Angry Children 2


Problem Statement :


Bill Gates is on one of his philanthropic journeys to a village in Utopia. He has brought a box of packets of candies and would like to distribute one packet to each of the children. Each of the packets contains a number of candies. He wants to minimize the cumulative difference in the number of candies in the packets he hands out. This is called the unfairness sum. Determine the minimum unfairness sum achievable.

For example, he brings n = 7 packets where the number of candies is packets = [3,3,4,5,7,9,10]. There are k=3 children. The minimum difference between all packets can be had with 3,3,4 from indices 0,1 and 2. We must get the difference in the following pairs: {(0,1),(0,2),(1,2)}. We calculate the unfairness sum as:

packets candies				
0	3		indices		difference	result
1	3		(0,1),(0,2)	|3-3| + |3-4|	1 
2	4		(1,2)		|3-4|		1

Total = 2
Function Description

Complete the angryChildren function in the editor below. It should return an integer that represents the minimum unfairness sum achievable.

angryChildren has the following parameter(s):

k: an integer that represents the number of children
packets: an array of integers that represent the number of candies in each packet

Input Format

The first line contains an integer n.
The second line contains an integer k.
Each of the next n lines contains an integer packets[i].

Constraints
2 <= n <=10^5
2 <= k <= n
0 <= packets[i]  <= 10^9

Output Format

A single integer representing the minimum achievable unfairness sum.


Solution :



title-img 


                            Solution in C :

In c++ :





#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;

const int N = 100005;
long long a[N], sum[N];
long long s(int a, int b) { return sum[b] - (a ? sum[a - 1] : 0LL); }

int main()
{
    //freopen("test.in", "r", stdin);
    int n, k;
    scanf("%i %i", &n, &k);
    for(int i = 0; i < n; i++)
        scanf("%lld", &a[i]);
    sort(a, a + n);

    sum[0] = a[0];
    for(int i = 1; i < n; i++)
        sum[i] = sum[i - 1] + a[i];

    long long curr = 0;
    for(int i = 0; i < k; i++)
        curr += (long long)(2 * i + 1) * a[i];

    long long res = curr;
    for(int i = 0; i + k - 1 < n; i++)
    {
        res = min(res, curr - k * s(i, i + k - 1));
        curr -= a[i] + 2LL * s(i + 1, i + k - 1);
        curr += (long long)(2 * k - 1) * a[i + k];
    }
    printf("%lld\n", res);

    return 0;
}








In Java :





import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    static long curTime;
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt(), k = in.nextInt();
        long[] x = new long[n];
        for(int i = 0; i < n; i++) x[i] = in.nextInt();
        Arrays.sort(x);
        curTime = System.currentTimeMillis();
        System.out.println(f(n, k, x));
    }
    
    private static long f(int n, int k, long[] x){
        long t = unFair(k, x), min = t, s = 0;
        //long s = sum(1, k-1, x);
        for(int i = 1; i < k; i++) s += x[i];
        
        for(int i = 1; i + k-1 < x.length; i++){
            //System.out.println(t);
            t += (k-1)*(x[i-1] + x[i+k-1]);
            t -= 2*s;
            if(t < min) min = t;
            s += x[i+k-1] -x[i];
        }
        return min;
    }
    
    private static long unFair(int k, long[] x){
        long s = 0;
        for(int i = 0; i < k; i++){
            s += x[i] *(2*i-k+1);
        }
        return s;
    }
    
    private static long sum(int i, int j, long[] x){
        long s = 0;
        for(int k = i; k <= j; k++) s += x[k];
        return s;
    }
    
    /*private static long aSum(int i, int j, long[] x){
        if(i == j) return x[i];
        return x[j] - aSum(i, j-1, x);
    }*/
}








 In C :





#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

long long int min(long long int a,long long int b){
    if(a<b)
        return a;
    return b;
}

int cmpr(const void * a,const void *b){
    return (*(long long int *)a)-(*(long long int *)b);
}

int main(){
    long long int n,k,i=0,j;
    scanf("%lld %lld",&n,&k);
    long long int a[n];
    while (i<n) {
        scanf("%lld",&a[i]);
        i++;
    }
    qsort(a, n, sizeof(long long int), cmpr);
    i=0;
    j=1-k;
    long long int sum=0,sum1=0;
    while(i<k){
        sum+=a[i]*j;
        sum1+=a[i];
        j+=2;
        i++;
    }
    i=0;
    j=k;
    long long int ans=sum;
    while (j<n) {
        sum1-=a[i];
        sum=sum-(2*sum1)-((1-k)*a[i])+(k-1)*a[j];
        sum1+=a[j];
        ans=min(ans,sum);
        i++;
        j++;
    }
    printf("%lld\n",ans);
    return 0;
}








In Python3 :





n = int(input())
k = int(input())
a = []
for i in range(n):
    a.append(int(input()))
a.sort()
ps = [0]
for i in range(n):
    ps.append(ps[-1] + a[i])
cur = 0
for i in range(k):
    cur += i * a[i] - ps[i]
ans = cur
for i in range(1, n - k + 1):
    cur -= ps[i + k - 1] - ps[i - 1] - k * a[i - 1]
    cur += k * a[i + k - 1] - ps[i + k] + ps[i]
    ans = min(ans, cur)
print(ans)








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