Given two arrays of integers of size m and n. The task is to find the minimum length subarray in the first array that contains all the elements of second array.
Note: element of second array may be present in the large array in non-contiguous but order must same. ( m < n )
Examples :
Input : A[] = {2, 2, 4, 5, 8, 9}
B[] = {2, 5, 9}
Output : 5
Smallest subarray of A[] that contains all
elements of B[] is {2, 4, 5, 8, 9} which is
of size 5.
Input : A[] = {5, 6, 5, 2, 7, 5, 6, 7, 5, 5, 7}
B[] = {5, 5, 7}
Output : 3
Method 1 (Naive): A simple solution is to generate all subarrays of given array and check for every sub-array if it contains elements of another array or not. In the end, return the minimum length of sub-array that contain another array.
Below is the implementation of above idea.
C++
// CPP program to find smallest length // subarray that contains all elements // of another array.#include <bits/stdc++.h>using namespace std;// function return the minimum length of sub_arrayint minimumSubArray(int A[], int n, int B[], int m){ int result = INT_MAX; // Pick starting point for (int i = 0; i < n; i++) { // Pick ending point for (int j = i; j < n; j++) { // k is index in first array and // 'index' is index in second array. int k, index = 0; for (k = i; k <= j; k++) { if (A[k] == B[index]) index++; if (index == m) break; } // update minimum length sub_array if (index == m && result > k - i + 1) result = (k == n) ? k - i : k - i + 1; } } // return minimum length subarray return result;}// driver program to test above functionint main(){ int A[] = { 5, 6, 5, 2, 7, 5, 6, 7, 5, 5, 7 }; int B[] = { 5, 5, 7 }; int n = sizeof(A)/sizeof(A[0]); int m = sizeof(B)/sizeof(B[0]); cout << minimumSubArray(A, n, B, m); return 0;} |
Java
// Java program to find smallest length // subarray that contains all elements // of another array.import java.io.*;class GFG {// function return the minimum length // of sub_arraystatic int minimumSubArray(int A[], int n, int B[], int m){ int result = Integer.MAX_VALUE; // Pick starting point for (int i = 0; i < n; i++) { // Pick ending point for (int j = i; j < n; j++) { // k is index in first array // and 'index' is index in // second array. int k, index = 0; for (k = i; k <= j; k++) { if (A[k] == B[index]) index++; if (index == m) break; } // update minimum length sub_array if (index == m && result > k - i + 1) result = (k == n) ? k - i : k - i + 1; } } // return minimum length subarray return result;}// driver program to test above functionpublic static void main(String[] args){ int A[] = { 5, 6, 5, 2, 7, 5, 6, 7, 5, 5, 7 }; int B[] = { 5, 5, 7 }; int n = A.length; int m = B.length; System.out.println(minimumSubArray(A, n, B, m));}}// This code is contributed by Prerna Saini |
Python3
# Python 3 program to find smallest # length subarray that contains all # elements of another array.# function return the minimum length # of sub_arraydef minimumSubArray(A, n, B, m) : result = 10000000 # Pick starting point for i in range(0, n) : # Pick ending point for j in range(0,n) : # k is index in first array and # 'index' is index in second array. index = 0 for k in range(i, j + 1) : if (A[k] == B[index]) : index=index + 1 if (index == m) : break # update minimum length sub_array if (index == m and result > k - i + 1) : if (k == n) : result = k - i else : result = k - i + 1 # return minimum length subarray return result # driver program to test above functionA = [ 5, 6, 5, 2, 7, 5, 6, 7, 5, 5, 7 ]B = [ 5, 5, 7 ]n = len(A)m = len(B)print(minimumSubArray(A, n, B, m))#This code is contributed by Nikita Tiwari |
C#
// C# program to find smallest length // subarray that contains all elements // of another array.using System;class GFG {// function return the minimum // length of sub_arraystatic int minimumSubArray(int []A, int n, int []B, int m){ int result = int.MaxValue; // Pick starting point for (int i = 0; i < n; i++) { // Pick ending point for (int j = i; j < n; j++) { // k is index in first array // and 'index' is index in // second array. int k, index = 0; for (k = i; k <= j; k++) { if (A[k] == B[index]) index++; if (index == m) break; } // update minimum length sub_array if (index == m && result > k - i + 1) result = (k == n) ? k - i : k - i + 1; } } // return minimum length subarray return result;} // Driver code public static void Main() { int []A = { 5, 6, 5, 2, 7, 5, 6, 7, 5, 5, 7 }; int []B = { 5, 5, 7 }; int n = A.Length; int m = B.Length; Console.Write(minimumSubArray(A, n, B, m)); }}// This code is contributed by nitin mittal. |
PHP
<?php// PHP program to find smallest length // subarray that contains all elements // of another array.// function return the minimum// length of sub_arrayfunction minimumSubArray($A, $n, $B, $m){ $result = PHP_INT_MAX; // Pick starting point for ($i = 0; $i < $n; $i++) { // Pick ending point for ( $j = $i; $j < $n; $j++) { // k is index in first array and // 'index' is index in second array. $k; $index = 0; for ($k = $i; $k <= $j; $k++) { if ($A[$k] == $B[$index]) $index++; if ($index == $m) break; } // update minimum length // sub_array if ($index == $m && $result > $k - $i + 1) $result = ($k == $n) ? $k - $i : $k - $i + 1; } } // return minimum length subarray return $result;} // Driver Code $A = array(5, 6, 5, 2, 7, 5, 6, 7, 5, 5, 7); $B = array(5, 5, 7); $n = count($A); $m = count($B); echo minimumSubArray($A, $n, $B, $m); // This code is contributed by anuj_67?> |
Javascript
<script>// Javascript program to find smallest length // subarray that contains all elements // of another array.// function return the minimum length of sub_arrayfunction minimumSubArray(A, n, B, m){ var result = 1000000000; // Pick starting point for (var i = 0; i < n; i++) { // Pick ending point for (var j = i; j < n; j++) { // k is index in first array and // 'index' is index in second array. var k, index = 0; for (k = i; k <= j; k++) { if (A[k] == B[index]) index++; if (index == m) break; } // update minimum length sub_array if (index == m && result > k - i + 1) result = (k == n) ? k - i : k - i + 1; } } // return minimum length subarray return result;}// driver program to test above functionvar A = [ 5, 6, 5, 2, 7, 5, 6, 7, 5, 5, 7 ];var B = [ 5, 5, 7 ];var n = A.length;var m = B.length;document.write( minimumSubArray(A, n, B, m));// This code is contributed by noob2000.</script> |
Output:
3
Time Complexity : O(n3)
Auxiliary Space : O(1)
Method 2 (Efficient): Method 2 is an optimized version of method 1. Here we consider only those subarrays whose first element match with the first element of second array. If first element matches, then we match the rest of the elements of second array in the Main_array and if all elements match then we update length if need. In the end, we return minimum length of sub_array.
Below is the implementation of above idea:
C++
// C++ program to find smallest length // subarray that contains all elements // of another array.#include <bits/stdc++.h>using namespace std;// Returns the minimum length of sub_array int minimumSubArray(int A[], int n, int B[], int m){ int result = INT_MAX; // Traverse main_array element for (int i = 0; i < n - m + 1; i++) { // Pick only those subarray of main_array // whose first element match with the // first element of second_array if (A[i] == B[0]) { // initialize starting of both // subarrays int j = 0, index = i; for (; index < n; index++) { if (A[index] == B[j]) j++; // if we found all elements of // second array if (j == m) break; } // update minimum length sub_array if (j == m && result > index - i + 1) result = (index == n) ? index - i : index - i + 1; }} // return minimum length subarray return result;}// driver program to test above functionint main(){ int A[] = { 5, 6, 5, 2, 7, 5, 6, 7, 5, 5, 7 }; int B[] = { 5, 5, 7 }; int n = sizeof(A)/sizeof(A[0]); int m = sizeof(B)/sizeof(B[0]); cout << minimumSubArray(A, n, B, m); return 0;} |
Java
// Java program to find smallest length // subarray that contains all elements // of another array.import java.io.*;class GFG {// Returns the minimum length of sub_array static int minimumSubArray(int A[], int n, int B[], int m){ int result = Integer.MAX_VALUE; // Traverse main_array element for (int i = 0; i < n - m + 1; i++) { // Pick only those subarray of // main_array whose first element // match with the first element // of second_array if (A[i] == B[0]) { // initialize starting of // both subarrays int j = 0, index = i; for (; index < n; index++) { if (A[index] == B[j]) j++; // if we found all elements // of second array if (j == m) break; } // update minimum length sub_array if (j == m && result > index - i + 1) result = (index == n) ? index - i : index - i + 1; } } // return minimum length subarray return result;}// driver program to test above functionpublic static void main(String[] args){ int A[] = { 5, 6, 5, 2, 7, 5, 6, 7, 5, 5, 7 }; int B[] = { 5, 5, 7 }; int n = A.length; int m = B.length; System.out.println(minimumSubArray(A, n, B, m));}}// This code is contributed by Prerna Saini |
Python3
# Python 3 program to find smallest length # subarray that contains all elements # of another array.# Returns the minimum length of sub_array def minimumSubArray( A,n, B, m) : result = 1000000 # Traverse main_array element for i in range(0, n - m + 1) : # Pick only those subarray of main_array # whose first element match with the # first element of second_array if (A[i] == B[0]) : # initialize starting of both # subarrays j = 0 index = i for index in range(i, n) : if (A[index] == B[j]) : j = j+1 # if we found all elements # of second array if (j == m) : break # update minimum length sub_array if (j == m and result > index - i + 1) : if(index == n) : result = index - i else: result = index - i + 1 # return minimum length subarray return result # driver program to test above functionA = [ 5, 6, 5, 2, 7, 5, 6, 7, 5, 5, 7 ]B = [ 5, 5, 7 ]n = len(A)m = len(B)print(minimumSubArray(A, n, B, m))# This code is contributed by Nikita Tiwari. |
C#
// C# program to find smallest length // subarray that contains all elements // of another array.using System;class GFG {// Returns the minimum length of sub_array static int minimumSubArray(int []A, int n, int []B, int m){ int result = int.MaxValue; // Traverse main_array element for (int i = 0; i < n - m + 1; i++) { // Pick only those subarray of // main_array whose first element // match with the first element // of second_array if (A[i] == B[0]) { // initialize starting of // both subarrays int j = 0, index = i; for (; index < n; index++) { if (A[index] == B[j]) j++; // if we found all elements // of second array if (j == m) break; } // update minimum length sub_array if (j == m && result > index - i + 1) result = (index == n) ? index - i : index - i + 1; } } // return minimum length subarray return result;} // Driver code public static void Main() { int []A = { 5, 6, 5, 2, 7, 5, 6, 7, 5, 5, 7 }; int []B = { 5, 5, 7 }; int n = A.Length; int m = B.Length; Console.Write(minimumSubArray(A, n, B, m)); }}// This code is contributed by nitin mittal. |
PHP
<?php// PHP program to find smallest length // subarray that contains all elements // of another array.// Returns the minimum length of sub_array function minimumSubArray(&$A, $n, &$B, $m){ $result = PHP_INT_MAX; // Traverse main_array element for ($i = 0; $i < $n - $m + 1; $i++) { // Pick only those subarray of main_array // whose first element match with the // first element of second_array if ($A[$i] == $B[0]) { // initialize starting of both // subarrays $j = 0; $index = $i; for ($index = $i; $index < $n; $index++) { if ($A[$index] == $B[$j]) $j++; // if we found all elements of // second array if ($j == $m) break; } // update minimum length sub_array if ($j == $m && $result > $index - $i + 1) $result = ($index == $n) ? $index - $i : $index - $i + 1; } } // return minimum length subarray return $result;}// Driver Code$A = array(5, 6, 5, 2, 7, 5, 6, 7, 5, 5, 7 );$B = array(5, 5, 7 );$n = sizeof($A);$m = sizeof($B);echo(minimumSubArray($A, $n, $B, $m));// This code is contributed // by Shivi_Aggarwal ?> |
Javascript
<script> // Javascript program to find smallest length // subarray that contains all elements // of another array. // Returns the minimum length of sub_array function minimumSubArray(A, n, B, m) { let result = Number.MAX_VALUE; // Traverse main_array element for (let i = 0; i < n - m + 1; i++) { // Pick only those subarray of // main_array whose first element // match with the first element // of second_array if (A[i] == B[0]) { // initialize starting of // both subarrays let j = 0, index = i; for (; index < n; index++) { if (A[index] == B[j]) j++; // if we found all elements // of second array if (j == m) break; } // update minimum length sub_array if (j == m && result > index - i + 1) result = (index == n) ? index - i : index - i + 1; } } // return minimum length subarray return result; } let A = [ 5, 6, 5, 2, 7, 5, 6, 7, 5, 5, 7 ]; let B = [ 5, 5, 7 ]; let n = A.length; let m = B.length; document.write(minimumSubArray(A, n, B, m)); // This code is contributed by divyesh072019.</script> |
Output:
3
Time Complexity : O(n*n)
Auxiliary Space : O(1)
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