Check whether an Array is Subarray of another Array

Given two arrays A[] and B[] consisting of and integers. The task is to check whether the array B[] is a subarray of the array A[] or not.

Examples: 

Input : A[] = {2, 3, 0, 5, 1, 1, 2}, B[] = {3, 0, 5, 1} 
Output : Yes

Input : A[] = {1, 2, 3, 4, 5}, B[] = {2, 5, 6} 
Output : No 

Source: Visa Interview Experience

Approach:

• Iterate through the array A[] using a loop that runs from 0 to n-m, where n and m are the sizes of arrays A[] and B[] respectively. This loop represents the starting index of the subarray of A[] that we are comparing with B[].
• For each starting index in the loop, iterate through the array B[] using another loop that runs from 0 to m-1. This loop represents the index of the elements in the subarray that we are comparing with the corresponding elements in B[].
• If the elements at the corresponding indices in the subarray of A[] and B[] are not equal, break out of the inner loop and move to the next starting index in the outer loop.
• If all the elements in the subarray of A[] match the elements in B[], return true.
• If none of the starting indices in the outer loop result in a matching subarray, return false.

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>
using namespace std;

// Function to check if an array is
// subarray of another array
bool isSubArray(int A[], int B[], int n, int m) {
    for (int i = 0; i <= n-m; i++) {  // Loop through possible starting indices of subarray
        int j;
        for (j = 0; j < m; j++) {  // Loop through elements of subarray and compare with B[]
            if (A[i+j] != B[j]) {
                break;
            }
        }
        if (j == m) {  // All elements of B[] match subarray starting at index i
            return true;
        }
    }
    return false;
}

// Driver Code
int main() {
    int A[] = { 2, 3, 0, 5, 1, 1, 2 };
    int n = sizeof(A) / sizeof(int);
    int B[] = { 3, 0, 5, 1 };
    int m = sizeof(B) / sizeof(int);

    if (isSubArray(A, B, n, m))
        cout << "YES\n";
    else
        cout << "NO\n";

    return 0;
}

Output

YES

Time Complexity: O(n*m), where n and m are the sizes of the arrays A[] and B[], respectively. 
Space Complexity: O(1) as we are not using any additional space to store the arrays or any other variables.

Efficient Approach : An efficient approach is to use two pointers to traverse both the array simultaneously. Keep the pointer of array B[] still and if any element of A[] matches with the first element of B[] then increase the pointer of both the array else set the pointer of A to the next element of the previous starting point and reset the pointer of B to 0. If all the elements of B are matched then print YES otherwise print NO.

Below is the implementation of the above approach: 

C++

// C++ program to check if an array is
// subarray of another array

#include <bits/stdc++.h>
using namespace std;

// Function to check if an array is
// subarray of another array
bool isSubArray(int A[], int B[], int n, int m)
{
    // Two pointers to traverse the arrays
    int i = 0, j = 0;

    // Traverse both arrays simultaneously
    while (i < n && j < m) {

        // If element matches
        // increment both pointers
        if (A[i] == B[j]) {

            i++;
            j++;

            // If array B is completely
            // traversed
            if (j == m)
                return true;
        }
        // If not,
        // increment i and reset j
        else {
            i = i - j + 1;
            j = 0;
        }
    }

    return false;
}

// Driver Code
int main()
{
    int A[] = { 2, 3, 0, 5, 1, 1, 2 };
    int n = sizeof(A) / sizeof(int);
    int B[] = { 3, 0, 5, 1 };
    int m = sizeof(B) / sizeof(int);

    if (isSubArray(A, B, n, m))
        cout << "YES\n";
    else
        cout << "NO\n";

    return 0;
}

Java

// Java program to check if an array is
// subarray of another array
class gfg
{

    // Function to check if an array is
    // subarray of another array
    static boolean isSubArray(int A[], int B[],
                                   int n, int m)
    {
        // Two pointers to traverse the arrays
        int i = 0, j = 0;

        // Traverse both arrays simultaneously
        while (i < n && j < m)
        {

            // If element matches
            // increment both pointers
            if (A[i] == B[j])
            {

                i++;
                j++;

                // If array B is completely
                // traversed
                if (j == m)
                    return true;
            }

            // If not,
            // increment i and reset j
            else
            {
                i = i - j + 1;
                j = 0;
            }
        }
        return false;
    }

    // Driver Code
    public static void main(String arr[])
    {
        int A[] = { 2, 3, 0, 5, 1, 1, 2 };
        int n = A.length;
        int B[] = { 3, 0, 5, 1 };
        int m = B.length;

        if (isSubArray(A, B, n, m))
            System.out.println("YES");
        else
            System.out.println("NO");
    }
}

// This code is contributed by gp6

Python3

# Python3 program to check if an array is
# subarray of another array

# Function to check if an array is
# subarray of another array
def isSubArray(A, B, n, m):

    # Two pointers to traverse the arrays
    i = 0; j = 0;

    # Traverse both arrays simultaneously
    while (i < n and j < m):

        # If element matches
        # increment both pointers
        if (A[i] == B[j]):

            i += 1;
            j += 1;

            # If array B is completely
            # traversed
            if (j == m):
                return True;

        # If not,
        # increment i and reset j
        else:
            i = i - j + 1;
            j = 0;

    return False;

# Driver Code
if __name__ == '__main__':
    A = [ 2, 3, 0, 5, 1, 1, 2 ];
    n = len(A);
    B = [ 3, 0, 5, 1 ];
    m = len(B);

    if (isSubArray(A, B, n, m)):
        print("YES");
    else:
        print("NO");

# This code is contributed by Rajput-Ji

C#

// C# program to check if an array is
// subarray of another array
using System;

public class GFG
{

    // Function to check if an array is
    // subarray of another array
    static bool isSubArray(int []A, int []B,
                                   int n, int m)
    {
        // Two pointers to traverse the arrays
        int i = 0, j = 0;

        // Traverse both arrays simultaneously
        while (i < n && j < m)
        {

            // If element matches
            // increment both pointers
            if (A[i] == B[j])
            {

                i++;
                j++;

                // If array B is completely
                // traversed
                if (j == m)
                    return true;
            }

            // If not,
            // increment i and reset j
            else
            {
                i = i - j + 1;
                j = 0;
            }
        }
        return false;
    }

    // Driver Code
    public static void Main(String []arr)
    {
        int []A = { 2, 3, 0, 5, 1, 1, 2 };
        int n = A.Length;
        int []B = { 3, 0, 5, 1 };
        int m = B.Length;

        if (isSubArray(A, B, n, m))
            Console.WriteLine("YES");
        else
            Console.WriteLine("NO");
    }
}

// This code is contributed by PrinciRaj1992

Javascript

<script>
// Javascript program to check if an array is
// subarray of another array


// Function to check if an array is
// subarray of another array
function isSubArray(A, B, n,m)
{
    // Two pointers to traverse the arrays
    var i = 0, j = 0;

    // Traverse both arrays simultaneously
    while (i < n && j < m) {

        // If element matches
        // increment both pointers
        if (A[i] == B[j]) {

            i++;
            j++;

            // If array B is completely
            // traversed
            if (j == m)
                return true;
        }
        // If not,
        // increment i and reset j
        else {
            i = i - j + 1;
            j = 0;
        }
    }

    return false;
}

var A = [ 2, 3, 0, 5, 1, 1, 2 ];
var n = A.length;
var B = [ 3, 0, 5, 1 ];
var m =B.length;
if (isSubArray(A, B, n, m))
        document.write( "YES<br>");
    else
        document.write( "NO<br>");

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

Output

YES

Time Complexity: O(N*N)
Auxiliary Space: O(1)

Related Topic: Subarrays, Subsequences, and Subsets in Array