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Arrays Interview Questions

Unique Color Shirt

Prepbytes

CONCEPTS USED:

Hashing

DIFFICULTY LEVEL:

Easy

PROBLEM STATEMENT(SIMPLIFIED):

Given an array A with N integers, find the count of unique integers in the array.

For Example:

Input : arr = [3 2 4 1 2 3]

Output : 2

Explanation: only 1 and 4 are unique in the array rest 2, 3 are occurring 2 times.

SOLVING APPROACH:

  1. Initialize a temporary Hash Array for storing the frequency of all the elements in the array.
  2. Traverse the array and keep incrementing the frequency of elements in the Hash Array.
  3. Finally, traverse and find the count of all such unique elements, hence print it.

SOLUTIONS:

#include 

int main()
{
  int n; scanf("%d", &n);
  int arr[n];

  /* hash array for storing frequency of same coloured shirts */
  int hash[1001] = {0};

  //count of unique shirts
  int count = 0;

  /* input the shirt and increment its frequency */
  for(int i=0; i





						 
#include 
using namespace std;

int main()
{
  int n; cin>>n;
  int arr[n];

  /* hash array for storing frequency of same coloured shirts */
  int hash[1001] = {0};

  //count of unique shirts
  int count = 0;

  /* input the shirt and increment its frequency */
  for(int i=0; i>arr[i];
    hash[arr[i]]++;
  }

  //count all unique shirts
  for(int i=0; i

						 
import java.util.*;
import java.io.*;

public class Main {
  public static void main(String args[]) throws IOException {

    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int arr[] = new int[n];

    /* hash array for storing frequency of same coloured shirts */
    int hash[] = new int[1001];

    //count of unique shirts
    int count = 0;

    /* input the shirt and increment its frequency */
    for(int i=0; i
						 
n=int(input())

# hash array for storing frequency of same coloured shirts
hash_=[0 for i in range(1001)]

# count of unique shirts
count=0

# input the shirt and increment its frequency
arr=list(map(int,input().split()))

for i in range(n):
	hash_[arr[i]]+=1

# count all unique shirts
for i in range(n):
	if hash_[arr[i]]==1:
		count+=1
print(count)

Space Complexity : O(N), due to additional Hash Array.

[forminator_quiz id="521"]

This article tried to discuss the concept of Hashing. Hope this blog helps you understand and solve the problem. To practice more problems on Hashing you can check out .

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