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# Check whether two strings can be made equal by reversing substring of equal length from both strings

Give two strings S1 and S2, the task is to check whether string S1 can be made equal to string S2 by reversing substring from both strings of equal length.

Note: A substring can be reversed any number of times.

Example:

Input: S1 = “abbca”, S2 = “acabb”
Output: Yes
Explanation:
The string S1 and S2 can be made equal by:
Reverse S1 in the range [2, 4] (length = 3), S1 = “abacb”
Reverse S2 in the range [1, 3] (length = 3), S2 = “abacb”
S1 = “abacb” and S2 = “abacb”, after reversing.
Hence, both can be made equal.

Input: “S1 = “abcd”, S2 = “abdc”
Output: No

Approach:

1. The idea is to make both the strings equal by sorting both of them.
2. First, check if both the strings have the same set of characters or not. If not, then answer is “No”.
3. Fix the length of substring to reverse to 2. Now, this means swapping adjacent characters.
4. Now, number of moves needed to sort a string by reversing substring of size 2 or in other words, swapping adjacent characters, is the Inversion Count of the string.
5. If both strings have the same inversion count, then the answer is “Yes”.
6. If they have different inversion counts, making them equal is only possible if at least one of the following conditions matches:
• First, if the parity of the inversion count is the same, that is both even or odd, then answer is “Yes”. We’ll continue swapping any pair of elements in the sorted string until the other one is sorted. Since, the difference of inversion count will be even, so swapping any pair twice doesn’t produce any change.
• Second, if the parity is not the same, then there has to be at least a character with a frequency greater than 1 in any of the string. We’ll just swap them until the other one gets sorted. Since swapping identical character doesn’t make any change.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ` `using` `namespace` `std; `   `// function to count inversion ` `// count of the string ` `int` `inversionCount(string& s) ` `{ `   `    ``// for storing frequency ` `    ``int` `freq[26] = { 0 }; ` `    ``int` `inv = 0; ` `    ``for` `(``int` `i = 0; i < s.length(); i++) { ` `        ``int` `temp = 0; `   `        ``// Add all the characters which ` `        ``// are less than the ith character ` `        ``// before i. ` `        ``for` `(``int` `j = 0; j < ``int``(s[i] - ``'a'``); j++) ` `            ``// adding the count to inversion ` `            ``// count ` `            ``temp += freq[j]; `   `        ``inv += (i - temp); ` `        ``// updating the character in ` `        ``// the frequency array `   `        ``freq[s[i] - ``'a'``]++; ` `    ``} ` `    ``return` `inv; ` `} `   `// function to check whether any ` `// of the string have a repeated ` `// character ` `bool` `haveRepeated(string& S1, ` `                ``string& S2) ` `{ `   `    ``int` `freq[26] = { 0 }; ` `    ``for` `(``char` `i : S1) { ` `        ``if` `(freq[i - ``'a'``] > 0) ` `            ``return` `true``; ` `        ``freq[i - ``'a'``]++; ` `    ``} `   `    ``for` `(``int` `i = 0; i < 26; i++) ` `        ``freq[i] = 0; `   `    ``for` `(``char` `i : S2) { ` `        ``if` `(freq[i - ``'a'``] > 0) ` `            ``return` `true``; ` `        ``freq[i - ``'a'``]++; ` `    ``} ` `    ``return` `false``; ` `} `   `// function to check whether the ` `// string S1 and S2 can be made ` `// equal by reversing sub strings ` `// of same size in both strings ` `void` `checkToMakeEqual(string S1, ` `                    ``string S2) ` `{ `   `    ``// frequency array to check ` `    ``// whether both string have ` `    ``// same character or not ` `    ``int` `freq[26] = { 0 }; `   `    ``for` `(``int` `i = 0; i < S1.length(); i++) { ` `        ``// adding the frequency; ` `        ``freq[S1[i] - ``'a'``]++; ` `    ``} `   `    ``bool` `flag = 0; ` `    ``for` `(``int` `i = 0; i < S2.length(); i++) { ` `        ``if` `(freq[S2[i] - ``'a'``] == 0) { ` `            ``// if the character is not in S1 ` `            ``flag = ``true``; ` `            ``break``; ` `        ``} ` `        ``// decrementing the frequency ` `        ``freq[S2[i] - ``'a'``]--; ` `    ``} `   `    ``if` `(flag == ``true``) { ` `        ``// If both string doesnot ` `        ``// have same characters or not ` `        ``cout << ``"No\n"``; ` `        ``return``; ` `    ``} `   `    ``// finding inversion count ` `    ``// of both strings ` `    ``int` `invCount1 = inversionCount(S1); ` `    ``int` `invCount2 = inversionCount(S2); `   `    ``if` `(invCount1 == invCount2 ` `        ``|| (invCount1 & 1) == (invCount2 & 1) ` `        ``|| haveRepeated(S1, S2)) { ` `        ``// If inversion count is same, ` `        ``// or have same parity or if ` `        ``// any of the string have a ` `        ``// repeated character then ` `        ``// the answer is Yes else No ` `        ``cout << ``"Yes\n"``; ` `    ``} ` `    ``else` `        ``cout << ``"No\n"``; ` `} `   `// driver code ` `int` `main() ` `{ ` `    ``string S1 = ``"abbca"``, S2 = ``"acabb"``; ` `    ``checkToMakeEqual(S1, S2); ` `    ``return` `0; ` `} `

## Python3

 `# Python3 program for the above approach`   `# function to count inversion ` `# count of the string ` `def` `inversionCount(s): ` `    `  `    ``# for storing frequency ` `    ``freq ``=` `[``0` `for` `_ ``in` `range``(``26``)] ` `    ``inv ``=` `0` `    ``for` `i ``in` `range``(``len``(s)): ` `        `  `        ``# we'll add all the characters ` `        ``# which are less than the ith ` `        ``# character before i. ` `        ``temp ``=` `0` `        ``for` `j ``in` `range``(``ord``(s[i]) ``-` `ord``(``'a'``)): ` `            `  `            ``# adding the count to ` `            ``# inversion count ` `            ``temp ``+``=` `freq[j] ` `            `  `        ``inv ``+``=` `(i ``-` `temp) ` `        `  `        ``# updating the character in ` `        ``# the frequency array ` `        ``freq[``ord``(s[i]) ``-` `ord``(``'a'``)] ``+``=` `1` `    ``return` `inv `   `# function to check whether ` `# any of the string have a ` `# repeated character ` `def` `haveRepeated(S1, S2): ` `    ``freq ``=` `[``0` `for` `_ ``in` `range``(``26``)] ` `    ``for` `i ``in` `range``(``len``(S1)): ` `        ``if` `freq[``ord``(S1[i]) ``-` `ord``(``'a'``)] > ``0``: ` `            ``return` `1` `        ``freq[``ord``(S1[i]) ``-` `ord``(``'a'``)] ``+``=` `1`   `    ``for` `i ``in` `range``(``26``): ` `        ``freq[i] ``=` `0`   `    ``for` `i ``in` `range``(``len``(S2)): ` `        ``if` `freq[``ord``(S2[i]) ``-` `ord``(``'a'``)] > ``0``: ` `            ``return` `1` `        ``freq[``ord``(S2[i]) ``-` `ord``(``'a'``)] ``+``=` `1`   `    ``return` `0`   `# function to check whether ` `# the string S1 and S2 can ` `# be made equal by reversing ` `# sub strings ofsame size in ` `# both strings ` `def` `checkToMakeEqual(S1, S2): `   `    ``# frequency array to check ` `    ``# whether both string have ` `    ``# same character or not ` `    ``freq ``=` `[``0` `for` `_ ``in` `range``(``26``)] `   `    ``for` `i ``in` `range``(``len``(S1)): ` `        `  `        ``# adding the frequency; ` `        ``freq[``ord``(S1[i]) ``-` `ord``(``'a'``)] ``+``=` `1`   `    ``flag ``=` `0` `    ``for` `i ``in` `range``(``len``(S2)): ` `        ``if` `freq[``ord``(S2[i]) ``-` `ord``(``'a'``)] ``=``=` `0``: ` `            `  `            ``# if the character is not in S1 ` `            ``flag ``=` `1` `            ``break` `            `  `        ``# decrementing the frequency ` `        ``freq[``ord``(S2[i]) ``-` `ord``(``'a'``)] ``-``=` `1` `    `  `    ``if` `flag ``=``=` `1``: ` `        `  `        ``# If both string does not ` `        ``# have same characters or not ` `        ``print``(``"No"``) ` `        ``return`   `    ``# finding inversion count of ` `    ``# both strings ` `    ``invCount1 ``=` `inversionCount(S1) ` `    ``invCount2 ``=` `inversionCount(S2) `   `    ``if` `((invCount1 ``=``=` `invCount2) ``or` `       ``((invCount1 ``%` `2``) ``=``=` `(invCount2 ``%` `2``)) ``or` `         ``haveRepeated(S1, S2) ``=``=` `1``): ` `             `  `        ``# If inversion count is same, ` `        ``# or have same parity ` `        ``# or if any of the string ` `        ``# have a repeated character ` `        ``# then the answer is Yes else No ` `        ``print``(``"Yes"``) ` `    ``else``: ` `        ``print``(``"No"``) `   `# Driver Code ` `S1 ``=` `"abbca"` `S2 ``=` `"acabb"`   `checkToMakeEqual(S1, S2) `

## Java

 `// Java program for the above approach` `import` `java.io.*;` `import` `java.util.*;`   `class` `GFG{` `    `  `// Function to count inversion` `// count of the string` `static` `int` `inversionCount(String s)` `{` `    `  `    ``// For storing frequency` `    ``int``[] freq = ``new` `int``[``26``];` `    ``int` `inv = ``0``;` `    ``for``(``int` `i = ``0``; i < s.length(); i++)` `    ``{` `        ``int` `temp = ``0``;` `        `  `        ``// Add all the characters which` `        ``// are less than the ith character` `        ``// before i.` `        ``for``(``int` `j = ``0``; ` `                ``j < (``int``)(s.charAt(i) - ``'a'``);` `                ``j++)` `        `  `        ``// Adding the count to inversion` `        ``// count` `        ``temp += freq[j];` `        ``inv += (i - temp);` `        `  `        ``// Updating the character in` `        ``// the frequency array` `        ``freq[s.charAt(i) - ``'a'``]++;` `    ``}` `    ``return` `inv;` `}`   `// Function to check whether any` `// of the string have a repeated` `// character` `static` `boolean` `haveRepeated(String S1,` `                            ``String S2)` `{` `    ``int``[] freq = ``new` `int``[``26``];` `    ``for``(``char` `i : S1.toCharArray())` `    ``{` `        ``if` `(freq[i - ``'a'``] > ``0``)` `            ``return` `true``;` `        ``freq[i - ``'a'``]++;` `    ``}` `    `  `    ``for``(``int` `i = ``0``; i < ``26``; i++)` `        ``freq[i] = ``0``;` `        `  `    ``for``(``char` `i : S2.toCharArray())` `    ``{` `        ``if` `(freq[i - ``'a'``] > ``0``)` `            ``return` `true``;` `        ``freq[i - ``'a'``]++;` `    ``}` `    ``return` `false``;` `}`   `// Function to check whether the` `// string S1 and S2 can be made` `// equal by reversing sub strings` `// of same size in both strings` `static` `void` `checkToMakeEqual(String S1,` `                             ``String S2)` `{` `    `  `    ``// Frequency array to check` `    ``// whether both string have` `    ``// same character or not` `    ``int``[] freq = ``new` `int``[``26``];` `    ``for``(``int` `i = ``0``; i < S1.length(); i++) ` `    ``{` `        `  `        ``// Adding the frequency;` `        ``freq[S1.charAt(i) - ``'a'``]++;` `    ``}` `    `  `    ``boolean` `flag = ``false``;` `    ``for``(``int` `i = ``0``; i < S2.length(); i++)` `    ``{` `        ``if` `(freq[S2.charAt(i) - ``'a'``] == ``0``) ` `        ``{` `            `  `            ``// If the character is not in S1` `            ``flag = ``true``;` `            ``break``;` `        ``}` `        `  `        ``// Decrementing the frequency` `        ``freq[S2.charAt(i) - ``'a'``]--;` `    ``}` `    `  `    ``if` `(flag == ``true``) ` `    ``{` `        `  `        ``// If both string doesnot` `        ``// have same characters or not` `        ``System.out.println(``"No"``);` `        ``return``;` `    ``}` `    `  `    ``// Finding inversion count` `    ``// of both strings` `    ``int` `invCount1 = inversionCount(S1);` `    ``int` `invCount2 = inversionCount(S2);` `    `  `    ``if` `(invCount1 == invCount2 || ` `       ``(invCount1 & ``1``) == (invCount2 & ``1``) || ` `        ``haveRepeated(S1, S2)) ` `    ``{` `        `  `        ``// If inversion count is same,` `        ``// or have same parity or if` `        ``// any of the string have a` `        ``// repeated character then` `        ``// the answer is Yes else No` `        ``System.out.println(``"Yes"``);` `    ``}` `    ``else` `    ``System.out.println(``"No"``);` `}`   `// Driver Code` `public` `static` `void` `main (String[] args)` `{` `    ``String S1 = ``"abbca"``, S2 = ``"acabb"``;` `    `  `    ``checkToMakeEqual(S1, S2);` `}` `}`   `// This code is contributed by offbeat`

## C#

 `// C# program for the above approach` `using` `System;` `class` `GFG{` `    `  `// Function to count inversion` `// count of the string` `static` `int` `inversionCount(String s)` `{` `    `  `    ``// For storing frequency` `    ``int``[] freq = ``new` `int``[26];` `    ``int` `inv = 0;` `    ``for``(``int` `i = 0; i < s.Length; i++)` `    ``{` `        ``int` `temp = 0;` `        `  `        ``// Add all the characters which` `        ``// are less than the ith character` `        ``// before i.` `        ``for``(``int` `j = 0; ` `                ``j < (``int``)(s[i] - ``'a'``);` `                ``j++)` `        `  `        ``// Adding the count to inversion` `        ``// count` `        ``temp += freq[j];` `        ``inv += (i - temp);` `        `  `        ``// Updating the character in` `        ``// the frequency array` `        ``freq[s[i] - ``'a'``]++;` `    ``}` `    ``return` `inv;` `}`   `// Function to check whether any` `// of the string have a repeated` `// character` `static` `bool` `haveRepeated(String S1,` `                            ``String S2)` `{` `    ``int``[] freq = ``new` `int``[26];` `    ``foreach``(``char` `i ``in` `S1.ToCharArray())` `    ``{` `        ``if` `(freq[i - ``'a'``] > 0)` `            ``return` `true``;` `        ``freq[i - ``'a'``]++;` `    ``}` `    `  `    ``for``(``int` `i = 0; i < 26; i++)` `        ``freq[i] = 0;` `        `  `    ``foreach``(``char` `i ``in` `S2.ToCharArray())` `    ``{` `        ``if` `(freq[i - ``'a'``] > 0)` `            ``return` `true``;` `        ``freq[i - ``'a'``]++;` `    ``}` `    ``return` `false``;` `}`   `// Function to check whether the` `// string S1 and S2 can be made` `// equal by reversing sub strings` `// of same size in both strings` `static` `void` `checkToMakeEqual(String S1,` `                             ``String S2)` `{` `    `  `    ``// Frequency array to check` `    ``// whether both string have` `    ``// same character or not` `    ``int``[] freq = ``new` `int``[26];` `    ``for``(``int` `i = 0; i < S1.Length; i++) ` `    ``{` `        `  `        ``// Adding the frequency;` `        ``freq[S1[i] - ``'a'``]++;` `    ``}` `    `  `    ``bool` `flag = ``false``;` `    ``for``(``int` `i = 0; i < S2.Length; i++)` `    ``{` `        ``if` `(freq[S2[i] - ``'a'``] == 0) ` `        ``{` `            `  `            ``// If the character is not in S1` `            ``flag = ``true``;` `            ``break``;` `        ``}` `        `  `        ``// Decrementing the frequency` `        ``freq[S2[i] - ``'a'``]--;` `    ``}` `    `  `    ``if` `(flag == ``true``) ` `    ``{` `        `  `        ``// If both string doesnot` `        ``// have same characters or not` `        ``Console.WriteLine(``"No"``);` `        ``return``;` `    ``}` `    `  `    ``// Finding inversion count` `    ``// of both strings` `    ``int` `invCount1 = inversionCount(S1);` `    ``int` `invCount2 = inversionCount(S2);` `    `  `    ``if` `(invCount1 == invCount2 || ` `       ``(invCount1 & 1) == (invCount2 & 1) || ` `        ``haveRepeated(S1, S2)) ` `    ``{` `        `  `        ``// If inversion count is same,` `        ``// or have same parity or if` `        ``// any of the string have a` `        ``// repeated character then` `        ``// the answer is Yes else No` `        ``Console.WriteLine(``"Yes"``);` `    ``}` `    ``else` `    ``Console.WriteLine(``"No"``);` `}`   `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `    ``String S1 = ``"abbca"``, S2 = ``"acabb"``;` `    `  `    ``checkToMakeEqual(S1, S2);` `}` `}`   `// This code is contributed by gauravrajput1`

## Javascript

 ``

Output:

`Yes`

Time Complexity:

Space Complexity:

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