Related Articles

# Check if characters of one string can be swapped to form other

• Last Updated : 07 May, 2021

Two strings are given, we need to find whether we can form second string by swapping the character of the first string.
Examples:

```Input : str1 = "geeksforgeeks"
str2 = "geegeeksksfor"
Output : YES

Input : str1 = "geeksfor"
str2 = "geeekfor"
Output : NO```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

First of all, we will find the length of strings and if lengths are not equal that means we can not form the target string by swapping characters of the first string. If lengths are equal then we iterate through the first string and create a map for it and after that, we will iterate through the second string and decrease the count of the map if any of the indexes go negative in the map that means we can not form the target string otherwise we can form the target string.

```Algorithm-
1- l1 = str1.length()  &&   l2 = str2.length()
2- if (l1 != l2)
print "NO"
3- Else
map = {0};
for i=0 to l1
map[str1[i]-'a']++;
for i=0 to l2
map[str2[i]-'a']--;
if (map[str-'a'<0)
print "NO"
4- if no index goes negative print "YES"
5- End```

## C++

 `#include ``using` `namespace` `std;``const` `int` `MAX = 26;` `bool` `targetstring(string str1, string str2)``{``    ``int` `l1 = str1.length();``    ``int` `l2 = str2.length();` `    ``// if length is not same print no``    ``if` `(l1 != l2)``        ``return` `false``;` `    ``int` `map[MAX] = { 0 };` `    ``// Count frequencies of character in``    ``// first string.``    ``for` `(``int` `i = 0; i < l1; i++)``        ``map[str1[i] - ``'a'``]++;` `    ``// iterate through the second string``    ``// decrement counts of characters in``    ``// second string``    ``for` `(``int` `i = 0; i < l2; i++) {``        ``map[str2[i] - ``'a'``]--;` `        ``// Since lengths are same, some``        ``// value would definitely become``        ``// negative if result is false.``        ``if` `(map[str2[i] - ``'a'``] < 0)``            ``return` `false``;``    ``}` `    ``return` `true``;``}` `// driver function``int` `main()``{``    ``string str1 = ``"geeksforgeeks"``;``    ``string str2 = ``"geegeeksksfor"``;``    ``if` `(targetstring(str1, str2))``        ``cout << ``"YES"``;``    ``else``        ``cout << ``"NO"``;` `    ``return` `0;``}`

## Java

 `// Java program to check if``// characters of one string``// can be swapped to form other``class` `GFG``{``static` `int` `MAX = ``26``;` `static` `boolean` `targetstring(String str1,``                            ``String str2)``{``    ``int` `l1 = str1.length();``    ``int` `l2 = str2.length();` `    ``// if length is not same print no``    ``if` `(l1 != l2)``        ``return` `false``;` `    ``int` `[]map = ``new` `int``[MAX];``    ` `    ``// Count frequencies of``    ``// character in first string.``    ``for` `(``int` `i = ``0``; i < l1; i++)``        ``map[str1.charAt(i) - ``'a'``]++;` `    ``// iterate through the second``    ``// string decrement counts of``    ``// characters in second string``    ``for` `(``int` `i = ``0``; i < l2; i++)``    ``{``        ``map[str2.charAt(i) - ``'a'``]--;` `        ``// Since lengths are same,``        ``// some value would definitely``        ``// become negative if result``        ``// is false.``        ``if` `(map[str2.charAt(i) - ``'a'``] < ``0``)``            ``return` `false``;``    ``}` `    ``return` `true``;``}` `// Driver Code``public` `static` `void` `main(String args[])``{``    ``String str1 = ``"geeksforgeeks"``;``    ``String str2 = ``"geegeeksksfor"``;``    ``if` `(targetstring(str1, str2))``        ``System.out.print(``"YES"``);``    ``else``        ``System.out.print(``"NO"``);``}``}` `// This code is contributed by``// Akanksha Rai`

## Python3

 `# Python3 program to check if``# characters of one string``# can be swapped to form other``MAX` `=` `26` `def` `targetstring(str1, str2):` `    ``l1 ``=` `len``(str1)``    ``l2 ``=` `len``(str2)` `    ``# if length is not same print no``    ``if` `(l1 !``=` `l2):``        ``return` `False` `    ``map` `=` `[``0``] ``*` `MAX` `    ``# Count frequencies of character``    ``# in first string.``    ``for` `i ``in` `range` `(l1):``        ``map``[``ord``(str1[i]) ``-` `ord``(``'a'``)] ``+``=` `1` `    ``# iterate through the second string``    ``# decrement counts of characters in``    ``# second string``    ``for` `i ``in` `range``(l2) :``        ``map``[``ord``(str2[i]) ``-` `ord``(``'a'``)] ``-``=` `1` `        ``# Since lengths are same, some``        ``# value would definitely become``        ``# negative if result is false.``        ``if` `(``map``[``ord``(str2[i]) ``-` `ord``(``'a'``)] < ``0``):``            ``return` `False` `    ``return` `True` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``str1 ``=` `"geeksforgeeks"``    ``str2 ``=` `"geegeeksksfor"``    ``if` `(targetstring(str1, str2)):``        ``print``(``"YES"``)``    ``else``:``        ``print``(``"NO"``)` `# This code is contributed by ita_c`

## C#

 `// C# program to check if``// characters of one string``// can be swapped to form other``using` `System;` `class` `GFG``{``    ``static` `int` `MAX = 26;``    ` `    ``static` `bool` `targetstring(``string` `str1,``                             ``string` `str2)``    ``{``        ``int` `l1 = str1.Length;``        ``int` `l2 = str2.Length;``    ` `        ``// if length is not``        ``// same print no``        ``if` `(l1 != l2)``            ``return` `false``;``    ` `        ``int` `[]map = ``new` `int``[MAX];``        ``Array.Clear(map, 0, 26);``    ` `        ``// Count frequencies of``        ``// character in first string.``        ``for` `(``int` `i = 0; i < l1; i++)``            ``map[str1[i] - ``'a'``]++;``    ` `        ``// iterate through the second``        ``// string decrement counts of``        ``// characters in second string``        ``for` `(``int` `i = 0; i < l2; i++)``        ``{``            ``map[str2[i] - ``'a'``]--;``    ` `            ``// Since lengths are same,``            ``// some value would definitely``            ``// become negative if result``            ``// is false.``            ``if` `(map[str2[i] - ``'a'``] < 0)``                ``return` `false``;``        ``}``    ` `        ``return` `true``;``    ``}``    ` `    ``// Driver Code``    ``static` `void` `Main()``    ``{``        ``string` `str1 = ``"geeksforgeeks"``;``        ``string` `str2 = ``"geegeeksksfor"``;``        ``if` `(targetstring(str1, str2))``            ``Console.Write(``"YES"``);``        ``else``            ``Console.Write(``"NO"``);``    ``}``}` `// This code is contributed by``// Manish Shaw(manishshaw1)`

## PHP

 ``

## Javascript

 ``
Output:
`YES`

Time Complexity – O(n)

My Personal Notes arrow_drop_up