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 ith 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 str2int 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 Codeint main(){ string str1 = "abcdef"; string str2 = "ffffff"; cout << minOperationsReq(str1, str2); return 0;} |
Java
// Java program to implement// the above approachimport 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 str2def 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 Codestr1 = "abcdef"str2 = "ffffff" print(minOperationsReq(str1, str2))# This code is contributed by code_hunt |
C#
// C# program to implement// the above approachusing 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)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
