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# Transform One String to Another using Minimum Number of Given Operation

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: 1Explanation: Pick B and insert it at front.Input:  A = "EACBD", B = "EABCD"Output: 3Explanation: Pick B and insert at front, EACBD => BEACD             Pick A and insert at front, BEACD => ABECD             Pick E and insert at front, ABECD => EABCD`
Recommended Practice

BRUTE METHOD: (Using HashMap)

Algorithm:

1. We declare a HashMap<Character,Integer> to store frequency map.
2. 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.
3. We make a variable res and point two pointer i and j to the last of both strings and start traversing from back.
4. 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.
5. Atlast we return res.

Implementation:

## C++

 `#include ``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 code``int` `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 string` `import` `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 m``            ``= ``new` `HashMap();``        ``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 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 code``A ``=` `"EACBD"``B ``=` `"EABCD"``print``(``"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 code``const 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 ``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 program``int` `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 ``#include ` `// 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 program``int` `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 other``import` `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 B``def` `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 program``A ``=` `"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

 ``

## PHP

 `=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.