# Interleaving of two given strings with no common characters

• Difficulty Level : Easy
• Last Updated : 17 Jun, 2022

Given three strings A, B and C. Write a function that checks whether C is an interleaving of A and B. It may be assumed that there is no common character between A and B (Please see this for an extended solution that handles common characters also), C is said to be interleaving A and B, if it contains all characters of A and B and order of all characters in individual strings is preserved. See previous post for examples.

Solution: Pick each character of C one by one and match it with the first character in A. If it doesn’t match then match it with first character of B. If it doesn’t even match first character of B, then return false. If the character matches with first character of A, then repeat the above process from second character of C, second character of A and first character of B. If first character of C matches with the first character of B (and doesn’t match the first character of A), then repeat the above process from the second character of C, first character of A and second character of B. If all characters of C match either with a character of A or a character of B and length of C is sum of lengths of A and B, then C is an interleaving A and B.

## C++

 `// C++ program to check if given string is ` `// an interleaving of the other two strings ` `#include ` `using` `namespace` `std;`   `// Returns true if C is an interleaving of A and B, ` `// otherwise returns false ` `bool` `isInterleaved (``char` `A[], ``char` `B[], ``char` `C[]) ` `{ ` `    ``// Iterate through all characters of C. ` `    ``while` `(*C != 0) ` `    ``{ ` `        ``// Match first character of C with first character ` `        ``// of A. If matches them move A to next ` `        ``if` `(*A == *C) ` `            ``A++; `   `        ``// Else Match first character of C with first ` `        ``// character of B. If matches them move B to next ` `        ``else` `if` `(*B == *C) ` `            ``B++; `   `        ``// If doesn't match with either A or B, then return ` `        ``// false ` `        ``else` `            ``return` `false``; ` `        `  `        ``// Move C to next for next iteration ` `        ``C++; ` `    ``} `   `    ``// If A or B still have some characters, then length of ` `    ``// C is smaller than sum of lengths of A and B, so ` `    ``// return false ` `    ``if` `(*A || *B) ` `        ``return` `false``; `   `    ``return` `true``; ` `} `   `// Driver program to test above functions ` `int` `main() ` `{ ` `    ``char` `A[] = ``"AB"``; ` `    ``char` `B[] = ``"CD"``; ` `    ``char` `C[] = ``"ACBG"``; ` `    ``if` `(isInterleaved(A, B, C) == ``true``) ` `        ``cout << C << ``" is interleaved of "` `<< A << ``" and "` `<< B; ` `    ``else` `        ``cout << C << ``" is not interleaved of "` `<< A << ``" and "` `<< B; `   `    ``return` `0; ` `} `   `// This is code is contributed by rathbhupendra`

## C

 `// C program to check if given string is an interleaving` `// of the other two strings` `#include`   `// Returns true if C is an interleaving of A and B,` `// otherwise returns false` `bool` `isInterleaved (``char` `*A, ``char` `*B, ``char` `*C)` `{` `    ``// Iterate through all characters of C.` `    ``while` `(*C != 0)` `    ``{` `        ``// Match first character of C with first character` `        ``// of A. If matches them move A to next ` `        ``if` `(*A == *C)` `            ``A++;`   `        ``// Else Match first character of C with first ` `        ``// character of B. If matches them move B to next ` `        ``else` `if` `(*B == *C)` `            ``B++;` ` `  `        ``// If doesn't match with either A or B, then return` `        ``// false` `        ``else` `            ``return` `false``;` `        `  `        ``// Move C to next for next iteration` `        ``C++;` `    ``}`   `    ``// If A or B still have some characters, then length of` `    ``// C  is smaller than sum of lengths of A and B, so ` `    ``// return false` `    ``if` `(*A || *B)` `        ``return` `false``;`   `    ``return` `true``;` `}`   `// Driver program to test above functions` `int` `main()` `{` `    ``char` `*A = ``"AB"``;` `    ``char` `*B = ``"CD"``;` `    ``char` `*C = ``"ACBG"``;` `    ``if` `(isInterleaved(A, B, C) == ``true``)` `        ``printf``(``"%s is interleaved of %s and %s"``, C, A, B);` `    ``else` `        ``printf``(``"%s is not interleaved of %s and %s"``, C, A, B);`   `    ``return` `0;` `}`   `// This code is contributed by Venkat`

## Java

 `// Java program to check if the given string is ` `// an interleaving of the other two strings ` `public` `class` `GfG{`   `    ``// Returns true if C is an interleaving ` `    ``// of A and B, otherwise returns false ` `    ``static` `boolean` `isInterleaved (String A, String B, String C) ` `    ``{` `        ``int` `i = ``0``, j = ``0``, k = ``0``;` `        `  `        ``// Iterate through all characters of C. ` `        ``while` `(k != C.length()) ` `        ``{ ` `            ``// Match first character of C with first character ` `            ``// of A. If matches them move A to next ` `            ``if` `(i

## Python3

 `# Python3 program to check if given string is an interleaving of` `# the other two strings`   `# Returns true if C is an interleaving of A and B, otherwise` `# returns false` `def` `isInterleaved(A, B, C):`   `    ``# Utility variables` `    ``i ``=` `0` `    ``j ``=` `0` `    ``k ``=` `0`   `    ``# Iterate through all characters of C.` `    ``while` `k !``=` `len``(C)``-``1``:`   `        ``# Match first character of C with first character of A,` `        ``# If matches them move A to next` `        ``if` `i<``len``(A) ``and` `A[i] ``=``=` `C[k]:` `            ``i``+``=``1`   `        ``# Else Match first character of C with first character` `        ``# of B. If matches them move B to next` `        ``elif` `j< ``len``(B) ``and` `B[j] ``=``=` `C[k]:` `            ``j``+``=``1`   `        ``# If doesn't match with either A or B, then return false` `        ``else``:` `            ``return` `0`   `        ``# Move C to next for next iteration` `        ``k``+``=``1`   `    ``# If A or B still have some characters, then length of C is` `    ``# smaller than sum of lengths of A and B, so return false` `    ``if` `A[i``-``1``] ``or` `B[j``-``1``]:` `        ``return` `0`   `    ``return` `1`   `# Driver program to test the above function` `A ``=` `"AB"` `B ``=` `"CD"` `C ``=` `"ACBG"` `if` `isInterleaved(A, B, C) ``=``=` `1``:` `    ``print` `(C ``+` `" is interleaved of "` `+` `A ``+` `" and "` `+` `B)` `else``:` `    ``print` `(C ``+` `" is not interleaved of "` `+` `A ``+` `" and "` `+` `B)`   `# This code is contributed by Bhavya Jain`

## C#

 `// C# program to check if the given string is ` `// an interleaving of the other two strings ` `using` `System;`   `class` `GfG` `{`   `    ``// Returns true if C is an interleaving ` `    ``// of A and B, otherwise returns false ` `    ``static` `bool` `isInterleaved (String A, String B, String C) ` `    ``{` `        ``int` `i = 0, j = 0, k = 0;` `        `  `        ``// Iterate through all characters of C. ` `        ``while` `(k != C.Length - 1) ` `        ``{ ` `            ``// Match first character of C with first character ` `            ``// of A. If matches them move A to next ` `            ``if` `(A[i] == C[k]) ` `                ``i++; ` `    `  `            ``// Else Match first character of C with first ` `            ``// character of B. If matches them move B to next ` `            ``else` `if` `(B[j] == C[k]) ` `                ``j++; ` `    `  `            ``// If doesn't match with either A or B, then return ` `            ``// false ` `            ``else` `                ``return` `false``; ` `            `  `            ``// Move C to next for next iteration ` `            ``k++; ` `        ``} ` `    `  `        ``// If A or B still have some characters, ` `        ``// then length of C is smaller than sum ` `        ``// of lengths of A and B, so return false ` `        ``if` `(i < A.Length || j < B.Length) ` `            ``return` `false``; ` `    `  `        ``return` `true``; ` `    ``} `   `    ``// Driver code` `    ``public` `static` `void` `Main(String []args)` `    ``{` `        `  `        ``String A = ``"AB"``; ` `        ``String B = ``"CD"``; ` `        ``String C = ``"ACBG"``; ` `        ``if` `(isInterleaved(A, B, C) == ``true``) ` `            ``Console.WriteLine(``"{0} is interleaved of {1} and {2}"``, C, A, B); ` `        ``else` `            ``Console.WriteLine(``"{0} is not interleaved of {1} and {2}"``, C, A, B);` `    ``}` `}`   `// This code contributed by Rajput-Ji`

## PHP

 ``

## Javascript

 ``

Output

`ACBG is not interleaved of AB and CD`

Time Complexity:

O(m+n) where m and n are the lengths of strings A and B respectively.
Note that the above approach doesn’t work if A and B have some characters in common. For example, if string A = “AAB”, string B = “AAC” and string C = “AACAAB”, then the above method will return false. We have discussed here an extended solution that handles common characters

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