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# Unique Color Shirt

Last Updated on March 23, 2022 by Ria Pathak

Hashing

Easy

### PROBLEM STATEMENT`(`SIMPLIFIED`)`:

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

See original problem statement here

#### 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 MYCODE | Competitive Programming.