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Check whether an Array is Subarray of another Array

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Given two arrays A[] and B[] consisting of n                and m                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;
}

                    

Java

public class Main {
 
    // Function to check if an array is
    // subarray of another array
    static boolean isSubArray(int[] A, int[] B, int n, int m) {
        // Loop through possible starting indices of subarray
        for (int i = 0; i <= n - m; i++) {
            int j;
            // Loop through elements of subarray and compare with B[]
            for (j = 0; j < m; j++) {
                if (A[i + j] != B[j]) {
                    break;
                }
            }
            // All elements of B[] match subarray starting at index i
            if (j == m) {
                return true;
            }
        }
        return false;
    }
 
    // Driver Code
    public static void main(String[] args) {
        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");
    }
}

                    

Python3

# function to check that B is subarray of A or not
def isSubArray(A, B):
    n = len(A)
    m = len(B)
 
    for i in range(n - m + 1):
        for j in range(m):
            if A[i + j] != B[j]:
                break
        else
            # If the inner loop did not break, all elements of B match the subarray starting at index i
            return True
 
    return False
 
# Driver code
A = [2, 3, 0, 5, 1, 1, 2]
B = [3, 0, 5, 1]
 
if isSubArray(A, B):
    print("YES")
else:
    print("NO")
 
# This code is contributed by Kirti Agarwal

                    

C#

using System;
 
class GFG {
    // Function to check if an array is a subarray of
    // another array
    static 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
    static void Main()
    {
        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");
        }
    }
}

                    

Javascript

// Function to check if an array is a subarray of another array
function isSubArray(A, B, n, m) {
// Loop through possible starting indices of subarray
    for (let i = 0; i <= n - m; i++) {
        let j;
        // Loop through elements of subarray and compare with B[]
        for (j = 0; j < m; j++) {
            if (A[i + j] !== B[j]) {
                break;
            }
        }
        // All elements of B[] match subarray starting at index i
        if (j === m) {
            return true;
        }
    }
    return false;
}
 
// Driver code to test above function
let A = [2, 3, 0, 5, 1, 1, 2];
let n = A.length;
let B = [3, 0, 5, 1];
let m = B.length;
 
if (isSubArray(A, B, n, m))
    console.log("YES");
else
    console.log("NO");
     
// THIS CODE IS CONTRIBUTED BY KIRTI AGARWAL

                    

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



Last Updated : 13 Sep, 2023
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