Given an sorted array of positive integers, count number of occurrences for each element in the array. Assume all elements in the array are less than some constant M.
Do this without traversing the complete array. i.e. expected time complexity is less than O(n).
Input: arr = [1, 1, 1, 2, 3, 3, 5, 5, 8, 8, 8, 9, 9, 10] Output: Element 1 occurs 3 times Element 2 occurs 1 times Element 3 occurs 2 times Element 5 occurs 2 times Element 8 occurs 3 times Element 9 occurs 2 times Element 10 occurs 1 times
Method 1 (Linear Search)
The idea is traverse the input array and for each distinct element of array, store its frequency in a map and finally print the map. This approach takes O(n) time.
Method 2 (Use Binary Search)
This problem can be solved in less than O(n) using a modified binary search. The idea is to recursively divide the array into two equal subarrays if its end elements are different. If both its end elements are same, that means that all elements in the subarray is also same as the array is already sorted. We then simply increment the count of the element by size of the subarray.
The time complexity of above approach is O(m log n), where m is number of distinct elements in the array of size n. Since m <= M (a constant), the time complexity of this solution is O(log n).
Below is the implementation of above idea –
Element 1 occurs 3 times Element 2 occurs 1 times Element 3 occurs 2 times Element 5 occurs 2 times Element 8 occurs 3 times Element 9 occurs 2 times Element 10 occurs 1 times
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