HackerRank ‘Max Min’ / ‘Angry Children’ Solution

Short Problem Definition:

Given a list of N integers, your task is to select K integers from the list such that itsĀ unfairness is minimized.

Link

Max Min

Complexity:

time complexity is O(N*log(N));

space complexity is O(N)

Execution:

The unfairness is the distance between K elements in a sorted array.

Solution:
#!/usr/bin/py
if __name__ == '__main__':
    n = input()
    k = input()
    candies = [input() for _ in range(0,n)]
    candies.sort()
    min_diff = 1000000000
    ## Write code here to compute the answer using (n, k, candies)

    for i in xrange(n - k + 1):
        min_diff = min(min_diff, candies[i+k-1] - candies[i])
    
    print min_diff

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <limits>
using namespace std;

int main() {
    /* The code required to enter n,k, candies is provided*/

    int N, K, unfairness = std::numeric_limits<int>::max();
    cin >> N >> K;
    int candies[N];
    for (int i=0; i<N; i++)
        cin >> candies[i];
    
    sort(candies, candies + N);
        
    for (int i=0; i < N - K + 1; i++){
        unfairness = min(unfairness, candies[i+K-1] - candies[i]);
    }
    
    cout << unfairness << "\n";
    return 0;
}

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