Given an array Arr of size N, it’s diminished by 1 aspect at every step. Most and Minimal parts can be eliminated in alternating order from the remaining array till a single aspect is remaining within the array. The duty is to search out the remaining aspect within the given array.
Examples:
Enter: arr[] = {1, 5, 4, 2}
Output: 2
Clarification:
Take away Max aspect i.e., 5 arr[] = {1, 4, 2}
Take away Min aspect i.e., 1 arr[] = {4, 2}
Take away Max aspect i.e., 4 arr[] = {2}Enter: arr[] = {5, 10}
Output: 5
Strategy:
Observe the under concept to resolve the issue:
The thought is to type the array and return the center aspect as the entire proper and left parts can be eliminated within the course of.
Observe the under steps to resolve this downside:
- If N =1, return arr[0]
- Kind the array arr[]
- Return the center aspect of the array i.e., arr[(N-1)/2]
Under is the implementation of the above method:
C++
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Time Complexity: O(N * log(N)), for sorting the given array of dimension N.
Auxiliary House: O(1), as fixed further house is required.
