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;

}

Accepted Answer

Floyd’s cycle finding algorithm

The idea is to maintain two pointers, fast and slow. The pointer slow goes forward one step, and the pointer fast goes forward two steps each time. The two pointers meet in a cycle when the pointer fast becomes equal to the pointer slow, and then the duplicate number must be the entry point of the cycle.

Our approach will be to find the cycle. We will set slow as ARR[0] and fast as ARR[ ARR[0] ]. We will iterate till fast is not equal to small. In each iteration, we will move the pointer slow one step forward, and the pointer fast two steps forward. So, we will update slow with ARR[slow] and fast with ARR[ ARR[fast] ].
After we have found the cycle, we will find the entry pointer of the cycle. We will set fast as 0. We will iterate till fast is not equal to small, and in each iteration, we will update slow with ARR[slow] and fast with ARR[fast].

In the end, the variable slow will contain the duplicate element present in the array. So, we will return the variable slow.

Algorithm:

We will set slow as ARR[0] and fast as ARR[ ARR[0] ].The variable slow and fast are the two-pointer that we are using to find the duplicate element in the array.
Iterate till fast is not equal to slow.
- Update slow with ARR[slow]. We are moving the pointer slow one step forward.
- Update fast with ARR[ ARR[fast] ]. We are moving the pointer fast two steps forward.
Assign fast as 0.
Iterate till fast is not equal to slow.
- Update slow with ARR[slow].
- Update fast with ARR[fast].
The variable slow will contain the duplicate element. So, we will return the variable slow.

Space Complexity: O(1)Explanation:

O(1).

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

Time Complexity: O(n)Explanation:

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

The Time Complexity to find the duplicate element in the array using Floyd’s cycle finding algorithm is O(N). Hence the overall Time Complexity is O(N).

Java (SE 1.8)

/*

    Time Complexity O(N)

    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) {



		// The variable slow and fast are the two pointers that we are using to find the duplicate element

		int slow = arr.get(0);

		int fast = arr.get(arr.get(0));



		// Iterate till fast is not equal to slow

		while (fast != slow) {



			// Update slow with ARR[slow]. We are moving the pointer slow one step forward

			slow = arr.get(slow);



			// Update fast with ARR[ ARR[fast] ]. We are moving the pointer fast two steps forward

			fast = arr.get(arr.get(fast));

		}



		fast = 0;



		// Loop to find the duplicate element in the array

		while (slow != fast) {

			slow = arr.get(slow);

			fast = arr.get(fast);

		}



		// Return the variable slow

		return slow;



	}

}

Python (3.5)

'''

   Time Complexity: O(N)

   Space Complexity: O(1)



   Where N is the number of elements in the array.

'''



def findDuplicate(arr):

    

    # The variable slow and fast are the two pointers that we are using to find the duplicate element  

    slow = arr[0]

    fast = arr[arr[0]]



    # Iterate till fast is not equal to slow

    while (fast != slow):



        # Update slow with ARR[slow]. We are moving the pointer slow one step forward 

        slow = arr[slow]



        # Update fast with ARR[ ARR[fast] ]. We are moving the pointer fast two steps forward  

        fast = arr[arr[fast]]



    fast = 0



    # Loop to find the duplicate element in the array

    while (slow != fast):

        slow = arr[slow]

        fast = arr[fast]



    # Return the variable slow

    return slow

C++ (g++ 5.4)

/*

   Time Complexity: O(N)

   Space Complexity: O(1)



   Where N is the number of elements in the array.

*/



int findDuplicate(vector &arr) 

{

    // The variable slow and fast are the two pointers that we are using to find the duplicate element  

    int slow = arr[0];

    int fast = arr[arr[0]];



    // Iterate till fast is not equal to slow

    while (fast != slow) 

    {

        // Update slow with ARR[slow]. We are moving the pointer slow one step forward 

        slow = arr[slow];



        // Update fast with ARR[ ARR[fast] ]. We are moving the pointer fast two steps forward  

        fast = arr[arr[fast]];

    }



    fast = 0;



    // Loop to find the duplicate element in the array

    while (slow != fast) 

    {

        slow = arr[slow];

        fast = arr[fast];

    }



    // Return the variable slow

    return slow;

}

Accepted Answer

XOR-based approach

The idea is to observe the fact that the XOR of two same numbers gives 0. The given array contains values from 1 to N - 1 once and one duplicate value. Using this idea, we can find the duplicate element val in the array ARR in the following way given below:

duplicate= (1 ^ 1) ^ (2 ^ 2) ^ (3 ^ 3)....((N - 1) ^ (N - 1)) ^ val

The XOR value of each number with itself gives 0, and the remaining value will be the duplicate value val that is present in the array ARR.

Our approach will be to initialize the variable duplicate, which will store the duplicate integer value present in the array. We will set duplicate as 0.

We will iterate currentNumber from 1 to N - 1, and in each iteration, we will update duplicate with XOR of duplicate and currentNumber.
We will iterate index from 0 to N-1, and we will update duplicate with XOR of duplicate and ARR[index].

In the end, the variable duplicate will contain the duplicate integer value present in the array ARR, and we will return the variable duplicate.

Algorithm:

We will initialize duplicate as 0. The variable duplicate stores the duplicate element in the array.
Iterate currentNumber from 1 to N - 1.
- Update duplicate with XOR of duplicate and currentNumber.
Iterate index from 0 to N - 1.
- Update duplicate with XOR of duplicate and ARR[index].
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)Explanation:

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

The Time Complexity to find the duplicate element in the array using the XOR-based approach is O(N). Hence the overall Time Complexity is O(N).

Python (3.5)

'''

   Time Complexity: O(N)

   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



    # Iterate currentNumber from 1 to N - 1

    for currentNumber in range(n):



        # Update duplicate with XOR of duplicate and currentNumber

        duplicate = duplicate ^ currentNumber



    # Traverse through the array ARR

    for index in range(n):



        # Update duplicate with XOR of duplicate and ARR[index]

        duplicate = duplicate ^ arr[index]



    # Return the variable duplicate

    return duplicate

Java (SE 1.8)

/*

    Time Complexity O(N)

    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;



		// Iterate currentNumber from 1 to N - 1

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



			// Update duplicate with XOR of duplicate and currentNumber

			duplicate = duplicate ^ currentNumber;

		}



		// Traverse through the array ARR

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



			// Update duplicate with XOR of duplicate and ARR[index]

			duplicate = duplicate ^ arr.get(index);

		}



		// Return the variable duplicate

		return duplicate;



	}

}

C++ (g++ 5.4)

/*

   Time Complexity: O(N)

   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 = 0;



    // Iterate currentNumber from 1 to N - 1

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

    {

        // Update duplicate with XOR of duplicate and currentNumber

        duplicate = duplicate ^ currentNumber;

    }



    // Traverse through the array ARR

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

    {

        // Update duplicate with XOR of duplicate and ARR[index]

        duplicate = duplicate ^ arr[index];

    }



    // Return the variable duplicate

    return duplicate;

}

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:

Note :

Input Format:

Output Format:

Constraints:

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