Find duplicates in a given array when elements are not limited to a range
Given an array of n integers. The task is to print the duplicates in the given array. If there are no duplicates then print -1.
Examples:
Input: {2, 10,10, 100, 2, 10, 11,2,11,2} Output: 2 10 11 Input: {5, 40, 1, 40, 100000, 1, 5, 1} Output: 5 40 1
Note: The duplicate elements can be printed in any order.
Simple Approach: The idea is to use nested loop and for each element check if the element is present in the array more than once or not. If present, then store it in a Hash-map. Otherwise, continue checking other elements.
Below is the implementation of the above approach:
C++
// C++ implementation of the // above approach #include <bits/stdc++.h> using namespace std; // Function to find the Duplicates, // if duplicate occurs 2 times or // more than 2 times in array so, // it will print duplicate value // only once at output void findDuplicates( int arr[], int len) { // Initialize ifPresent as false bool ifPresent = false ; // ArrayList to store the output vector< int > al; for ( int i = 0; i < len - 1; i++) { for ( int j = i + 1; j < len; j++) { if (arr[i] == arr[j]) { // Checking if element is // present in the ArrayList // or not if present then break auto it = std::find(al.begin(), al.end(), arr[i]); if (it != al.end()) { break ; } // If element is not present in the // ArrayList then add it to ArrayList // and make ifPresent at true else { al.push_back(arr[i]); ifPresent = true ; } } } } // If duplicates is present // then print ArrayList if (ifPresent == true ) { cout << "[" << al[0] << ", " ; for ( int i = 1; i < al.size() - 1; i++) { cout << al[i] << ", " ; } cout << al[al.size() - 1] << "]" ; } else { cout << "No duplicates present in arrays" ; } } // Driver code int main() { int arr[] = { 12, 11, 40, 12, 5, 6, 5, 12, 11 }; int n = sizeof (arr) / sizeof (arr[0]); findDuplicates(arr, n); return 0; } // This code is contributed by divyeshrabadiya07 |
Java
// Java implementation of the // above approach import java.util.ArrayList; public class GFG { // Function to find the Duplicates, // if duplicate occurs 2 times or // more than 2 times in // array so, it will print duplicate // value only once at output static void findDuplicates( int arr[], int len) { // initialize ifPresent as false boolean ifPresent = false ; // ArrayList to store the output ArrayList<Integer> al = new ArrayList<Integer>(); for ( int i = 0 ; i < len - 1 ; i++) { for ( int j = i + 1 ; j < len; j++) { if (arr[i] == arr[j]) { // checking if element is // present in the ArrayList // or not if present then break if (al.contains(arr[i])) { break ; } // if element is not present in the // ArrayList then add it to ArrayList // and make ifPresent at true else { al.add(arr[i]); ifPresent = true ; } } } } // if duplicates is present // then print ArrayList if (ifPresent == true ) { System.out.print(al + " " ); } else { System.out.print( "No duplicates present in arrays" ); } } // Driver Code public static void main(String[] args) { int arr[] = { 12 , 11 , 40 , 12 , 5 , 6 , 5 , 12 , 11 }; int n = arr.length; findDuplicates(arr, n); } } |
Python3
# Python3 implementation of the # above approach # Function to find the Duplicates, # if duplicate occurs 2 times or # more than 2 times in array so, # it will print duplicate value # only once at output def findDuplicates(arr, Len ): # Initialize ifPresent as false ifPresent = False # ArrayList to store the output a1 = [] for i in range ( Len - 1 ): for j in range (i + 1 , Len ): # Checking if element is # present in the ArrayList # or not if present then break if (arr[i] = = arr[j]): if arr[i] in a1: break # If element is not present in the # ArrayList then add it to ArrayList # and make ifPresent at true else : a1.append(arr[i]) ifPresent = True # If duplicates is present # then print ArrayList if (ifPresent): print (a1, end = " " ) else : print ( "No duplicates present in arrays" ) # Driver Code arr = [ 12 , 11 , 40 , 12 , 5 , 6 , 5 , 12 , 11 ] n = len (arr) findDuplicates(arr, n) # This code is contributed by rag2127 |
C#
// C# implementation of the // above approach using System; using System.Collections.Generic; class GFG{ // Function to find the Duplicates, // if duplicate occurs 2 times or // more than 2 times in array so, // it will print duplicate value // only once at output static void findDuplicates( int [] arr, int len) { // Initialize ifPresent as false bool ifPresent = false ; // ArrayList to store the output List< int > al = new List< int >(); for ( int i = 0; i < len - 1; i++) { for ( int j = i + 1; j < len; j++) { if (arr[i] == arr[j]) { // Checking if element is // present in the ArrayList // or not if present then break if (al.Contains(arr[i])) { break ; } // If element is not present in the // ArrayList then add it to ArrayList // and make ifPresent at true else { al.Add(arr[i]); ifPresent = true ; } } } } // If duplicates is present // then print ArrayList if (ifPresent == true ) { Console.Write( "[" + al[0] + ", " ); for ( int i = 1; i < al.Count - 1; i++) { Console.Write(al[i] + ", " ); } Console.Write(al[al.Count - 1] + "]" ); } else { Console.Write( "No duplicates present in arrays" ); } } // Driver code static void Main() { int [] arr = { 12, 11, 40, 12, 5, 6, 5, 12, 11 }; int n = arr.Length; findDuplicates(arr, n); } } // This code is contributed by divyesh072019 |
Javascript
<script> // JavaScript implementation of the // above approach // Function to find the Duplicates, // if duplicate occurs 2 times or // more than 2 times in // array so, it will print duplicate // value only once at output function findDuplicates(arr, len) { // initialize ifPresent as false let ifPresent = false ; // ArrayList to store the output let al = new Array(); for (let i = 0; i < len - 1; i++) { for (let j = i + 1; j < len; j++) { if (arr[i] == arr[j]) { // checking if element is // present in the ArrayList // or not if present then break if (al.includes(arr[i])) { break ; } // if element is not present in the // ArrayList then add it to ArrayList // and make ifPresent at true else { al.push(arr[i]); ifPresent = true ; } } } } // if duplicates is present // then print ArrayList if (ifPresent == true ) { document.write(`[${al}]`); } else { document.write( "No duplicates present in arrays" ); } } // Driver Code let arr = [12, 11, 40, 12, 5, 6, 5, 12, 11]; let n = arr.length; findDuplicates(arr, n); </script> |
[12, 11, 5]
Time Complexity: O(N2)
Auxiliary Space: O(N)
Efficient Approach: Use unordered_map for hashing. Count frequency of occurrence of each element and the elements with frequency more than 1 is printed. unordered_map is used as range of integers is not known. For Python, Use Dictionary to store number as key and it’s frequency as value. Dictionary can be used as range of integers is not known.
Below is the implementation of the above approach:
C++
// C++ program to find // duplicates in the given array #include <bits/stdc++.h> using namespace std; // function to find and print duplicates void printDuplicates( int arr[], int n) { // unordered_map to store frequencies unordered_map< int , int > freq; for ( int i=0; i<n; i++) freq[arr[i]]++; bool dup = false ; unordered_map< int , int >:: iterator itr; for (itr=freq.begin(); itr!=freq.end(); itr++) { // if frequency is more than 1 // print the element if (itr->second > 1) { cout << itr->first << " " ; dup = true ; } } // no duplicates present if (dup == false ) cout << "-1" ; } // Driver program to test above int main() { int arr[] = {12, 11, 40, 12, 5, 6, 5, 12, 11}; int n = sizeof (arr) / sizeof (arr[0]); printDuplicates(arr, n); return 0; } |
Java
// Java program to find // duplicates in the given array import java.util.HashMap; import java.util.Map; import java.util.Map.Entry; public class FindDuplicatedInArray { // Driver program public static void main(String[] args) { int arr[] = { 12 , 11 , 40 , 12 , 5 , 6 , 5 , 12 , 11 }; int n = arr.length; printDuplicates(arr, n); } // function to find and print duplicates private static void printDuplicates( int [] arr, int n) { Map<Integer,Integer> map = new HashMap<>(); int count = 0 ; boolean dup = false ; for ( int i = 0 ; i < n; i++){ if (map.containsKey(arr[i])){ count = map.get(arr[i]); map.put(arr[i], count + 1 ); } else { map.put(arr[i], 1 ); } } for (Entry<Integer,Integer> entry : map.entrySet()) { // if frequency is more than 1 // print the element if (entry.getValue() > 1 ){ System.out.print(entry.getKey()+ " " ); dup = true ; } } // no duplicates present if (!dup){ System.out.println( "-1" ); } } } |
Python3
# Python3 program to find duplicates # using dictionary approach. def printDuplicates(arr): dict = {} for ele in arr: try : dict [ele] + = 1 except : dict [ele] = 1 for item in dict : # if frequency is more than 1 # print the element if ( dict [item] > 1 ): print (item, end = " " ) print ( "\n" ) # Driver Code if __name__ = = "__main__" : list = [ 12 , 11 , 40 , 12 , 5 , 6 , 5 , 12 , 11 ] printDuplicates( list ) # This code is contributed # by Sushil Bhile |
C#
// C# program to find // duplicates in the given array using System; using System.Collections.Generic; class GFG { // function to find and print duplicates static void printDuplicates( int [] arr, int n) { Dictionary< int , int > map = new Dictionary< int , int >(); int count = 0; bool dup = false ; for ( int i = 0; i < n; i++) { if (map.ContainsKey(arr[i])) { count = map[arr[i]]; map[arr[i]]++; } else map.Add(arr[i], 1); } foreach (KeyValuePair< int , int > entry in map) { // if frequency is more than 1 // print the element if (entry.Value > 1) Console.Write(entry.Key + " " ); dup = true ; } // no duplicates present if (!dup) Console.WriteLine( "-1" ); } // Driver Code public static void Main(String[] args) { int [] arr = { 12, 11, 40, 12, 5, 6, 5, 12, 11 }; int n = arr.Length; printDuplicates(arr, n); } } // This code is contributed by // sanjeev2552 |
Javascript
<script> // JavaScript program to find // duplicates in the given array // Function to find and print duplicates function printDuplicates(arr, n) { // unordered_map to store frequencies var freq = new Map(); for (let i = 0; i < n; i++) { if (freq.has(arr[i])) { freq.set(arr[i], freq.get(arr[i]) + 1); } else { freq.set(arr[i], 1); } } var dup = false ; for (let [key, value] of freq) { // If frequency is more than 1 // print the element if (value > 1) { document.write(key + " " ); dup = true ; } } // No duplicates present if (dup == false ) document.write( "-1" ); } // Driver code var arr = [ 12, 11, 40, 12, 5, 6, 5, 12, 11 ]; var n = arr.length; printDuplicates(arr, n); // This code is contributed by lokeshpotta20 </script> |
5 12 11
Time Complexity: O(N)
Auxiliary Space: O(N)
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