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)
View More Similar Problems
Array and simple queries
Given two numbers N and M. N indicates the number of elements in the array A[](1-indexed) and M indicates number of queries. You need to perform two types of queries on the array A[] . You are given queries. Queries can be of two types, type 1 and type 2. Type 1 queries are represented as 1 i j : Modify the given array by removing elements from i to j and adding them to the front. Ty
View Solution →Median Updates
The median M of numbers is defined as the middle number after sorting them in order if M is odd. Or it is the average of the middle two numbers if M is even. You start with an empty number list. Then, you can add numbers to the list, or remove existing numbers from it. After each add or remove operation, output the median. Input: The first line is an integer, N , that indicates the number o
View Solution →Maximum Element
You have an empty sequence, and you will be given N queries. Each query is one of these three types: 1 x -Push the element x into the stack. 2 -Delete the element present at the top of the stack. 3 -Print the maximum element in the stack. Input Format The first line of input contains an integer, N . The next N lines each contain an above mentioned query. (It is guaranteed that each
View Solution →Balanced Brackets
A bracket is considered to be any one of the following characters: (, ), {, }, [, or ]. Two brackets are considered to be a matched pair if the an opening bracket (i.e., (, [, or {) occurs to the left of a closing bracket (i.e., ), ], or }) of the exact same type. There are three types of matched pairs of brackets: [], {}, and (). A matching pair of brackets is not balanced if the set of bra
View Solution →Equal Stacks
ou have three stacks of cylinders where each cylinder has the same diameter, but they may vary in height. You can change the height of a stack by removing and discarding its topmost cylinder any number of times. Find the maximum possible height of the stacks such that all of the stacks are exactly the same height. This means you must remove zero or more cylinders from the top of zero or more of
View Solution →Game of Two Stacks
Alexa has two stacks of non-negative integers, stack A = [a0, a1, . . . , an-1 ] and stack B = [b0, b1, . . . , b m-1] where index 0 denotes the top of the stack. Alexa challenges Nick to play the following game: In each move, Nick can remove one integer from the top of either stack A or stack B. Nick keeps a running sum of the integers he removes from the two stacks. Nick is disqualified f
View Solution →