# Minimum Distinct Labels

Posted: 27 Feb, 2021
Difficulty: Moderate

## PROBLEM STATEMENT

#### For example:

``````Consider M = 2 and the array ARR = [ 3, 4, 5, 3 ]
If we remove the second and the third box, then all the boxes remaining on the table will have label 3. Hence, the minimum number of distinct labels left will be 1 in this case.
``````
##### Input Format:
``````The first line of the input contains an integer, 'T,’ denoting the number of test cases.

The first line of each test case contains two space-separated integers, 'N' and 'M', denoting the number of boxes and the number of boxes to be removed respectively.

The second line of each test case contains 'N' space-separated integers denoting the labels of each box.
``````
##### Output Format:
``````For each test case, return the minimum number of distinct labels left on the table after removing exactly 'M' boxes.
``````
##### Note :
``````You do not need to print anything. It has already been taken care of. Just implement the given function.
``````
##### Constraints:
``````1 <= T <= 10
1 <= N <= 10^4
0 <= M <= N
1 <= ARR[i] <= 10^9

Time Limit: 1 sec
`````` Approach 1

The idea is to generate all the possible combinations of the ‘M’ boxes that can be removed, and for each combination, find out the number of distinct labels that are left on the table and compare it with results obtained from all other combinations. In the end we will return minimum possible labels obtained.

To check the number of distinct labels left, we will insert all the labels on the boxes which are not removed into a set and the number of distinct labels will be equal to the number of elements in the set.

To generate all the combinations of removals, we will take help of the next permutation function. We will create an array having ‘M’ zeros and ‘N’ - ‘M’ ones to store the current combination. The ‘M’ zeros will represent the indices of the boxes that will be removed, and the ‘N’ - ‘M’ ones will represent the indices of the boxes left on the table. We will place all the ‘M’ zeros at the start of the array and the ones at the end. To move to the next combination we will call the next permutation function for the current permutation.

Steps:

1. Let 'ANS' be the minimum distinct labels left. Initialize it as the number of distinct labels initially present on the table.
2. Create an array 'PERM' having ‘M’ zeros followed by ‘N’ - ‘M’ ones to store the current permutation.
3. While a greater permutation for the current permutation exists,
• Change current permutation to the next permutation.
• Create an empty set ‘LABS’ to store the distinct labels.
• Iterate from ‘j’ = 0 to ‘N - 1’
• If 'PERM'[j] is equal to 1, then insert ARR[i] to the set ‘LABELS’.
• Let the number of elements in the set ‘LABELS’ be ‘K’
• Set 'ANS' to the minimum value among 'ANS' and ‘K’.
4. Return the variable 'ANS'.