# C++ program to print all the Non-repeated Numbers in an Array

In this article, we are going to see how to print all the non-repeating numbers in an array with use of STL?
Submitted by Radib Kar, on December 19, 2018

Problem statement: Write a C++ program to print all the non-repeated numbers in an array in minimum time complexity.

Input Example:

```    Array length: 10
Array input: 2 5 3 2 4 5 3 6 7 3
Output:
Non-repeated numbers are: 7, 6, 4
```

Solution

Data structures used:

```    Unordered_map <int, int>
```
• Key in the map is array value
• Value of key is frequency

Algorithm:

1. Declare a map hash to store array elements as keys and to associate their frequencies with them.
2. ```    Unordered_map <int, int>hash;
```
3. For each array element
Insert it as key & increase frequencies. (0 ->1)
For same key it will only increase frequencies.
4. ```For i=0: n-1
hash[array [i]]++;
End For
```
5. Now to print the non-repeated character we need to print the keys (array elements) having value (frequency) exactly 1. (Non-repeating)
Set an iterator to hash.begin().
iterator->first is the key (array element) & iterator->second is the value( frequency of corresponding array value)
6. ```IF
Iterator->second > 1
Print iterator->first (the array element)
END IF
```

Time complexity: O(n)

Explanation with example:

For this array: 2 5 3 2 4 5 3 6 7 3

The code:

```for(int i=0;i<n;i++){//creating the map
hash[a[i]]++;//for same key increase frequency
}
```

Actually does the following

```    At i=0
array[i]=2
Insert 2 & increase frequency

Hash:
Key(element)	Value(frequency)
2	            1

At i=1
array[i]=5
Insert 5 & increase frequency

Hash:
Key(element)	Value(frequency)
2	            1
5	            1

At i=2
array[i]=3
Insert 3 & increase frequency

Hash:
Key(element)	Value(frequency)
2	            1
5	            1
3	            1

At i=3
array[i]=2
Insert 2 increase frequency
'2' is already there, thus frequency increase.

Hash:
Key(element)	Value(frequency)
2	            2
5	            1
3	            1

At i=4
array[i]=4
Insert 4 &increase frequency

Hash:
Key(element)	Value(frequency)
2	            2
5	            1
3	            1
4	            1

At i=5
array[i]=5
'5' is already there, thus frequency increase.

Hash:
Key(element)	Value(frequency)
2	            2
5	            2
3	            1
4	            1

At i=6
array[i]=3
'3' is already there, thus frequency increase.

Hash:
Key(element)	Value(frequency)
2	            2
5	            2
3	            2
4	            1

At i=7
array[i]=6
Insert 6, increase frequency.

Hash:
Key(element)	Value(frequency)
2	            2
5	            2
3	            2
4	            1
6	            1

At i=8
array[i]=7
Insert 7, increase frequency.

Hash:
Key(element)	Value(frequency)
2	            2
5	            2
3	            2
4	            1
6	            1
7	            1

At i=9
array[i]=3
'3' is already there, thus frequency increase.

Hash:
Key(element)	Value(frequency)
2	            2
5	            2
3	            3
4	            1
6	            1
7	            1
```

Thus, Elements with frequency 1 are: 7, 6, 4

## C++ implementation to print all the Non-Repeated Numbers with Frequency in an Array

```#include <bits/stdc++.h>

using namespace std;

void findNonRepeat(int* a, int n){
//Declare the map
unordered_map<int,int> hash;

for(int i=0;i<n;i++){//creating the map
hash[a[i]]++;//for same key increase frequency
}

cout<<"the nonrepeating numbers are: ";
//iterator->first == key(element value)
//iterator->second == value(frequency)

for(auto it=hash.begin();it!=hash.end();it++)
if(it->second==1)//frequency==1 means non-repeating element
printf("%d ",it->first);

printf("\n");

}

int main()
{
int n;
cout<<"enter array length\n";
cin>>n;
int* a=(int*)(malloc(sizeof(int)*n));

cout<<"input array elements...\n";

for(int i=0;i<n;i++)
scanf("%d",&a[i]);

//function to print repeating elements with their frequencies
findNonRepeat(a,n);

return 0;
}
```

Output

```enter array length
10
input array elements...
2 5 3 2 4 5 3 6 7 3
the nonrepeating numbers are: 7 6 4
```

Preparation

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