Given two strings A and B, the task is to convert A to B if possible. The only operation allowed is to put any character from A and insert it at front. Find if it’s possible to convert the string. If yes, then output minimum no. of operations required for transformation.
Examples:
Input: A = "ABD", B = "BAD"
Output: 1
Explanation: Pick B and insert it at front.
Input: A = "EACBD", B = "EABCD"
Output: 3
Explanation: Pick B and insert at front, EACBD => BEACD
Pick A and insert at front, BEACD => ABECD
Pick E and insert at front, ABECD => EABCD
BRUTE METHOD: (Using HashMap)
Algorithm:
- We declare a HashMap<Character,Integer> to store frequency map.
- We store the character of string 1 in the map and then while traversing string 2 ,we erase the characters and if the map is empty at the end that means the characters in both the string are same and we can continue,else we return -1.
- We make a variable res and point two pointer i and j to the last of both strings and start traversing from back.
- As soon as see a ith character that doesn’t match with jth character ,we start increasing res by 1 until again both the characters are same.
- Atlast we return res.
Implementation:
C++
#include <bits/stdc++.h>using namespace std;int transform(string A, string B){ if (A.length() != B.length()) { return -1; } // create a map to store the frequency of characters in string A unordered_map<char, int> m; int n = A.length(); for (int i = 0; i < n; i++) { if (m.count(A[i])) // if the character already exists in the map m[A[i]]++; // increment its frequency else m[A[i]] = 1; // add the character to the map with a frequency of 1 } // subtract the frequency of characters in string B from the map for (int i = 0; i < n; i++) { if (m.count(B[i])) m[B[i]]--; } // check if all the frequencies in the map are 0, indicating equal frequency of characters in both strings for (auto it : m) { if (it.second != 0) // if frequency is not zero return -1; // strings cannot be transformed into each other, return -1 } // calculate the minimum number of operations required to transform string A into string B int i = n - 1, j = n - 1; int res = 0; while (i >= 0 && j >= 0) { while (i >= 0 && A[i] != B[j]) { res++; // increment the number of operations required i--; // move the pointer i to the left } i--; j--; } return res; // returning result}// Driver codeint main(){ string A = "EACBD"; string B = "EABCD"; cout << "Minimum number of operations required is " << transform(A, B) << endl; return 0;} |
Java
// Java proram to transform the stringimport java.io.*;import java.util.*;class GFG { public static int transform(String A, String B) { // code here if (A.length() != B.length()) { return -1; } HashMap<Character, Integer> m = new HashMap<Character, Integer>(); int n = A.length(); for (int i = 0; i < n; i++) { if (m.containsKey(A.charAt(i))) m.put(A.charAt(i), m.get(A.charAt(i)) + 1); else m.put(A.charAt(i), 1); } for (int i = 0; i < n; i++) { if (m.containsKey(B.charAt(i))) m.put(B.charAt(i), m.get(B.charAt(i)) - 1); } for (Map.Entry<Character, Integer> entry : m.entrySet()) { if (entry.getValue() != 0) return -1; } int i = n - 1, j = n - 1; int res = 0; while (i >= 0 && j >= 0) { while (i >= 0 && A.charAt(i) != B.charAt(j)) { res++; i--; } i--; j--; } return res; } public static void main(String[] args) { String A = "EACBD"; String B = "EABCD"; System.out.println( "Minimum number of operations required is " + transform(A, B)); }}// This code is contributed by Raunak Singh |
Python3
def transform(A, B): if len(A) != len(B): return -1 # create a dictionary to store the frequency of characters in string A m = {} n = len(A) for i in range(n): if A[i] in m: # if the character already exists in the dictionary m[A[i]] += 1 # increment its frequency else: m[A[i]] = 1 # add the character to the dictionary with a frequency of 1 # subtract the frequency of characters in string B from the dictionary for i in range(n): if B[i] in m: m[B[i]] -= 1 # check if all the frequencies in the dictionary are 0, indicating equal frequency of characters in both strings for key in m: if m[key] != 0: # if frequency is not zero return -1 # strings cannot be transformed into each other, return -1 # calculate the minimum number of operations required to transform string A into string B i, j = n-1, n-1 res = 0 while i >= 0 and j >= 0: while i >= 0 and A[i] != B[j]: res += 1 # increment the number of operations required i -= 1 # move the pointer i to the left i -= 1 j -= 1 return res # returning result# Driver codeA = "EACBD"B = "EABCD"print("Minimum number of operations required is", transform(A, B)) |
C#
using System;using System.Collections.Generic;class GFG{ public static int Transform(string A, string B) { if (A.Length != B.Length) { return -1; } // create a dictionary to store the frequency of characters in string A Dictionary<char, int> m = new Dictionary<char, int>(); int n = A.Length; for (int i = 0; i < n; i++) { if (m.ContainsKey(A[i])) // if the character already exists in the dictionary { m[A[i]]++; // increment its frequency } else { m[A[i]] = 1; // add the character to the dictionary with a frequency of 1 } } // subtract the frequency of characters in string B from the dictionary for (int i = 0; i < n; i++) { if (m.ContainsKey(B[i])) { m[B[i]]--; } } // check if all the frequencies in the dictionary are 0, indicating equal frequency of characters in both strings foreach (var entry in m) { if (entry.Value != 0) { return -1; } } // calculate the minimum number of operations required to transform string A into string B int it = n - 1, j = n - 1; int res = 0; while (it >= 0 && j >= 0) { while (it >= 0 && A[it] != B[j]) { res++; it--; } it--; j--; } return res; } // Driver code public static void Main(string[] args) { string A = "EACBD"; string B = "EABCD"; Console.WriteLine("Minimum number of operations required is " + Transform(A, B)); }} |
Javascript
function transform(A, B) { if (A.length !== B.length) { return -1; } // Create an object to store the frequency of characters in string A const m = {}; const n = A.length; for (let i = 0; i < n; i++) { if (m[A[i]]) { // if the character already exists in the object m[A[i]]++; // increment its frequency } else { m[A[i]] = 1; // add the character to the object with a frequency of 1 } } // Subtract the frequency of characters in string B from the object for (let i = 0; i < n; i++) { if (m[B[i]]) { m[B[i]]--; } } // Check if all the frequencies in the object are 0, // indicating equal frequency of characters in both strings for (const char in m) { if (m[char] !== 0) { // if frequency is not zero return -1; // strings cannot be transformed into each other, return -1 } } // Calculate the minimum number of operations // required to transform string A into string B let i = n - 1, j = n - 1; let res = 0; while (i >= 0 && j >= 0) { while (i >= 0 && A[i] !== B[j]) { res++; // increment the number of operations required i--; // move the pointer i to the left } i--; j--; } return res; // returning result}// Driver codeconst A = "EACBD";const B = "EABCD";console.log("Minimum number of operations required is " + transform(A, B)); |
Output
Minimum number of operations required is 3
- Time Complexity: O(N)
- Auxiliary Space: O(N), since we are using a HashMap.
Checking whether a string can be transformed to another is simple. We need to check whether both strings have same number of characters and same set of characters. This can be easily done by creating a count array for first string and checking if second string has same count of every character.
How to find minimum number of operations when we are sure that we can transform A to B? The idea is to start matching from last characters of both strings. If last characters match, then our task reduces to n-1 characters. If last characters don’t match, then find the position of B’s mismatching character in A. The difference between two positions indicates that these many characters of A must be moved before current character of A.
Below is complete algorithm.
1) Find if A can be transformed to B or not by first creating a count array for all characters of A, then checking with B if B has same count for every character.
2) Initialize result as 0.
3) Start traversing from end of both strings.
……a) If current characters of A and B match, i.e., A[i] == B[j]
………then do i = i-1 and j = j-1
b) If current characters don’t match, then search B[j] in remaining
………A. While searching, keep incrementing result as these characters
………must be moved ahead for A to B transformation.
Below are the implementations based on this idea.
C++
// C++ program to find minimum number of operations required// to transform one string to other#include <bits/stdc++.h>using namespace std;// Function to find minimum number of operations required to// transform A to B.int minOps(string& A, string& B){ int m = A.length(), n = B.length(); // This parts checks whether conversion is possible or not if (n != m) return -1; int count[256]; memset(count, 0, sizeof(count)); // count characters in A for (int i = 0; i < n; i++) count[A[i]]++; // subtract count for every character in B for (int i = 0; i < n; i++) count[B[i]]--; // Check if all counts become 0 for (int i = 0; i < 256; i++) if (count[i]) return -1; // This part calculates the number of operations // required int res = 0; for (int i = n - 1, j = n - 1; i >= 0;) { // If there is a mismatch, then keep incrementing // result 'res' until B[j] is not found in A[0..i] while (i >= 0 && A[i] != B[j]) { i--; res++; } // If A[i] and B[j] match if (i >= 0) { i--; j--; } } return res;}// Driver programint main(){ string A = "EACBD"; string B = "EABCD"; cout << "Minimum number of operations required is " << minOps(A, B); return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
C
// C program to find minimum number of operations required// to transform one string to other#include <stdio.h>#include <string.h>// Function to find minimum number of operations required to// transform A to B.int minOps(char A[], char B[]){ int m = strlen(A), n = strlen(B); // This parts checks whether conversion is // possible or not if (n != m) return -1; int count[256]; for (int i = 0; i < 256; i++) count[i] = 0; // count characters in A for (int i = 0; i < n; i++) count[A[i]]++; // subtract count for every character in B for (int i = 0; i < n; i++) count[B[i]]--; // Check if all counts become 0 for (int i = 0; i < 256; i++) if (count[i]) return -1; // This part calculates the number of operations // required int res = 0; for (int i = n - 1, j = n - 1; i >= 0;) { // If there is a mismatch, then keep incrementing // result 'res' until B[j] is not found in A[0..i] while (i >= 0 && A[i] != B[j]) { i--; res++; } // If A[i] and B[j] match if (i >= 0) { i--; j--; } } return res;}// Driver programint main(){ char A[] = "EACBD"; char B[] = "EABCD"; printf("Minimum number of operations required is %d", minOps(A, B)); return 0;}// This code is contributed by Aditya Kumar (adityakumar129) |
Java
// Java program to find minimum number of operations// required to transform one string to otherimport java.io.*;import java.util.*;public class GFG { // Function to find minimum number of operations // required to transform A to B. public static int minOps(String A, String B) { // This parts checks whether conversion is possible // or not if (A.length() != B.length()) return -1; int i, j, res = 0; int count[] = new int[256]; // count characters in A // subtract count for every character in B for (i = 0; i < A.length(); i++) { count[A.charAt(i)]++; count[B.charAt(i)]--; } // Check if all counts become 0 for (i = 0; i < 256; i++) if (count[i] != 0) return -1; i = A.length() - 1; j = B.length() - 1; while (i >= 0) { // If there is a mismatch, then keep // incrementing result 'res' until B[j] is not // found in A[0..i] if (A.charAt(i) != B.charAt(j)) res++; else j--; i--; } return res; } // Driver code public static void main(String[] args) { String A = "EACBD"; String B = "EABCD"; System.out.println( "Minimum number of operations required is " + minOps(A, B)); }}// This code is contributed by Aditya Kumar (adityakumar129) |
Python3
# Python program to find the minimum number of# operations required to transform one string to other# Function to find minimum number of operations required# to transform A to Bdef minOps(A, B): m = len(A) n = len(B) # This part checks whether conversion is possible or not if n != m: return -1 count = [0] * 256 for i in range(n): # count characters in A count[ord(B[i])] += 1 for i in range(n): # subtract count for every char in B count[ord(A[i])] -= 1 for i in range(256): # Check if all counts become 0 if count[i]: return -1 # This part calculates the number of operations required res = 0 i = n-1 j = n-1 while i >= 0: # if there is a mismatch, then keep incrementing # result 'res' until B[j] is not found in A[0..i] while i>= 0 and A[i] != B[j]: i -= 1 res += 1 # if A[i] and B[j] match if i >= 0: i -= 1 j -= 1 return res# Driver programA = "EACBD"B = "EABCD"print ("Minimum number of operations required is " + str(minOps(A,B)))# This code is contributed by Bhavya Jain |
C#
// C# program to find minimum number of // operations required to transform one // string to other using System;class GFG{// Function to find minimum number of // operations required to transform // A to B. public static int minOps(string A, string B){ // This parts checks whether // conversion is possible or not if (A.Length != B.Length) { return -1; } int i, j, res = 0; int[] count = new int [256]; // count characters in A // subtract count for every // character in B for (i = 0; i < A.Length; i++) { count[A[i]]++; count[B[i]]--; } // Check if all counts become 0 for (i = 0; i < 256; i++) { if (count[i] != 0) { return -1; } } i = A.Length - 1; j = B.Length - 1; while (i >= 0) { // If there is a mismatch, then // keep incrementing result 'res' // until B[j] is not found in A[0..i] if (A[i] != B[j]) { res++; } else { j--; } i--; } return res;}// Driver code public static void Main(string[] args){ string A = "EACBD"; string B = "EABCD"; Console.WriteLine("Minimum number of " + "operations required is " + minOps(A, B));}}// This code is contributed by Shrikant13 |
Javascript
<script>// Javascript program to find minimum number // of operations required to transform one // string to other// Function to find minimum number of// operations required to transform// A to B.function minOps(A, B){ // This parts checks whether conversion // is possible or not if (A.length != B.length) return -1; let i, j, res = 0; let count = new Array(256); for(let i = 0; i < 256; i++) { count[i] = 0; } // count characters in A // Subtract count for every character in B for(i = 0; i < A.length; i++) { count[A[i].charCodeAt(0)]++; count[B[i].charCodeAt(0)]--; } // Check if all counts become 0 for(i = 0; i < 256; i++) if (count[i] != 0) return -1; i = A.length - 1; j = B.length - 1; while(i >= 0) { // If there is a mismatch, then // keep incrementing result 'res' // until B[j] is not found in A[0..i] if (A[i] != B[j]) res++; else j--; i--; } return res; }// Driver codelet A = "EACBD";let B = "EABCD"; document.write("Minimum number of " + "operations required is " + minOps(A, B));// This code is contributed by avanitrachhadiya2155</script> |
PHP
<?php // PHP program to find minimum number of// operations required to transform one string to other // Function to find minimum number of operations required to transform// A to B.function minOps($A, $B){ $m = strlen($A); $n = strlen($B); // This parts checks whether conversion is // possible or not if ($n != $m) return -1; $count = array_fill(0,256,NULL); for ($i=0; $i<$n; $i++) // count characters in A $count[ord($B[$i])]++; for ($i=0; $i<$n; $i++) // subtract count for $count[ord($A[$i])]--; // every character in B for ($i=0; $i<256; $i++) // Check if all counts become 0 if ($count[$i]) return -1; // This part calculates the number of operations required $res = 0; for ($i=$n-1, $j=$n-1; $i>=0; ) { // If there is a mismatch, then keep incrementing // result 'res' until B[j] is not found in A[0..i] while ($i>=0 && $A[$i] != $B[$j]) { $i--; $res++; } // If A[i] and B[j] match if ($i >= 0) { $i--; $j--; } } return $res;} // Driver program$A = "EACBD";$B = "EABCD";echo "Minimum number of operations ". "required is ". minOps($A, $B);return 0;?> |
Output:
Minimum number of operations required is 3
Time Complexity: O(n), please note that i is always decremented (in while loop and in if), and the for loop starts from n-1 and runs while i >= 0.
Auxiliary Space: O(1).
Thanks to Gaurav Ahirwar for above solution.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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