Interleaving of two given strings with no common characters

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 <bits/stdc++.h>
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<stdio.h>

// 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<A.length()&&A.charAt(i) == C.charAt(k))
                i++;

            // Else Match first character of C with first
            // character of B. If matches them move B to next
            else if (j<B.length()&&B.charAt(j) == C.charAt(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;
    }

    public static void main(String []args){

        String A = "AB";
        String B = "CD";
        String C = "ACBG";
        if (isInterleaved(A, B, C) == true)
            System.out.printf("%s is interleaved of %s and %s", C, A, B);
        else
            System.out.printf("%s is not interleaved of %s and %s", C, A, B);
    }
}

// This code is contributed by Rituraj Jain

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

<?php
// PHP 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
function isInterleaved ($A, $B, $C)
{
    // Iterate through all characters of C.
    while ($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 Code
$A = "AB";
$B = "CD";
$C = "ACBG";
if (isInterleaved($A, $B, $C) == true)
    echo $C . " is interleaved of " .
         $A . " and " . $B;
else
    echo $C . " is not interleaved of " .
         $A . " and " . $B;

// This code is contributed by ita_c
?>

Javascript

<script>
// Javascript program to check if the given string is
// an interleaving of the other two strings

    // Returns true if C is an interleaving
    // of A and B, otherwise returns false
    function isInterleaved (A,B,C)
    {
        let 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<A.length && 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 (j < B.length && 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;
    }

    let A = "AB";
    let B = "CD";
    let C = "ACBG";
    if (isInterleaved(A, B, C) == true)
    {
        document.write(C+ " is interleaved of "+ A+" and ", B);
    }
    else
    {
        document.write(C + " is not interleaved of "+ A +" and ", B);
    }

    // This code is contributed by avanitrachhadiya2155
</script>

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.