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] ).
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:
Minimum number of Visible Bottles are: 2
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Improved By : mohit kumar 29