Priyanka and Toys
Problem Statement :
Priyanka works for an international toy company that ships by container. Her task is to the determine the lowest cost way to combine her orders for shipping. She has a list of item weights. The shipping company has a requirement that all items loaded in a container must weigh less than or equal to 4 units plus the weight of the minimum weight item. All items meeting that requirement will be shipped in one container. What is the smallest number of containers that can be contracted to ship the items based on the given list of weights? Function Description Complete the toys function in the editor below. It should return the minimum number of containers required to ship. toys has the following parameter(s): w: an array of integers that represent the weights of each order to ship Input Format The first line contains an integer , the number of orders to ship. The next line contains space-separated integers, , representing the orders in a weight array. Output Format Return the integer value of the number of containers Priyanka must contract to ship all of the toys.
Solution :
Solution in C :
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int cmpfunc (const void * a, const void * b)
{
return ( *(int*)a - *(int*)b );
}
int main() {
long int n,i,ct=0,f;
scanf("%ld",&n);
long int*a=(long int*)calloc(sizeof(long int),n+10);
for(i=0;i<n;i++)scanf("%ld",&a[i]);
qsort(a, n, sizeof(long int), cmpfunc);
ct=1;f=a[0]+4;
for(i=1;i<n;i++){
if(a[i]<=f)continue;
else{
f=a[i]+4;
ct++;
}
}
printf("%ld",ct);
return 0;
}
Solution in C++ :
In C++ :
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cctype>
#include <cstring>
#include <string>
#define ll long long
#define PB push_back
#define Maxn 100010
using namespace std;
int N , a[Maxn];
int main(){
scanf("%d",&N);
for (int i=1;i<=N;i++)
scanf("%d",&a[i]);
sort(a+1,a+N+1);
int pre = -100 ,ans = 0;
for (int i=1;i<=N;i++){
if (a[i] <= pre + 4) continue;
ans ++;
pre = a[i];
}
cout << ans << endl;
}
Solution in Java :
In Java :
import java.io.*;
import java.util.*;
public class Solution {
private static Reader in;
private static PrintWriter out;
public static void main(String[] args) throws IOException {
in = new Reader();
out = new PrintWriter(System.out, true);
int N = in.nextInt();
int[] arr = new int[N];
for (int i = 0; i < N; i++)arr[i] = in.nextInt();
Arrays.sort(arr);
int last = -10000;
int count = 0;
for (int i = 0; i < N; i++) {
if (last+4 < arr[i]) {
last = arr[i];
count++;
}
}
out.println(count);
out.close();
System.exit(0);
}
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader() {
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException {
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException {
byte[] buf = new byte[1024];
int cnt = 0;
byte c = read();
while (c <= ' ')
c = read();
do {
buf[cnt++] = c;
} while ((c = read()) != '\n');
return new String(buf, 0, cnt);
}
public String next() throws IOException {
byte[] buf = new byte[1024];
int cnt = 0;
byte c = read();
while (c <= ' ')
c = read();
do {
buf[cnt++] = c;
} while ((c = read()) > ' ');
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException {
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException {
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException {
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.')
while ((c = read()) >= '0' && c <= '9')
ret += (c - '0') / (div *= 10);
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException {
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException {
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException {
if (din == null)
return;
din.close();
}
}
}
Solution in Python :
In Python3 :
#!/usr/bin/env python
import sys
if __name__ == '__main__':
N = int(sys.stdin.readline())
W = sorted(map(int, sys.stdin.readline().split()))
cost = 0
last_chosen = -5
for w in W:
if w > last_chosen + 4:
cost += 1
last_chosen = w
print(cost)
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