Duplicate In Array
You are given an array ‘ARR’ of size ‘N’ containing each number between 1 and ‘N’ - 1 at least once. There is a single integer value that is present in the array twice. Your task is to find the duplicate integer value present in the array.

For example:

Consider ARR = [1, 2, 3, 4, 4], the duplicate integer value present in the array is 4. Hence, the answer is 4 in this case.

Note :

A duplicate number is always present in the given array.

Input Format:

The first line of the input contains an integer, 'T,’ denoting the number of test cases. The first line of each test case contains a single integer, 'N', denoting the number of elements in the array. The second line of each test case contains 'N' space-separated integers denoting the elements of the array 'ARR'.

Output Format:

For each test case, print a single integer - the duplicate element in the array. Print the output of each test case in a separate line.

Constraints:

1

Question

Duplicate In Array

You are given an array ‘ARR’ of size ‘N’ containing each number between 1 and ‘N’ - 1 at least once. There is a single integer value that is present in the array twice. Your task is to find the duplicate integer value present in the array.

For example:

Consider ARR = [1, 2, 3, 4, 4], the duplicate integer value present in the array is 4. Hence, the answer is 4 in this case.

Note :

A duplicate number is always present in the given array.

Input Format:

The first line of the input contains an integer, 'T,’ denoting the number of test cases.
The first line of each test case contains a single integer, 'N', denoting the number of elements in the array.
The second line of each test case contains 'N' space-separated integers denoting the elements of the array 'ARR'.

Output Format:

For each test case, print a single integer - the duplicate element in the array.

Print the output of each test case in a separate line.

Constraints:

1 <=  T  <= 10
2 <=  N <= 10 ^ 5
1 <=  ARR[i]  <= N - 1

Where 'T' denotes the number of test cases, 'N' denotes the number of elements in the array, and 'ARR[i]' denotes the 'i-th' element of the array 'ARR'.

Time limit: 1 sec

Accepted Answer

Tip 1 : Use the filter() method in Javascript
Tip 2 : Using a Set() Method in Javascript
Tip 3 : Using the reduce() Method in Javascript

Accepted Answer

Brute Force

A simple method is to traverse through the array ARR to find the frequency of each number in the given array, and we will check if the frequency of the number is more than 1.

Therefore, our approach will be to iterate currentNumber from 1 to N - 1. In each iteration, we will traverse through the array ARR to find the frequency of currentNumber in the array. We will check if the frequency is more than 1, then there is a duplicate of the number currentNumber in the array ARR. In the end, we will return the duplicate integer value present in the array.

Algorithm:

We will initialize duplicate as 0. The variable duplicate stores the duplicate element in the array.
Iterate currentNumber from 1 to N - 1.
- We will set count as 0. The variable count stores the count of currentNumber in the array ARR.
- Iterate index from 0 to N - 1.
  - We will check if ARR[index] is equal to currentNumber,
    - We will Increment count by 1.
- We will check if count is more than 1,
  - Update the value of duplicate with currentNumber.
Return the variable duplicate.

Space Complexity: O(1)Explanation:

O(1).

Constant space is being used. Hence, the overall Space Complexity is O(1).

Time Complexity: O(n^2)Explanation:

O(N ^ 2), where N denotes the number of elements in the array.

We are doing N - 1 iteration, and in each iteration, we are traversing through the array ARR to find the frequency of the number in the given array that takes O(N) time. Hence, the overall Time Complexity = O(N) * (N) = O(N ^ 2).

Java (SE 1.8)

/*

    Time Complexity O(N^2)

    Space Complexity: O(1)



   Where N is the number of elements in the array.

*/



import java.util.ArrayList;



public class Solution {



	public static int findDuplicate(ArrayList arr) {



		int n = arr.size();



		// The variable duplicate stores the duplicate element in the array

		int duplicate = 0;



		for (int currentNumber = 1; currentNumber <= n - 1; currentNumber++) {

			// The variable count stores the number of currentNumber in the array ARR

			int count = 0;



			// Traverse through the array ARR

			for (int index = 0; index < n; index++) {

				// Check if ARR[index] is equal to currentNumber

				if (arr.get(index) == currentNumber) {

					count++;

				}

			}



			// Check if count is more than 1

			if (count > 1) {

				duplicate = currentNumber;

			}

		}



		// Return the variable duplicate

		return duplicate;



	}

}

C++ (g++ 5.4)

/*

   Time Complexity: O(N ^ 2)

   Space Complexity: O(1)



   Where N is the number of elements in the array.

*/



int findDuplicate(vector &arr) 

{

    int n = arr.size();



    // The variable duplicate stores the duplicate element in the array

    int duplicate;



    for (int currentNumber = 1; currentNumber <= n - 1; currentNumber++) 

    {

        // The variable count stores the number of currentNumber in the array ARR

        int count = 0;



        // Traverse through the array ARR

        for (int index = 0; index < n; index++) 

        {

            // Check if ARR[index] is equal to currentNumber

            if (arr[index] == currentNumber) 

            {

                count++;

            }

        }



        // Check if count is more than 1

        if (count > 1) 

        {

            duplicate = currentNumber;

        }

    }



    // Return the variable duplicate

    return duplicate;

}

Python (3.5)

'''

   Time Complexity: O(N ^ 2)

   Space Complexity: O(1)



   Where N is the number of elements in the array.

'''



def findDuplicate(arr):



    n = len(arr)



    # The variable duplicate stores the duplicate element in the array

    duplicate = 0



    for currentNumber in range(1, n):



        # The variable count stores the number of currentNumber in the array ARR

        count = 0



        # Traverse through the array ARR

        for index in range(n):



            # Check if ARR[index] is equal to currentNumber

            if (arr[index] == currentNumber) :

                count += 1

            

        # Check if count is more than 1

        if (count > 1):

            duplicate = currentNumber



    # Return the variable duplicate

    return duplicate

Accepted Answer

Dynamic Programming

The idea is to observe the fact that all array elements contain a value between 1 to N - 1. Our approach will be to construct an array to store the frequency of each element in the given array.

We will construct array frequency, which will store the frequency of each element in the array.

We will iterate index from 0 to N - 1, and we will increment the count of ARR[index] in the array frequency. So, we will increment frequency[ARR[index]] by 1.
We will iterate currentNumber from 1 to N - 1, and we will check if frequency[currentNumber] is more than 1, then there is a duplicate of the number currentNumber in the array ARR.

In the end, we will return the duplicate integer value present in the array.

Algorithm:

We will initialize duplicate as 0. The variable duplicate stores the duplicate element in the array.
Create an array frequency of size N, which will store the count of each element present in the array ARR.
Iterate index from 0 to N - 1.
- We will set the variable currentNumber as ARR[index].
- Increment frequency[currentNumber] by 1.
Iterate currentNumber from 1 to N - 1.
- We will check if frequency[currentNumber] is more than 1.
  - Update the value of duplicate with currentNumber.
Return the variable duplicate.

Space Complexity: O(n)Explanation:

O(N), where N denotes the number of elements in the array.

The Space Complexity O(N) is used to store the frequency of each element in the array ARR. Hence, the overall Space Complexity is O(N).

Time Complexity: O(n)Explanation:

O(N), where N denotes the number of elements in the array.

The Time Complexity to store the frequency of each element in the array ARR is O(N). The Time Complexity to find the duplicate element in the array is O(N). Hence the overall Time Complexity is O(N).

Java (SE 1.8)

/*

    Time Complexity O(N)

    Space Complexity: O(N)



    Where N is the number of elements in the array.

*/



import java.util.ArrayList;



public class Solution {



	public static int findDuplicate(ArrayList arr) {



		int n = arr.size();



		// The variable duplicate stores the duplicate element in the array

		int duplicate = 1;



		// Create an array frequency to store the count of each element present in the array ARR

		ArrayList frequency = new ArrayList(n);



		for (int i = 0; i < n; i++) {

			frequency.add( 0);

		}



		// Traverse through the array ARR

		for (int index = 0; index < n; index++) {



			int currentNumber = arr.get(index);

			frequency.set(currentNumber, frequency.get(currentNumber) + 1);



		}



		// Iterate currentNumber from 1 to N - 1

		for (int currentNumber = 1; currentNumber <= n - 1; currentNumber++) {



			// Check if frequency[currentNumber] is more than 1

			if ( frequency.get(currentNumber) > 1) {

				duplicate = currentNumber;

			}

		}



		// Return the variable duplicate

		return duplicate;



	}

}

Python (3.5)

'''

   Time Complexity: O(N)

   Space Complexity: O(N)



   Where N is the number of elements in the array.

'''



def findDuplicate(arr) :

    n = len(arr)



    # The variable duplicate stores the duplicate element in the array

    duplicate = 0



    # Create an array frequency to store the count of each element present in the array ARR

    frequency = [0 for i in range(n)]



    # Traverse through the array ARR

    for index in range(n):



        currentNumber = arr[index]

        frequency[currentNumber] += 1



    # Iterate currentNumber from 1 to N - 1

    for currentNumber in range(n):



        # Check if frequency[currentNumber] is more than 1

        if (frequency[currentNumber] > 1):

            duplicate = currentNumber



    # Return the variable duplicate

    return duplicate

C++ (g++ 5.4)

/*

   Time Complexity: O(N)

   Space Complexity: O(N)



   Where N is the number of elements in the array.

*/



int findDuplicate(vector &arr) 

{

    int n = arr.size();



    // The variable duplicate stores the duplicate element in the array

    int duplicate;



    // Create an array frequency to store the count of each element present in the array ARR

    vector frequency(n, 0);



    // Traverse through the array ARR

    for (int index = 0; index < n; index++) 

    {

        int currentNumber = arr[index];

        frequency[currentNumber]++;

    }



    // Iterate currentNumber from 1 to N - 1

    for (int currentNumber = 1; currentNumber <= n - 1; currentNumber++) 

    {

        // Check if frequency[currentNumber] is more than 1

        if (frequency[currentNumber] > 1) 

        {

            duplicate = currentNumber;

        }

    }



    // Return the variable duplicate

    return duplicate;

}

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