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# Discover the Array factor left after alternate elimination of minimal and most

Given an array Arr of size N, it’s lowered by 1 factor at every step. Most and Minimal parts shall be eliminated in alternating order from the remaining array till a single factor is remaining within the array. The duty is to seek out the remaining factor within the given array.

Examples:

Enter: arr[] = {1, 5, 4, 2}
Output: 2
Clarification
Take away Max factor i.e., 5 arr[] = {1, 4, 2}
Take away Min factor i.e., 1 arr[] = {4, 2}
Take away Max factor i.e., 4 arr[] = {2}

Enter: arr[] = {5, 10}
Output: 5

Strategy:

Comply with the under concept to unravel the issue:

The concept is to type the array and return the center factor as the entire proper and left parts shall be eliminated within the course of.

Comply with the under steps to unravel this downside:

• If N =1, return arr
• Kind the array arr[]
• Return the center factor of the array i.e., arr[(N-1)/2]

Beneath is the implementation of the above strategy:

## C++

 ` `  `#embody ` `utilizing` `namespace` `std;` ` `  `int` `lastRemaining(``int` `arr[], ``int` `N)` `{` `    ` `    ``if` `(N == 1)` `        ``return` `arr;` ` `  `    ` `    ``type(arr, arr + N);` ` `  `    ` `    ``return` `arr[(N - 1) / 2];` `}` ` `  `int` `primary()` `{` `    ``int` `arr[] = { 1, 5, 4, 2 };` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` ` `  `    ` `    ``cout << lastRemaining(arr, N) << endl;` `    ``return` `0;` `}`

Time Complexity: O(N * log(N)), for sorting the given array of dimension N.
Auxiliary House: O(1), as fixed additional house is required.

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