# Minimum number of Bottles visible when a bottle can be enclosed inside another Bottle

• Difficulty Level : Medium
• Last Updated : 28 May, 2022

Given N bottles. The ith bottle has A[i] radius. Once a bottle is enclosed inside another bottle, it ceases to be visible. The task is to minimize the number of visible bottles. You can put the ith bottle into a jth bottle if the following condition is fulfilled.

1. ith bottle itself is not enclosed in another bottle.
2. jth bottle does not enclose any other bottle.
3. Radius of bottle i is smaller than bottle j ( i.e. A[i] < A[j] ).

Examples:

```Input :
8
1 1 2 3 4 5 5 4
Output :
2
Explanation:
1 -> 2 [1, 2, 3, 4, 5, 5, 4]
2 -> 3 [1, 3, 4, 5, 5, 4]
3 -> 4 [1, 4, 5, 5, 4]
4 -> 5 [1, 5, 5, 4]
1 -> 4 [5, 5, 4]
4 -> 5 [5, 5]

Finally, there are 2 bottles
left which are visible.
Hence the answer is 2. ```

Approach: If you carefully observe, you will find that the number of minimum visible bottles will be equal to the maximum number of repeated bottles. Here intuition is, as these repeated bottles cannot be fit in single bigger bottle hence we require at least as many bigger bottles as the number of repeated bottles.
Below is the implementation of the above approach:

## C++

 `#include ``using` `namespace` `std;` `void` `min_visible_bottles(``int``* arr, ``int` `n)``{``    ``map<``int``, ``int``> m;``    ``int` `ans = 0;``    ``for` `(``int` `i = 0; i < n; i++) {``        ``m[arr[i]]++;``        ``ans = max(ans, m[arr[i]]);``    ``}` `    ``cout << ``"Minimum number of "``         ``<< ``"Visible Bottles are: "``         ``<< ans << endl;``}` `// Driver code``int` `main()``{``    ``int` `n = 8;``    ``int` `arr[] = { 1, 1, 2, 3, 4, 5, 5, 4 };` `    ``// Find the solution``    ``min_visible_bottles(arr, n);` `    ``return` `0;``}`

## Java

 `// Java code for the above approach``import` `java.util.*;` `class` `GFG``{``    ``static` `void` `min_visible_bottles(``int``[] arr, ``int` `n)``    ``{``        ``HashMap mp = ``new` `HashMap();``        ``int` `ans = ``0``;``        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{``            ``if` `(mp.containsKey(arr[i]))``            ``{``                ``mp.put(arr[i], mp.get(arr[i]) + ``1``);``            ``}``            ``else``            ``{``                ``mp.put(arr[i], ``1``);``            ``}``            ``ans = Math.max(ans, mp.get(arr[i]));``        ``}` `        ``System.out.print(``"Minimum number of "` `+``                      ``"Visible Bottles are: "` `+``                                   ``ans + ``"\n"``);``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `n = ``8``;``        ``int` `arr[] = { ``1``, ``1``, ``2``, ``3``, ``4``, ``5``, ``5``, ``4` `};` `        ``// Find the solution``        ``min_visible_bottles(arr, n);``    ``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 code for the above approach``def` `min_visible_bottles(arr, n):` `    ``m ``=` `dict``()``    ``ans ``=` `0``    ``for` `i ``in` `range``(n):``        ``m[arr[i]] ``=` `m.get(arr[i], ``0``) ``+` `1``        ``ans ``=` `max``(ans, m[arr[i]])` `    ``print``(``"Minimum number of"``,``          ``"Visible Bottles are: "``, ans)` `# Driver code``n ``=` `8``arr ``=` `[``1``, ``1``, ``2``, ``3``, ``4``, ``5``, ``5``, ``4``]` `# Find the solution``min_visible_bottles(arr, n)` `# This code is contributed``# by Mohit Kumar`

## C#

 `// C# code for the above approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{``    ``static` `void` `min_visible_bottles(``int``[] arr, ``int` `n)``    ``{``        ``Dictionary<``int``,``                   ``int``> mp = ``new` `Dictionary<``int``,``                                            ``int``>();``        ``int` `ans = 0;``        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ``if` `(mp.ContainsKey(arr[i]))``            ``{``                ``mp[arr[i]] = mp[arr[i]] + 1;``            ``}``            ``else``            ``{``                ``mp.Add(arr[i], 1);``            ``}``            ``ans = Math.Max(ans, mp[arr[i]]);``        ``}` `        ``Console.Write(``"Minimum number of "` `+``                   ``"Visible Bottles are: "` `+``                                ``ans + ``"\n"``);``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int` `n = 8;``        ``int` `[]arr = { 1, 1, 2, 3, 4, 5, 5, 4 };` `        ``// Find the solution``        ``min_visible_bottles(arr, n);``    ``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output:

`Minimum number of Visible Bottles are: 2`

Time Complexity: O(nlogn)

Auxiliary Space: O(n)

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