# Find whether an array is subset of another array using Map

Given two arrays: arr1[0..m-1] and arr2[0..n-1]. Find whether arr2[] is a subset of arr1[] or not. Both the arrays are not in sorted order. It may be assumed that elements in both arrays are distinct.

**Examples:**

Input:arr1[] = {11, 1, 13, 21, 3, 7}, arr2[] = {11, 3, 7, 1}Output:arr2[] is a subset of arr1[]Input:arr1[] = {1, 2, 3, 4, 5, 6}, arr2[] = {1, 2, 4}Output:arr2[] is a subset of arr1[]Input:arr1[] = {10, 5, 2, 23, 19}, arr2[] = {19, 5, 3}Output:arr2[] is not a subset of arr1[]

**Simple Approach:** A simple approach is to run two nested loops. The outer loop picks all the elements of B[] one by one. The inner loop linearly searches for the element picked by the outer loop in A[]. If all elements are found, then print Yes, else print No. You can check the solution here.

**Efficient Approach:** Create a map to store the frequency of each distinct number present in A[]. Then we will check if each number of B[] is present in map or not. If present in the map, we will decrement the frequency value for that number by one and check for the next number. If map value for any number becomes zero, we will erase it from the map. If any number of B[] is not found in the map, we will set the flag value and break the loops and print No. Otherwise, we will print Yes.

Below is the C++ implementation of the above approach:

`// C++ program to check if an array is ` `// subset of another array ` ` ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to check if an array is ` `// subset of another array ` ` ` `int` `isSubset(` `int` `a[], ` `int` `b[], ` `int` `m, ` `int` `n) ` `{ ` ` ` ` ` `// map to store the values of array a[] ` ` ` `map<` `int` `, ` `int` `> mp1; ` ` ` ` ` `for` `(` `int` `i = 0; i < m; i++) ` ` ` `mp1[a[i]]++; ` ` ` ` ` `// flag value ` ` ` `int` `f = 0; ` ` ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` `// if b[i] is not present in map ` ` ` `// then array b[] can not be a ` ` ` `// subset of array a[] ` ` ` ` ` `if` `(mp1.find(b[i]) == mp1.end()) { ` ` ` `f = 1; ` ` ` `break` `; ` ` ` `} ` ` ` ` ` `// if if b[i] is present in map ` ` ` `// decrement by one ` ` ` `else` `{ ` ` ` `mp1[b[i]]--; ` ` ` ` ` `if` `(mp1[b[i]] == 0) ` ` ` `mp1.erase(mp1.find(b[i])); ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `f; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `arr1[] = { 11, 1, 13, 21, 3, 7 }; ` ` ` `int` `arr2[] = { 11, 3, 7, 1 }; ` ` ` ` ` `int` `m = ` `sizeof` `(arr1) / ` `sizeof` `(arr1[0]); ` ` ` `int` `n = ` `sizeof` `(arr2) / ` `sizeof` `(arr2[0]); ` ` ` ` ` `if` `(!isSubset(arr1, arr2, m, n)) ` ` ` `cout<<` `"arr2[] is subset of arr1[] "` `; ` ` ` `else` ` ` `cout<<` `"arr2[] is not a subset of arr1[]"` `; ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

arr2[] is subset of arr1[]

**Time Complexity:** O (n)

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