Array elements that appear more than once using Hashing
The idea is to use Hashing to solve this in O(n) time on average. We store elements and their counts in a hash table. After storing counts, we traverse input array again and print those elements whose counts are more than once. To make sure that every output element is printed only once, we set count as 0 after printing the element.
Below is the implementation of the above approach:
// C++ program to print all repeating elements
#include <bits/stdc++.h>
using namespace std;
void printRepeating(int arr[], int n)
{
// Store elements and their counts in
// hash table
unordered_map<int, int> mp;
for (int i = 0; i < n; i++)
mp[arr[i]]++;
// Since we want elements in same order,
// we traverse array again and print
// those elements that appear more than
// once.
for (int i = 0; i < n; i++) {
if (mp[arr[i]] > 1) {
cout << arr[i] << " ";
// This is tricky, this is done
// to make sure that the current
// element is not printed again
mp[arr[i]] = 0;
}
}
}
// Driver code
int main()
{
int arr[] = { 12, 10, 9, 45, 2, 10, 10, 45 };
int n = sizeof(arr) / sizeof(arr[0]);
printRepeating(arr, n);
return 0;
}
// Java program to print all repeating elements
import java.util.*;
import java.util.Map.Entry;
import java.io.*;
import java.lang.*;
public class GFG {
static void printRepeating(int arr[], int n)
{
// Store elements and their counts in
// hash table
Map<Integer, Integer> map
= new LinkedHashMap<Integer, Integer>();
for (int i = 0; i < n; i++) {
try {
map.put(arr[i], map.get(arr[i]) + 1);
}
catch (Exception e) {
map.put(arr[i], 1);
}
}
// Since we want elements in the same order,
// we traverse array again and print
// those elements that appear more than once.
for (Entry<Integer, Integer> e : map.entrySet()) {
if (e.getValue() > 1) {
System.out.print(e.getKey() + " ");
}
}
}
// Driver code
public static void main(String[] args) throws IOException
{
int arr[] = { 12, 10, 9, 45, 2, 10, 10, 45 };
int n = arr.length;
printRepeating(arr, n);
}
}
// This code is contributed by Wrick
# Python3 program to print
# all repeating elements
def printRepeating(arr, n):
# Store elements and
# their counts in
# hash table
mp = [0] * 100
for i in range(0, n):
mp[arr[i]] += 1
# Since we want elements
# in same order, we
# traverse array again
# and print those elements
# that appear more than once.
for i in range(0, n):
if (mp[arr[i]] > 1):
print(arr[i], end = " ")
# This is tricky, this
# is done to make sure
# that the current element
# is not printed again
mp[arr[i]] = 0
# Driver code
arr = [12, 10, 9, 45,
2, 10, 10, 45]
n = len(arr)
printRepeating(arr, n)
# This code is contributed
# by Smita
// C# program to print all repeating elements
using System;
using System.Collections.Generic;
class GFG
{
static void printRepeating(int []arr, int n)
{
// Store elements and their counts in
// hash table
Dictionary<int,
int> map = new Dictionary<int,
int>();
for (int i = 0 ; i < n; i++)
{
if(map.ContainsKey(arr[i]))
{
var val = map[arr[i]];
map.Remove(arr[i]);
map.Add(arr[i], val + 1);
}
else
{
map.Add(arr[i], 1);
}
}
// Since we want elements in the same order,
// we traverse array again and print
// those elements that appear more than once.
foreach(KeyValuePair<int, int> e in map)
{
if (e.Value > 1)
{
Console.Write(e.Key + " ");
}
}
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 12, 10, 9, 45, 2, 10, 10, 45 };
int n = arr.Length;
printRepeating(arr, n);
}
}
// This code is contributed by PrinciRaj1992
// JavaScript program to print all
// repeating elements
function printRepeating(arr, n)
{
// Store elements and their counts in
// hash table
var mp = new Map();
for (var i = 0; i < n; i++)
{
if(mp.has(arr[i]))
mp.set(arr[i], mp.get(arr[i])+1)
else
mp.set(arr[i], 1)
}
// Since we want elements in same order,
// we traverse array again and print
// those elements that appear more than
// once.
for (var i = 0; i < n; i++) {
if (mp.get(arr[i]) > 1) {
console.log( arr[i] + " ");
// This is tricky, this is done
// to make sure that the current
// element is not printed again
mp.set(arr[i], 0);
}
}
}
// Driver code
var arr = [ 12, 10, 9, 45, 2, 10, 10, 45 ];
var n = arr.length;
printRepeating(arr, n);
Output
10 45
Time Complexity: O(n) under the assumption that hash insert and search functions work in O(1) time.
Auxiliary Space: O(n), where n represents the size of the given array.
Array elements that appear more than once
Given an integer array, print all repeating elements (Elements that appear more than once) in the array. The output should contain elements according to their first occurrences.
Examples:
Input: arr[] = {12, 10, 9, 45, 2, 10, 10, 45}
Output: 10 45Input: arr[] = {1, 2, 3, 4, 2, 5}
Output:2Input: arr[] = {1, 1, 1, 1, 1}
Output: 1