Find duplicates in O(n) time and O(1) extra space | Set 1

Given an array of n elements that contains elements from 0 to n-1, with any of these numbers appearing any number of times. Find these repeating numbers in O(n) and use only constant memory space.

Note: The repeating element should be printed only once.

Example:

Input: n=7 , array[]={1, 2, 3, 6, 3, 6, 1}
Output: 1, 3, 6
Explanation: The numbers 1 , 3 and 6 appears more than once in the array.

Input : n = 5 and array[] = {1, 2, 3, 4 ,3}
Output: 3
Explanation: The number 3 appears more than once in the array.

This problem is an extended version of the following problem.
Find the two repeating elements in a given array 

Approach 1( Using hashmap):

Use the input array to store the frequency of each element. While Traversing the array, if an element x is encountered then check if its frequency is greater than 1 , then put it in the result array. if result array is empty return -1 else sort the array and return array.

Follow the steps to implement the approach:

1. Create an empty unordered map (hashmap) to store the frequency of each element in the array.
2. Iterate through the given array.
3. For each element in the array, increment its frequency count in the hashmap.
4. Iterate through the hashmap.
5. For each key-value pair in the hashmap, if the value (frequency count) is greater than 1, add the corresponding key (element) to the result vector.
6. If the result vector is empty after step 3, it means no duplicates were found. In this case, add -1 to the result vector.
7. Return the result vector containing duplicate elements or -1 if no duplicates were found.

C++

#include <bits/stdc++.h>

using namespace std;

vector<int> duplicates(long long arr[], int n) {
    // Step 1: Create an empty unordered map to store element frequencies
    unordered_map<long long, int> freqMap;
    vector<int> result;

    // Step 2: Iterate through the array and count element frequencies
    for (int i = 0; i < n; i++) {
        freqMap[arr[i]]++;
    }

    // Step 3: Iterate through the hashmap to find duplicates
    for (auto& entry : freqMap) {
        if (entry.second > 1) {
            result.push_back(entry.first);
        }
    }

    // Step 4: If no duplicates found, add -1 to the result
    if (result.empty()) {
        result.push_back(-1);
    }
    //step 5: sort the result
    sort(result.begin(),result.end());
    // Step 6: Return the result vector containing duplicate elements or -1
    return result;
}

int main() {
    long long a[] = {1, 6, 5, 2, 3, 3, 2};
    int n = sizeof(a) / sizeof(a[0]);
    
    vector<int> duplicates_found = duplicates(a, n);
    
    cout << "Duplicate elements: ";
    for (int element : duplicates_found) {
        cout << element << " ";
    }
    cout << endl;
    
    return 0;
}

Java

import java.util.*;

public class Main {
    public static List<Integer> duplicates(long[] arr) {
        // Step 1: Create an empty hashmap to store element frequencies
        Map<Long, Integer> freqMap = new HashMap<>();
        List<Integer> result = new ArrayList<>();

        // Step 2: Iterate through the array and count element frequencies
        for (long num : arr) {
            freqMap.put(num, freqMap.getOrDefault(num, 0) + 1);
        }

        // Step 3: Iterate through the hashmap to find duplicates
        for (Map.Entry<Long, Integer> entry : freqMap.entrySet()) {
            if (entry.getValue() > 1) {
                result.add(Math.toIntExact(entry.getKey()));
            }
        }

        // Step 4: If no duplicates found, add -1 to the result
        if (result.isEmpty()) {
            result.add(-1);
        }
        // Step 5: Sort the result
        Collections.sort(result);
        // Step 6: Return the result list containing duplicate elements or -1
        return result;
    }

    public static void main(String[] args) {
        long[] a = {1, 6, 5, 2, 3, 3, 2};
        List<Integer> duplicatesFound = duplicates(a);
        
        System.out.print("Duplicate elements: ");
        for (int element : duplicatesFound) {
            System.out.print(element + " ");
        }
        System.out.println();
    }
}

Python3

def duplicates(arr):
    # Step 1: Create an empty dictionary to store element frequencies
    freq_map = {}
    result = []

    # Step 2: Iterate through the array and count element frequencies
    for num in arr:
        freq_map[num] = freq_map.get(num, 0) + 1

    # Step 3: Iterate through the dictionary to find duplicates
    for key, value in freq_map.items():
        if value > 1:
            result.append(key)

    # Step 4: If no duplicates found, add -1 to the result
    if not result:
        result.append(-1)
    # Step 5: Sort the result
    result.sort()
    # Step 6: Return the result list containing duplicate elements or -1
    return result

if __name__ == "__main__":
    a = [1, 6, 5, 2, 3, 3, 2]
    duplicates_found = duplicates(a)
    
    print("Duplicate elements:", end=" ")
    for element in duplicates_found:
        print(element, end=" ")
    print()

C#

using System;
using System.Collections.Generic;
using System.Linq;

class Program
{
    static List<int> Duplicates(long[] arr)
    {
        // Step 1: Create an empty dictionary to store element frequencies
        Dictionary<long, int> freqMap = new Dictionary<long, int>();
        List<int> result = new List<int>();

        // Step 2: Iterate through the array and count element frequencies
        foreach (long num in arr)
        {
            if (freqMap.ContainsKey(num))
                freqMap[num]++;
            else
                freqMap[num] = 1;
        }

        // Step 3: Iterate through the dictionary to find duplicates
        foreach (var entry in freqMap)
        {
            if (entry.Value > 1)
                result.Add((int)entry.Key);
        }

        // Step 4: If no duplicates found, add -1 to the result
        if (result.Count == 0)
            result.Add(-1);
        // Step 5: Sort the result
        result.Sort();
        // Step 6: Return the result list containing duplicate elements or -1
        return result;
    }

    static void Main(string[] args)
    {
        long[] a = { 1, 6, 5, 2, 3, 3, 2 };
        List<int> duplicatesFound = Duplicates(a);

        Console.Write("Duplicate elements: ");
        foreach (int element in duplicatesFound)
        {
            Console.Write(element + " ");
        }
        Console.WriteLine();
    }
}

JavaScript

function duplicates(arr) {
    // Step 1: Create an empty object to store element frequencies
    const freqMap = {};
    const result = [];

    // Step 2: Iterate through the array and count element frequencies
    for (let num of arr) {
        freqMap[num] = (freqMap[num] || 0) + 1;
    }

    // Step 3: Iterate through the object to find duplicates
    for (let key in freqMap) {
        if (freqMap[key] > 1) {
            result.push(Number(key));
        }
    }

    // Step 4: If no duplicates found, add -1 to the result
    if (result.length === 0) {
        result.push(-1);
    }
    // Step 5: Sort the result
    result.sort((a, b) => a - b);
    // Step 6: Return the result array containing duplicate elements or -1
    return result;
}

const a = [1, 6, 5, 2, 3, 3, 2];
const duplicatesFound = duplicates(a);

console.log("Duplicate elements: " + duplicatesFound.join(" "));

Output

Duplicate elements: 2 3 

Time Complexity: O(n), Only two traversals are needed. So the time complexity is O(n).

Auxiliary Space: O(n), The extra space is used for the array to be returned .

Approach 2( by modification in the array):

Iterate through the array, for each element, increment its frequency marker at arr[element % n] by n. Traverse the modified array, if an element’s value divided by n is greater than 1, add its index to result indicating a duplicate. If no duplicates found, append -1 to result. Return the result vector.

Follow the steps to implement the approach:

1. Iterate through the array, calculating the index for each element as `arr[i] % n`, where `arr[i]` is the current element and` n` is the size of the array.
2. Increment the value at the calculated index by `n` to mark the frequency of the corresponding element.
3. Traverse the modified array again after completing the initial traversal.
4. For each element, calculate its frequency by dividing the value at the current index by `n`.
5. If the calculated frequency is greater than 1, add the index to the result vector, indicating a duplicate element.
6. If no duplicates are found during the traversal, append -1 to the result vector.
7. Return the result vector containing duplicate elements or -1 if no duplicates were found.

C++

#include <iostream>
#include <vector>

using namespace std;

vector<int> duplicates(long long arr[], int n) {
    vector<int> result;

    // Step 1: Iterate through the array to mark frequencies
    for (int i = 0; i < n; i++) {
        // Calculate the index where the frequency needs to be stored
        int index = arr[i] % n;
        // Increment the value at the calculated index by n, marking the frequency
        arr[index] += n;
    }

    // Step 2: Traverse the modified array to find duplicate elements
    bool found = false;
    for (int i = 0; i < n; i++) {
        // Calculate the frequency of the current element
        if (arr[i] / n > 1) {
            // If frequency is greater than 1, add the index to the result vector
            result.push_back(i);
            found = true;
        }
    }

    // Step 3: If no duplicates found, add -1 to the result
    if (!found) {
        result.push_back(-1);
    }

    return result;
}

int main() {
    long long a[] = {1,2,3,3,2 ,6, 5, 2, 3, 3, 2};
    int n = sizeof(a) / sizeof(a[0]);
    
    // Call the duplicates function to find duplicate elements
    vector<int> duplicates_found = duplicates(a, n);
    
    // Output the results
    cout << "Duplicate elements: ";
    for (int element : duplicates_found) {
        cout << element << " ";
    }
    cout << endl;
    
    return 0;
}

Java

import java.util.ArrayList;
import java.util.List;

public class Main {
    public static List<Integer> duplicates(long[] arr) {
        List<Integer> result = new ArrayList<>();

        // Step 1: Iterate through the array to mark frequencies
        for (int i = 0; i < arr.length; i++) {
            // Calculate the index where the frequency needs to be stored
            int index = (int) (arr[i] % arr.length);
            // Increment the value at the calculated index by n, marking the frequency
            arr[index] += arr.length;
        }

        // Step 2: Traverse the modified array to find duplicate elements
        boolean found = false;
        for (int i = 0; i < arr.length; i++) {
            // Calculate the frequency of the current element
            if (arr[i] / arr.length > 1) {
                // If frequency is greater than 1, add the index to the result list
                result.add(i);
                found = true;
            }
        }

        // Step 3: If no duplicates found, add -1 to the result
        if (!found) {
            result.add(-1);
        }

        return result;
    }

    public static void main(String[] args) {
        long[] a = {1, 6, 5, 2, 3, 3, 2};

        // Call the duplicates function to find duplicate elements
        List<Integer> duplicatesFound = duplicates(a);

        // Output the results
        System.out.print("Duplicate elements: ");
        for (int element : duplicatesFound) {
            System.out.print(element + " ");
        }
        System.out.println();
    }
}

Python3

def duplicates(arr):
    result = []

    # Step 1: Iterate through the array to mark frequencies
    for i in range(len(arr)):
        # Calculate the index where the frequency needs to be stored
        index = arr[i] % len(arr)
        # Increment the value at the calculated index by n, marking the frequency
        arr[index] += len(arr)

    # Step 2: Traverse the modified array to find duplicate elements
    found = False
    for i in range(len(arr)):
        # Calculate the frequency of the current element
        if arr[i] // len(arr) > 1:
            # If frequency is greater than 1, add the index to the result list
            result.append(i)
            found = True

    # Step 3: If no duplicates found, add -1 to the result
    if not found:
        result.append(-1)

    return result

a = [1, 6, 5, 2, 3, 3, 2]

# Call the duplicates function to find duplicate elements
duplicates_found = duplicates(a)

# Output the results
print("Duplicate elements:", end=" ")
for element in duplicates_found:
    print(element, end=" ")
print()

C#

using System;
using System.Collections.Generic;

class MainClass {
    public static List<int> Duplicates(long[] arr) {
        List<int> result = new List<int>();

        // Step 1: Iterate through the array to mark frequencies
        for (int i = 0; i < arr.Length; i++) {
            // Calculate the index where the frequency needs to be stored
            int index = (int)(arr[i] % arr.Length);
            // Increment the value at the calculated index by n, marking the frequency
            arr[index] += arr.Length;
        }

        // Step 2: Traverse the modified array to find duplicate elements
        bool found = false;
        for (int i = 0; i < arr.Length; i++) {
            // Calculate the frequency of the current element
            if (arr[i] / arr.Length > 1) {
                // If frequency is greater than 1, add the index to the result list
                result.Add(i);
                found = true;
            }
        }

        // Step 3: If no duplicates found, add -1 to the result
        if (!found) {
            result.Add(-1);
        }

        return result;
    }

    public static void Main(string[] args) {
        long[] a = {1, 6, 5, 2, 3, 3, 2};

        // Call the duplicates function to find duplicate elements
        List<int> duplicatesFound = Duplicates(a);

        // Output the results
        Console.Write("Duplicate elements: ");
        foreach (int element in duplicatesFound) {
            Console.Write(element + " ");
        }
        Console.WriteLine();
    }
}

JavaScript

function duplicates(arr) {
    let result = [];

    // Step 1: Iterate through the array to mark frequencies
    for (let i = 0; i < arr.length; i++) {
        // Calculate the index where the frequency needs to be stored
        let index = arr[i] % arr.length;
        // Increment the value at the calculated index by n, marking the frequency
        arr[index] += arr.length;
    }

    // Step 2: Traverse the modified array to find duplicate elements
    let found = false;
    for (let i = 0; i < arr.length; i++) {
        // Calculate the frequency of the current element
        if (arr[i] / arr.length > 1) {
            // If frequency is greater than 1, add the index to the result list
            result.push(i);
            found = true;
        }
    }

    // Step 3: If no duplicates found, add -1 to the result
    if (!found) {
        result.push(-1);
    }

    return result;
}

let a = [1, 6, 5, 2, 3, 3, 2];

// Call the duplicates function to find duplicate elements
let duplicatesFound = duplicates(a);

// Output the results
process.stdout.write("Duplicate elements: ");
for (let element of duplicatesFound) {
    process.stdout.write(element + " ");
}
console.log();

Output

Duplicate elements: 2 3 

Time Complexity: O(n), Only two traversals are needed. If the answer to be return should in ascending order, then in that case we will have to sort the list and complexity will become O(n logn).

Auxiliary Space: O(1). The extra space is used only for the array to be returned.