HackerRank Cut the sticks Solution

ByBrokenprogrammers

Dec 12, 2022

Hello Programmers, In this post, you will know how to solve the HackerRank Cut the sticks Solution. This problem is a part of the HackerRank Algorithms Series.

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Hackerrank by yourself. If you find any difficulty after trying several times, then you can look for solutions.

HackerRank Cut the sticks Solution

You are given a number of sticks of varying lengths. You will iteratively cut the sticks into smaller sticks, discarding the shortest pieces until there are none left. At each iteration you will determine the length of the shortest stick remaining, cut that length from each of the longer sticks and then discard all the pieces of that shortest length. When all the remaining sticks are the same length, they cannot be shortened so discard them.

Given the lengths of n sticks, print the number of sticks that are left before each iteration until there are none left.

Example
arr = [1, 2, 3]

The shortest stick length is 1, so cut that length from the longer two and discard the pieces of length 1. Now the lengths are arr = [1, 2]. Again, the shortest stick is of length 1, so cut that amount from the longer stick and discard those pieces. There is only one stick left, arr = [1], so discard that stick. The number of sticks at each iteration are answer = [3, 2, 1].

Function Description

Complete the cutTheSticks function in the editor below. It should return an array of integers representing the number of sticks before each cut operation is performed.

cutTheSticks has the following parameter(s):

• int arr[n]: the lengths of each stick

Returns

• int[]: the number of sticks after each iteration

Input Format

The first line contains a single integer n, the size of arr.
The next line contains n space-separated integers, each an arr[i], where each value represents the length of the ith stick.

Constraints

• 1 <= n <= 1000
• 1 <= arr[i] <= 1000

Sample Input 0

STDIN Function
—– ——–
6 arr[] size n = 6
5 4 4 2 2 8 arr = [5, 4, 4, 2, 2, 8]

Sample Output 0

6
4
2
1

Explanation 0

```sticks-length        length-of-cut   sticks-cut
5 4 4 2 2 8             2               6
3 2 2 _ _ 6             2               4
1 _ _ _ _ 4             1               2
_ _ _ _ _ 3             3               1
_ _ _ _ _ _           DONE            DONE
```

Sample Input 1

8
1 2 3 4 3 3 2 1

Sample Output 1

8
6
4
1

Explanation 1

```sticks-length         length-of-cut   sticks-cut
1 2 3 4 3 3 2 1         1               8
_ 1 2 3 2 2 1 _         1               6
_ _ 1 2 1 1 _ _         1               4
_ _ _ 1 _ _ _ _         1               1
_ _ _ _ _ _ _ _       DONE            DONE```

HackerRank Cut the sticks Solution

Cut the sticks Solution in C

```#include<stdio.h>
int main()
{
int n;
scanf("%d",&n);
int a[1002]={0};
int i;
for(i=0;i<n;i++)
{
int c;
scanf("%d",&c);
a[c]++;
}
int t=n;
for(i=0;i<1001;i++)
{
if(a[i]>0)
{
printf("%d\n",t);
t=t-a[i];
}
}
return  0;
}```

Cut the sticks Solution in Cpp

```#include <iostream>
#include <stdio.h>
#include <cmath>
#include <string>
#include <vector>
#include <stack>
#include <cassert>
#include <algorithm>
#define Pi 3.14159
typedef long long int ll;
using namespace std;
int main ()
{
int n;
cin >> n;
int a[n];
int b[1001]={0};
for (int i = 0 ; i<n;i++){cin >>a[i];b[a[i]] ++;}
cout<<n<<endl;
for (int i = 1; i <= 1000; i++){n-=b[i];if(b[i] && n)cout<<n<<endl;}
}
```

Cut the sticks Solution 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) {
int n = in.nextInt();
int[] counts = new int[1001];
for (int i = 0; i < n; i++) counts[in.nextInt()]++;
for (int i = 0; i <= 1000; i++) {
if (counts[i] > 0) {
System.out.println(n);
n -= counts[i];
}
}
}
}```

Cut the sticks Solution in Python

```#!/usr/bin/env python
import collections, sys
if __name__ == '__main__':

c = collections.Counter(a)
count = [c[k] for k in sorted(c)]

for i in range(len(count)):
print(sum(count[i:]))```

Cut the sticks Solution using JavaScript

```'use strict';
function processData(input) {
var parse_fun = function (s) { return parseInt(s, 10); };
var lines = input.split('\n');
var N = parse_fun(lines.shift());
var data = lines[0].split(' ').splice(0, N).map(parse_fun);
data.sort(function (n1, n2) { return n1 - n2; });
var res = [];
while (data.length > 0) {
res.push(data.length);
var min = data[0];
data = data.map(function (n) { return n - min; });
data = data.filter(function (n) {return n > 0; });
}
console.log(res.join('\n'));
}
process.stdin.resume();
process.stdin.setEncoding("ascii");
var _input = "";
process.stdin.on("data", function (input) { _input += input; });
process.stdin.on("end", function () { processData(_input); });```

Cut the sticks Solution in Scala

```object Solution {
def main(args: Array[String]) {
if (n < 1 || n > 1000) println("Invalid value for N")
else {
if (xs.length != n) println("Invalid number of items")
else {
val ls = cutTheSticks(xs, Nil)
ls.foreach(println)
}
}
}

def cutTheSticks(xs: List[Int], ls: List[Int]): List[Int] = xs match {
case Nil => ls.reverse
case h :: t =>
val cutLen = xs.min
val xs1 = xs.map(_ - cutLen)
cutTheSticks(xs1.filter(_ > 0), xs.length :: ls)
}
}```

Cut the sticks Solution in Pascal

```(* Enter your code here. Read input from STDIN. Print output to STDOUT *)
program cut;
var
test : integer;
length : array[1..1000] of integer;
short : integer;
count : integer;
cutted : integer;
i,j : integer;
begin
cutted := test;
for i := 1 to test do
for i := 1 to 1000 do
begin
count := 0;
for j := 1 to test do
begin
if length[j]=i then
begin
count := count+1;
end;
end;
if count <> 0 then
begin
writeln(cutted);
cutted := cutted - count;
end;
end;
end.
```

Disclaimer: This problem (Cut the sticks) is generated by HackerRank but the Solution is Provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.