Counting Sort 1


Problem Statement :


Comparison Sorting
Quicksort usually has a running time of n x log( n ) , but is there an algorithm that can sort even faster? In general, this is not possible. Most sorting algorithms are comparison sorts, i.e. they sort a list just by comparing the elements to one another. A comparison sort algorithm cannot beat n x log( n ) (worst-case) running time, since n x log( n )  represents the minimum number of comparisons needed to know where to place each element. For more details, you can see these notes (PDF).

Alternative Sorting
Another sorting method, the counting sort, does not require comparison. Instead, you create an integer array whose index range covers the entire range of values in your array to sort. Each time a value occurs in the original array, you increment the counter at that index. At the end, run through your counting array, printing the value of each non-zero valued index that number of times.

Note
For this exercise, always return a frequency array with 100 elements. The example above shows only the first 4 elements, the remainder being zeros.

Challenge
Given a list of integers, count and return the number of times each value appears as an array of integers.

Function Description

Complete the countingSort function in the editor below.

countingSort has the following parameter(s):

arr[n]: an array of integers
Returns

int[100]: a frequency array
Input Format

The first line contains an integer n, the number of items in arr.
Each of the next n lines contains an integer arr[i] where 0 <= i <  n.

Constraints

100  <=  n  <=  10^6
0  <=  arr[ i [  <  100


Solution :



title-img 


                            Solution in C :

In  C++  :






#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

vector<int>List;
auto main()->int
{
	int size;
	cin >> size;
	//Initialize
	List.resize(100,0);
	for (int i = 0; i != size; i++)
	{
		int n;
		cin >> n;
		List[n] = List[n] + 1;
	}
	for (int r = 0; r != List.size(); r++)
	{
		cout << List[r]<<" ";
	}
	return 0;
}









In   Java  :





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

public class Solution {
	
	static void printArray(int[] ar) {
		for (int n : ar) {
			System.out.print(n + " ");
		}
		System.out.println("");
	}
	
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int[] ar = new int[100];
		for (int i = 0; i < n; i++) {
			ar[in.nextInt()]++;
		}
		
		printArray(ar);
	}

}










In   C  :






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

int main() {


    int n,i;
    scanf("%d",&n);
    int b[100],a;
    for(i=0;i<100;i++)
    {   
         b[i]=0;
    }
    for(i=0;i<n;i++)
    {    scanf("%d",&a);
         b[a]++;
    }
    for(i=0;i<100;i++)
    {   
        printf("%d ", b[i]);
    }
   
    return 0;
}









In   Python3  :






N = int(input())

arr = list(map(int, input().split()))

freq = [0]*100

for i in arr:
    freq[i] = freq[i]+1


print (' '.join(map(str, freq)))








View More Similar Problems

Fibonacci Numbers Tree

Shashank loves trees and math. He has a rooted tree, T , consisting of N nodes uniquely labeled with integers in the inclusive range [1 , N ]. The node labeled as 1 is the root node of tree , and each node in is associated with some positive integer value (all values are initially ). Let's define Fk as the Kth Fibonacci number. Shashank wants to perform 22 types of operations over his tree, T

View Solution →

Pair Sums

Given an array, we define its value to be the value obtained by following these instructions: Write down all pairs of numbers from this array. Compute the product of each pair. Find the sum of all the products. For example, for a given array, for a given array [7,2 ,-1 ,2 ] Note that ( 7 , 2 ) is listed twice, one for each occurrence of 2. Given an array of integers, find the largest v

View Solution →

Lazy White Falcon

White Falcon just solved the data structure problem below using heavy-light decomposition. Can you help her find a new solution that doesn't require implementing any fancy techniques? There are 2 types of query operations that can be performed on a tree: 1 u x: Assign x as the value of node u. 2 u v: Print the sum of the node values in the unique path from node u to node v. Given a tree wi

View Solution →

Ticket to Ride

Simon received the board game Ticket to Ride as a birthday present. After playing it with his friends, he decides to come up with a strategy for the game. There are n cities on the map and n - 1 road plans. Each road plan consists of the following: Two cities which can be directly connected by a road. The length of the proposed road. The entire road plan is designed in such a way that if o

View Solution →

Heavy Light White Falcon

Our lazy white falcon finally decided to learn heavy-light decomposition. Her teacher gave an assignment for her to practice this new technique. Please help her by solving this problem. You are given a tree with N nodes and each node's value is initially 0. The problem asks you to operate the following two types of queries: "1 u x" assign x to the value of the node . "2 u v" print the maxim

View Solution →

Number Game on a Tree

Andy and Lily love playing games with numbers and trees. Today they have a tree consisting of n nodes and n -1 edges. Each edge i has an integer weight, wi. Before the game starts, Andy chooses an unordered pair of distinct nodes, ( u , v ), and uses all the edge weights present on the unique path from node u to node v to construct a list of numbers. For example, in the diagram below, Andy

View Solution →