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.
- ith bottle itself is not enclosed in another bottle.
- jth bottle does not enclose any other bottle.
- 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 <bits/stdc++.h>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 codeint 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 approachimport java.util.*;class GFG { static void min_visible_bottles(int[] arr, int n) { HashMap<Integer, Integer> mp = new HashMap<Integer, Integer>(); 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 approachdef 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 coden = 8arr = [1, 1, 2, 3, 4, 5, 5, 4]# Find the solutionmin_visible_bottles(arr, n)# This code is contributed # by Mohit Kumar |
C#
// C# code for the above approachusing 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
<script>function min_visible_bottles(arr, n) { let m = new Map(); let ans = 0; for (let i = 0; i < n; i++) { if(m.has(arr[i])){ m.set(arr[i], m.get(arr[i]) + 1) }else{ m.set(arr[i], 1) } ans = Math.max(ans, m.get(arr[i])); } document.write("Minimum number of " + "Visible Bottles are: " + ans + "<br>");}// Driver code let n = 8; let arr = [1, 1, 2, 3, 4, 5, 5, 4]; // Find the solution min_visible_bottles(arr, n);// This code is contributed by _saurabh_jaiswal</script> |
Output:
Minimum number of Visible Bottles are: 2
Time Complexity: O(nlogn)
Auxiliary Space: O(n)
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