Max Min
Problem Statement :
You will be given a list of integers, arr , and a single integer k. You must create an array of length k from elements of arr such that its unfairness is minimized. Call that array arr' . Unfairness of an array is calculated as max( arr' ) - min( arr ' ) Where: - max denotes the largest integer in arr' . - min denotes the smallest integer in arr'. Note: Integers in may not be unique. Function Description Complete the maxMin function in the editor below. maxMin has the following parameter(s): int k: the number of elements to select int arr[n]:: an array of integers Returns int: the minimum possible unfairness Input Format The first line contains an integer n, the number of elements in array arr. The second line contains an integer k . Each of the next n lines contains an integer arr[ i ] where 0 <= i < n. Constraints 2 <= n <= 10^5 2 <= k <= n 0 < = arr[ i ] <= 10^9 Sample Input 0 7 3 10 100 300 200 1000 20 30 Sample Output 0 20
Solution :
Solution in C :
In C :
#include <stdio.h>
#include<stdlib.h>
int compare(int *a,int *b)
{
return *(int*)a-*(int*)b;
}
int main(void)
{
int n,k,i,j,min,a[100010];
scanf("%d",&n);
scanf("%d",&k);
for(i=0;i<n;i++)
scanf("%d",&a[i]);
qsort(a,n,sizeof(int),compare);
min=a[k-1]-a[0];
for(i=0;i<=n-k;i++)
{
if( (j=(a[i+k-1]-a[i]) ) <min)
min=j;
}
printf("%d",min);
return 0;
}
Solution in C++ :
In C ++ :
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int arr[100010];
int main() {
int n,k;
cin>>n>>k;
for(int i=0;i<n;i++)cin>>arr[i];
sort(arr,arr+n);
int ans=1e9;
for(int i=k-1;i<n;i++){
ans=min(arr[i]-arr[i-k+1],ans);
}
cout<<ans<<endl;
return 0;
}
Solution in Java :
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt(), k = in.nextInt();
int[] x = new int[n];
for(int i = 0; i < n; i++) x[i] = in.nextInt();
Arrays.sort(x);
System.out.println(f(n, k, x));
}
private static int f(int n, int k, int[] x){
int min = 100000000;
for(int i = 0; i + k-1 < x.length; i++){
if(x[i+k-1] - x[i] < min) min = x[i+k-1]-x[i];
}
return min;
}
}
Solution in Python :
In Python3 :
def unfairness(candies, i, j):
res = candies[j-1] - candies[i]
return res
n = int(input())
kids = int(input())
candies = []
for i in range(n):
candies.append(int(input()))
candies = sorted(candies)
min_uf = unfairness(candies, 0, kids)
for i in range(1, len(candies)-kids):
this_uf = unfairness(candies, i, i+kids)
min_uf = min(min_uf, this_uf)
print(min_uf)
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