }

    return cntOp;
    }

    // Driver Code

         str1 = "abcdef";
         str2 = "ffffff";
        document.write(min_Operations(str1, str2));

// This code is contributed by todaysgaurav

</script>

Output:

2

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