Find whether an array is subset of another array | Added Method 3 - GeeksforGeeks
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.
Input: arr1[] = {11, 1, 13, 21, 3, 7}, arr2[] = {11, 3, 7, 1}
Output: arr2[] is a subset of arr1[]
Method 3 (Use Sorting and Merging )
1) Sort both arrays: arr1[] and arr2[] O(mLogm + nLogn)
2) Use Merge type of process to see if all elements of sorted arr2[] are present in sorted arr1[].
Time Complexity: O(mLogm + nLogn) which is better than method 2
Method 1 (Simple)
Use two loops: The outer loop picks all the elements of arr2[] one by one. The inner loop linearly searches for the element picked by outer loop. If all elements are found then return 1, else return 0.
Read full article from Find whether an array is subset of another array | Added Method 3 - GeeksforGeeks
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.
Input: arr1[] = {11, 1, 13, 21, 3, 7}, arr2[] = {11, 3, 7, 1}
Output: arr2[] is a subset of arr1[]
Method 3 (Use Sorting and Merging )
1) Sort both arrays: arr1[] and arr2[] O(mLogm + nLogn)
2) Use Merge type of process to see if all elements of sorted arr2[] are present in sorted arr1[].
Time Complexity: O(mLogm + nLogn) which is better than method 2
bool
isSubset(
int
arr1[],
int
arr2[],
int
m,
int
n)
{
int
i = 0, j = 0;
if
(m < n)
return
0;
quickSort(arr1, 0, m-1);
quickSort(arr2, 0, n-1);
while
( i < n && j < m )
{
if
( arr1[j] <arr2[i] )
j++;
else
if
( arr1[j] == arr2[i] )
{
j++;
i++;
}
else
if
( arr1[j] > arr2[i] )
return
0;
}
if
( i < n )
return
0;
else
return
1;
}
Method 4 (Use Hashing)
1) Create a Hash Table for all the elements of arr1[].
2) Traverse arr2[] and search for each element of arr2[] in the Hash Table. If element is not found then return 0.
3) If all elements are found then return 1.
1) Create a Hash Table for all the elements of arr1[].
2) Traverse arr2[] and search for each element of arr2[] in the Hash Table. If element is not found then return 0.
3) If all elements are found then return 1.
Note that method 1, method 2 and method 4 don’t handle the cases when we have duplicates in arr2[]. For example, {1, 4, 4, 2} is not a subset of {1, 4, 2}, but these methods will print it as a subset.
Method 2 (Use Sorting and Binary Search)
1) Sort arr1[] O(mLogm) 2) For each element of arr2[], do binary search for it in sorted arr1[]. a) If the element is not found then return 0. 3) If all elements are present then return 1.Time Complexity: O(mLogm + nLogm).
Method 1 (Simple)
Use two loops: The outer loop picks all the elements of arr2[] one by one. The inner loop linearly searches for the element picked by outer loop. If all elements are found then return 1, else return 0.
Time Complexity: O(m*n)
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