Given an array arr[] of N+2 elements. All elements of the array are in the range of 1 to N. And all elements occur once except two numbers which occur twice. Find the two repeating numbers.
Examples:
Input: arr = [4, 2, 4, 5, 2, 3, 1], N = 5
Output: 4 2
Explanation: The above array has n + 2 = 7 elements with all elements occurring once except 2 and 4 which occur twice. So the output should be 4 2.
Input: arr = [2, 1, 2, 1, 3], N = 3
Output: 1 2
Explanation: The above array has n + 2 = 5 elements with all elements occurring once except 1 and 2 which occur twice. So the output should be 1 2.
Naive Approach:
The idea is to use two loops.
Follow the steps below to solve the problem:
- In the outer loop, pick elements one by one and count the number of occurrences of the picked element using a nested loop.
- Whenever an element is found to be equal to the picked element in nested then print that element.
Below is the implementation of the above approach.
C++
#include <bits/stdc++.h>
using namespace std;
void printTwoRepeatNumber(int arr[], int size)
{
int i, j;
cout << "Repeating elements are ";
for (i = 0; i < size; i++) {
for (j = i + 1; j < size; j++) {
if (arr[i] == arr[j]) {
cout << arr[i] << " ";
break;
}
}
}
}
int main()
{
int arr[] = { 4, 2, 4, 5, 2, 3, 1 };
int arr_size = sizeof(arr) / sizeof(arr[0]);
printTwoRepeatNumber(arr, arr_size);
return 0;
}
|
C
#include <stdio.h>
#include <stdlib.h>
void printRepeating(int arr[], int size)
{
int i, j;
printf(" Repeating elements are ");
for (i = 0; i < size - 1; i++)
for (j = i + 1; j < size; j++)
if (arr[i] == arr[j])
printf("%d ", arr[i]);
}
int main()
{
int arr[] = { 4, 2, 4, 5, 2, 3, 1 };
int arr_size = sizeof(arr) / sizeof(arr[0]);
printRepeating(arr, arr_size);
getchar();
return 0;
}
|
Java
import java.io.*;
class RepeatElement {
void printRepeating(int arr[], int size)
{
int i, j;
System.out.print("Repeating Elements are ");
for (i = 0; i < size - 1; i++) {
for (j = i + 1; j < size; j++) {
if (arr[i] == arr[j])
System.out.print(arr[i] + " ");
}
}
}
public static void main(String[] args)
{
RepeatElement repeat = new RepeatElement();
int arr[] = { 4, 2, 4, 5, 2, 3, 1 };
int arr_size = arr.length;
repeat.printRepeating(arr, arr_size);
}
}
|
Python3
def printRepeating(arr, size):
print("Repeating elements are", end=' ')
for i in range(0, size-1):
for j in range(i + 1, size):
if arr[i] == arr[j]:
print(arr[i], end=' ')
arr = [4, 2, 4, 5, 2, 3, 1]
arr_size = len(arr)
printRepeating(arr, arr_size)
|
C#
using System;
class GFG {
static void printRepeating(int[] arr, int size)
{
int i, j;
Console.Write("Repeating Elements are ");
for (i = 0; i < size - 1; i++) {
for (j = i + 1; j < size; j++) {
if (arr[i] == arr[j])
Console.Write(arr[i] + " ");
}
}
}
public static void Main()
{
int[] arr = { 4, 2, 4, 5, 2, 3, 1 };
int arr_size = arr.Length;
printRepeating(arr, arr_size);
}
}
|
PHP
<?php
function printRepeating($arr, $size)
{
$i;
$j;
echo " Repeating elements are ";
for($i = 0; $i < $size-1; $i++)
for($j = $i + 1; $j < $size; $j++)
if($arr[$i] == $arr[$j])
echo $arr[$i], " ";
}
$arr = array(4, 2, 4, 5, 2, 3, 1);
$arr_size = sizeof($arr, 0);
printRepeating($arr, $arr_size);
?>
|
Javascript
<script>
function printRepeating(arr , size)
{
var i, j;
document.write("Repeating Elements are ");
for (i = 0; i < size-1; i++)
{
for (j = i + 1; j < size; j++)
{
if (arr[i] == arr[j])
document.write(arr[i] + " ");
}
}
}
var arr = [4, 2, 4, 5, 2, 3, 1];
var arr_size = arr.length;
printRepeating(arr, arr_size);
</script>
|
Output
Repeating elements are 4 2
Time Complexity: O(N*N), Iterating over the array of size N for all the N elements.
Auxiliary Space: O(1)
Find the two repeating elements in a given array using Visited Array:
The idea is to keep track of elements. Whenever an element is encountered that is already present then print that element.
Follow the steps below to solve the problem:
- Create a count array of size N to keep track of elements that are visited.
- Traverse the array once. While traversing, keep track of the count of all elements in the array using a temp array count[] of size n,
- When an element is encountered that is already present, print that element.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void printRepeating(int arr[], int size)
{
int* count = new int[sizeof(int) * (size - 2)];
int i;
cout << " Repeating elements are ";
for (i = 0; i < size; i++) {
if (count[arr[i]] == 1)
cout << arr[i] << " ";
else
count[arr[i]]++;
}
}
int main()
{
int arr[] = { 4, 2, 4, 5, 2, 3, 1 };
int arr_size = sizeof(arr) / sizeof(arr[0]);
printRepeating(arr, arr_size);
return 0;
}
|
C
#include <stdio.h>
#include <stdlib.h>
void printRepeating(int arr[], int size)
{
int* count = (int*)calloc(sizeof(int), (size - 2));
int i;
printf(" Repeating elements are ");
for (i = 0; i < size; i++) {
if (count[arr[i]] == 1)
printf("%d ", arr[i]);
else
count[arr[i]]++;
}
}
int main()
{
int arr[] = { 4, 2, 4, 5, 2, 3, 1 };
int arr_size = sizeof(arr) / sizeof(arr[0]);
printRepeating(arr, arr_size);
getchar();
return 0;
}
|
Java
import java.io.*;
class RepeatElement {
void printRepeating(int arr[], int size)
{
int count[] = new int[size];
int i;
System.out.print("Repeating elements are ");
for (i = 0; i < size; i++) {
if (count[arr[i]] == 1)
System.out.print(arr[i] + " ");
else
count[arr[i]]++;
}
}
public static void main(String[] args)
{
RepeatElement repeat = new RepeatElement();
int arr[] = { 4, 2, 4, 5, 2, 3, 1 };
int arr_size = arr.length;
repeat.printRepeating(arr, arr_size);
}
}
|
Python3
def printRepeating(arr, size):
count = [0] * size
print(" Repeating elements are ", end="")
for i in range(0, size):
if(count[arr[i]] == 1):
print(arr[i], end=" ")
else:
count[arr[i]] = count[arr[i]] + 1
arr = [4, 2, 4, 5, 2, 3, 1]
arr_size = len(arr)
printRepeating(arr, arr_size)
|
C#
using System;
class GFG {
static void printRepeating(int[] arr, int size)
{
int[] count = new int[size];
int i;
Console.Write("Repeating elements are ");
for (i = 0; i < size; i++) {
if (count[arr[i]] == 1)
Console.Write(arr[i] + " ");
else
count[arr[i]]++;
}
}
public static void Main()
{
int[] arr = { 4, 2, 4, 5, 2, 3, 1 };
int arr_size = arr.Length;
printRepeating(arr, arr_size);
}
}
|
PHP
<?php
function printRepeating($arr, $size)
{
$count = array_fill(0, $size, 0);
echo "Repeating elements are ";
for ($i = 0; $i < $size; $i++)
{
if ($count[$arr[$i]] == 1)
echo $arr[$i]." ";
else
$count[$arr[$i]]++;
}
}
$arr = array(4, 2, 4, 5, 2, 3, 1);
$arr_size = count($arr);
printRepeating($arr, $arr_size);
?>
|
Javascript
<script>
function printRepeating(arr, size)
{
let count = new Array(size);
count.fill(0);
let i;
document.write("Repeating elements are ");
for (i = 0; i < size; i++)
{
if (count[arr[i]] == 1)
document.write(arr[i] + " ");
else
count[arr[i]]++;
}
}
let arr = [4, 2, 4, 5, 2, 3, 1];
let arr_size = arr.length;
printRepeating(arr, arr_size);
</script>
|
Output
Repeating elements are 4 2
Time Complexity: O(N), Traversing the array of size N
Auxiliary Space: O(N), Count array of size N to keep track of elements
Find the two repeating elements in a given array using Mathematics:
The idea is to calculate the sum and product of elements that are repeating in the array and using those two equations find those repeating elements.
Follow the steps below to solve the problem:
- To find the sum of repeating elements (let’s say X and Y) subtract the sum of the first N natural numbers from the total sum of the array i.e. X + Y = sum(arr) – N*(N + 1) / 2
- Now, finding the product of repeating elements that is X*Y = P / N!, where P is the product of all elements in the array.
- For X – Y = sqrt((X+Y)2 – 4*XY)
- By solving these two equations X and Y will be calculated.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int fact(int n);
void printRepeating(int arr[], int size)
{
int S = 0;
int P = 1;
int x, y;
int D;
int n = size - 2, i;
for (i = 0; i < size; i++) {
S = S + arr[i];
P = P * arr[i];
}
S = S - n * (n + 1) / 2;
P = P / fact(n);
D = sqrt(S * S - 4 * P);
x = (D + S) / 2;
y = (S - D) / 2;
cout << "Repeating elements are " << x << " " << y;
}
int fact(int n) { return (n == 0) ? 1 : n * fact(n - 1); }
int main()
{
int arr[] = { 4, 2, 4, 5, 2, 3, 1 };
int arr_size = sizeof(arr) / sizeof(arr[0]);
printRepeating(arr, arr_size);
return 0;
}
|
Java
import java.io.*;
class RepeatElement {
void printRepeating(int arr[], int size)
{
int S = 0;
int P = 1;
int x, y;
int D;
int n = size - 2, i;
for (i = 0; i < size; i++) {
S = S + arr[i];
P = P * arr[i];
}
S = S - n * (n + 1) / 2;
P = P / fact(n);
D = (int)Math.sqrt(S * S - 4 * P);
x = (D + S) / 2;
y = (S - D) / 2;
System.out.println(
"The two repeating elements are :");
System.out.print(x + " " + y);
}
int fact(int n)
{
return (n == 0) ? 1 : n * fact(n - 1);
}
public static void main(String[] args)
{
RepeatElement repeat = new RepeatElement();
int arr[] = { 4, 2, 4, 5, 2, 3, 1 };
int arr_size = arr.length;
repeat.printRepeating(arr, arr_size);
}
}
|
Python3
import math
def printRepeating(arr, size):
S = 0
P = 1
n = size - 2
for i in range(0, size):
S = S + arr[i]
P = P * arr[i]
S = S - n * (n + 1) // 2
P = P // fact(n)
D = math.sqrt(S * S - 4 * P)
x = (D + S) // 2
y = (S - D) // 2
print("Repeating elements are ",
(int)(x), " ", (int)(y))
def fact(n):
if(n == 0):
return 1
else:
return(n * fact(n - 1))
arr = [4, 2, 4, 5, 2, 3, 1]
arr_size = len(arr)
printRepeating(arr, arr_size)
|
C#
using System;
class GFG {
static void printRepeating(int[] arr, int size)
{
int S = 0;
int P = 1;
int x, y;
int D;
int n = size - 2, i;
for (i = 0; i < size; i++) {
S = S + arr[i];
P = P * arr[i];
}
S = S - n * (n + 1) / 2;
P = P / fact(n);
D = (int)Math.Sqrt(S * S - 4 * P);
x = (D + S) / 2;
y = (S - D) / 2;
Console.Write("Repeating elements are ");
Console.Write(x + " " + y);
}
static int fact(int n)
{
return (n == 0) ? 1 : n * fact(n - 1);
}
public static void Main()
{
int[] arr = { 4, 2, 4, 5, 2, 3, 1 };
int arr_size = arr.Length;
printRepeating(arr, arr_size);
}
}
|
PHP
<?php
function fact($n){
return ($n == 0)? 1 : $n*fact($n-1);
}
function printRepeating($arr, $size)
{
$S = 0;
$P = 1;
$x; $y;
$D;
$n = $size - 2;
for($i = 0; $i < $size; $i++)
{
$S = $S + $arr[$i];
$P = $P*$arr[$i];
}
$S = $S - $n*($n+1)/2;
$P = $P/fact($n);
$D = sqrt($S*$S - 4*$P);
$x = ($D + $S)/2;
$y = ($S - $D)/2;
echo "Repeating elements are ".$x." ".$y;
}
$arr = array(4, 2, 4, 5, 2, 3, 1);
$arr_size = count($arr);
printRepeating($arr, $arr_size);
?>
|
Javascript
<script>
function printRepeating(arr , size)
{
var S = 0;
var P = 1;
var x, y;
var D;
var n = size - 2, i;
for (i = 0; i < size; i++)
{
S = S + arr[i];
P = P * arr[i];
}
S = S - n * parseInt((n + 1) / 2);
P = parseInt(P / fact(n));
D = parseInt( Math.sqrt(S * S - 4 * P));
x = parseInt((D + S) / 2);
y = parseInt((S - D) / 2);
document.write("Repeating elements are " + x + " " + y);
}
function fact(n)
{ var ans = false;
if(n == 0)
return 1;
else
return(n * fact(n - 1));
}
var arr = [4, 2, 4, 5, 2, 3, 1];
var arr_size = arr.length;
printRepeating(arr, arr_size);
</script>
|
Output
Repeating elements are 4 2
Time Complexity: O(N), Traversing over the array of size N to find the sum and product of the array
Auxiliary Space: O(1)
Find the two repeating elements in a given array using XOR:
The idea is to use find the XOR of the repeating elements and then find the repeating elements.
Follow the steps below to solve the problem:
- To find the XOR of repeating elements XOR all the elements of the array and then XOR that result with XOR of the first N natural numbers.
- Now, find the rightmost set bit of X^Y to divide the array on the basis of the rightmost set bit.
- Let’s say x = 0 and y = 0, XOR all the numbers in the array with x whose bit is set at the rightmost position of X^Y and those numbers whose bit is not set at that position then XOR those numbers with y.
- after this run a loop from 1 to N and check whose bit is set at that rightmost position of X^Y then XOR that number with x, otherwise XOR the number with y.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void printRepeating(int arr[], int size)
{
int Xor = arr[0];
int set_bit_no;
int i;
int n = size - 2;
int x = 0, y = 0;
for (i = 1; i < size; i++)
Xor ^= arr[i];
for (i = 1; i <= n; i++)
Xor ^= i;
set_bit_no = Xor & ~(Xor - 1);
for (i = 0; i < size; i++) {
if (arr[i] & set_bit_no)
x = x ^ arr[i];
else
y = y ^ arr[i];
}
for (i = 1; i <= n; i++) {
if (i & set_bit_no)
x = x ^ i;
else
y = y ^ i;
}
cout << "Repeating elements are " << y << " " << x;
}
int main()
{
int arr[] = { 4, 2, 4, 5, 2, 3, 1 };
int arr_size = sizeof(arr) / sizeof(arr[0]);
printRepeating(arr, arr_size);
return 0;
}
|
C
#include <stdio.h>
void printRepeating(int arr[], int size)
{
int xor = arr[0];
int set_bit_no;
int i;
int n = size - 2;
int x = 0, y = 0;
for (i = 1; i < size; i++)
xor ^= arr[i];
for (i = 1; i <= n; i++)
xor ^= i;
set_bit_no = xor&~(xor-1);
for (i = 0; i < size; i++) {
if (arr[i] & set_bit_no)
x = x ^ arr[i];
else
y = y ^ arr[i];
}
for (i = 1; i <= n; i++) {
if (i & set_bit_no)
x = x ^ i;
else
y = y ^ i;
}
printf("Repeating elements are %d %d ", y, x);
}
int main()
{
int arr[] = { 4, 2, 4, 5, 2, 3, 1 };
int arr_size = sizeof(arr) / sizeof(arr[0]);
printRepeating(arr, arr_size);
getchar();
return 0;
}
|
Java
import java.io.*;
class RepeatElement {
void printRepeating(int arr[], int size)
{
int xor = arr[0];
int set_bit_no;
int i;
int n = size - 2;
int x = 0, y = 0;
for (i = 1; i < size; i++)
xor ^= arr[i];
for (i = 1; i <= n; i++)
xor ^= i;
set_bit_no = (xor & ~(xor - 1));
for (i = 0; i < size; i++) {
int a = arr[i] & set_bit_no;
if (a != 0)
x = x
^ arr[i];
else
y = y ^ arr[i];
}
for (i = 1; i <= n; i++) {
int a = i & set_bit_no;
if (a != 0)
x = x ^ i;
else
y = y ^ i;
}
System.out.print("Repeating elements are ");
System.out.println(y + " " + x);
}
public static void main(String[] args)
{
RepeatElement repeat = new RepeatElement();
int arr[] = { 4, 2, 4, 5, 2, 3, 1 };
int arr_size = arr.length;
repeat.printRepeating(arr, arr_size);
}
}
|
Python3
def printRepeating(arr, size):
xor = arr[0]
n = size - 2
x = 0
y = 0
for i in range(1, size):
xor ^= arr[i]
for i in range(1, n + 1):
xor ^= i
set_bit_no = xor & ~(xor-1)
for i in range(0, size):
if(arr[i] & set_bit_no):
x = x ^ arr[i]
else:
y = y ^ arr[i]
for i in range(1, n + 1):
if(i & set_bit_no):
x = x ^ i
else:
y = y ^ i
print("Repeating elements are", y, x)
arr = [4, 2, 4,
5, 2, 3, 1]
arr_size = len(arr)
printRepeating(arr, arr_size)
|
C#
using System;
class GFG {
static void printRepeating(int[] arr, int size)
{
int xor = arr[0];
int set_bit_no;
int i;
int n = size - 2;
int x = 0, y = 0;
for (i = 1; i < size; i++)
xor ^= arr[i];
for (i = 1; i <= n; i++)
xor ^= i;
set_bit_no = (xor & ~(xor - 1));
for (i = 0; i < size; i++) {
int a = arr[i] & set_bit_no;
if (a != 0)
x = x ^ arr[i];
else
y = y ^ arr[i];
}
for (i = 1; i <= n; i++) {
int a = i & set_bit_no;
if (a != 0)
x = x ^ i;
else
y = y ^ i;
}
Console.Write("Repeating elements are ");
Console.Write(y + " " + x);
}
public static void Main()
{
int[] arr = { 4, 2, 4, 5, 2, 3, 1 };
int arr_size = arr.Length;
printRepeating(arr, arr_size);
}
}
|
PHP
<?php
function printRepeating( $arr, $size)
{
$xor = $arr[0];
$set_bit_no;
$i;
$n = $size - 2;
$x = 0; $y = 0;
for($i = 1; $i < $size; $i++)
$xor ^= $arr[$i];
for($i = 1; $i <= $n; $i++)
$xor ^= $i;
$set_bit_no = $xor & ~($xor-1);
for($i = 0; $i < $size; $i++)
{
if($arr[$i] & $set_bit_no)
$x = $x ^ $arr[$i];
else
$y = $y ^ $arr[$i];
}
for($i = 1; $i <= $n; $i++)
{
if($i & $set_bit_no)
$x = $x ^ $i;
else
$y = $y ^ $i;
}
echo "Repeating elements are ";
echo $y. " ".$x;
}
$arr = array(4, 2, 4, 5, 2, 3, 1);
$arr_size = count($arr);
printRepeating($arr, $arr_size);
?>
|
Javascript
<script>
function printRepeating( arr, size)
{
let Xor = arr[0];
let set_bit_no;
let i;
let n = size - 2;
let x = 0, y = 0;
for(i = 1; i < size; i++)
Xor ^= arr[i];
for(i = 1; i <= n; i++)
Xor ^= i;
set_bit_no = Xor & ~(Xor-1);
for(i = 0; i < size; i++)
{
if(arr[i] & set_bit_no)
x = x ^ arr[i];
else
y = y ^ arr[i];
}
for(i = 1; i <= n; i++)
{
if(i & set_bit_no)
x = x ^ i;
else
y = y ^ i;
}
document.write("Repeating elements are "+y+" "+x);
}
let arr = [4, 2, 4, 5, 2, 3, 1];
let arr_size = arr.length;
printRepeating(arr, arr_size);
</script>
|
Output
Repeating elements are 4 2
Time Complexity: O(N), Linear Traversals are performed over the array of size N.
Auxiliary Space: O(1)
Find the two repeating elements in a given array using Array Elements as an Index:
The idea is to use the original array to mark the elements in the array by making them negative and when an index is found which is already marked then that index would be our possible answer.
Follow the steps below to solve the problem:
- Traverse over the array and go to index arr[i] and make it negative.
- If that number is already negative then that index(i.e. arr[i]) is a repeating element.
- Take absolute value while marking indexes may be the current number is negative.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void printRepeating(int arr[], int size)
{
int i;
cout << "Repeating elements are ";
for (i = 0; i < size; i++) {
if (arr[abs(arr[i])] > 0)
arr[abs(arr[i])] = -arr[abs(arr[i])];
else
cout << abs(arr[i]) << " ";
}
}
int main()
{
int arr[] = { 4, 2, 4, 5, 2, 3, 1 };
int arr_size = sizeof(arr) / sizeof(arr[0]);
printRepeating(arr, arr_size);
return 0;
}
|
C
#include <stdio.h>
#include <stdlib.h>
void printRepeating(int arr[], int size)
{
int i;
printf("Repeating elements are");
for (i = 0; i < size; i++) {
if (arr[abs(arr[i])] > 0)
arr[abs(arr[i])] = -arr[abs(arr[i])];
else
printf(" %d", abs(arr[i]));
}
}
int main()
{
int arr[] = { 4, 2, 4, 5, 2, 3, 1 };
int arr_size = sizeof(arr) / sizeof(arr[0]);
printRepeating(arr, arr_size);
getchar();
return 0;
}
|
Java
import java.io.*;
class RepeatElement {
void printRepeating(int arr[], int size)
{
int i;
System.out.print("Repeating elements are ");
for (i = 0; i < size; i++) {
int abs_val = Math.abs(arr[i]);
if (arr[abs_val] > 0)
arr[abs_val] = -arr[abs_val];
else
System.out.print(abs_val + " ");
}
}
public static void main(String[] args)
{
RepeatElement repeat = new RepeatElement();
int arr[] = { 4, 2, 4, 5, 2, 3, 1 };
int arr_size = arr.length;
repeat.printRepeating(arr, arr_size);
}
}
|
Python3
def printRepeating(arr, size):
print("Repeating elements are", end=" ")
for i in range(0, size):
if(arr[abs(arr[i])] > 0):
arr[abs(arr[i])] = (-1) * arr[abs(arr[i])]
else:
print(abs(arr[i]), end=" ")
arr = [4, 2, 4, 5, 2, 3, 1]
arr_size = len(arr)
printRepeating(arr, arr_size)
|
C#
using System;
class GFG {
static void printRepeating(int[] arr, int size)
{
int i;
Console.Write("Repeating elements are ");
for (i = 0; i < size; i++) {
int abs_val = Math.Abs(arr[i]);
if (arr[abs_val] > 0)
arr[abs_val] = -arr[abs_val];
else
Console.Write(abs_val + " ");
}
}
public static void Main()
{
int[] arr = { 4, 2, 4, 5, 2, 3, 1 };
int arr_size = arr.Length;
printRepeating(arr, arr_size);
}
}
|
PHP
<?php
function printRepeating($arr, $size)
{
$i;
echo "Repeating elements are"," ";
for($i = 0; $i < $size; $i++)
{
if($arr[abs($arr[$i])] > 0)
$arr[abs($arr[$i])] = -$arr[abs($arr[$i])];
else
echo abs($arr[$i])," ";
}
}
$arr = array (4, 2, 4, 5, 2, 3, 1);
$arr_size = sizeof($arr);
printRepeating($arr, $arr_size);
#This code is contributed by aj_36
?>
|
Javascript
<script>
function printRepeating(arr , size)
{
var i;
document.write("Repeating elements are ");
for(i = 0; i < size; i++)
{
var abs_val = Math.abs(arr[i]);
if(arr[abs_val] > 0)x
arr[abs_val] = -arr[abs_val];
else
document.write(abs_val + " ");
}
}
var arr = [4, 2, 4, 5, 2, 3, 1];
var arr_size = arr.length;
printRepeating(arr, arr_size);
</script>
|
Output
Repeating elements are 4 2
Time Complexity: O(N), Traversing over the array of size N.
Auxiliary Space: O(1)
Note: This method modifies the original array and may not be a recommended method if it’s not allowed to modify the array.
Find the two repeating elements in a given array by Modifying Original Array:
The idea is to increment every element at (arr[i] – 1)th index by N-1 (as the elements are present up to N-2 only)
Follow the steps below to solve the problem:
- incrementing each element at (arr[i]th-1) index by N-1
- so when this incremented element is divided by N-1 it will give a number of times we have added N-1 to it,
- check if the element at that index when divided by (N-1) gives 2
- If this is true then it means that the element has appeared twice
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void twoRepeated(int arr[], int N)
{
int m = N - 1;
for (int i = 0; i < N; i++) {
arr[arr[i] % m - 1] += m;
if ((arr[arr[i] % m - 1] / m) == 2)
cout << arr[i] % m << " ";
}
}
int main()
{
int arr[] = { 4, 2, 4, 5, 2, 3, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Repeating elements are ";
twoRepeated(arr, n);
return 0;
}
|
C
#include <stdio.h>
void twoRepeated(int arr[], int N)
{
int m = N - 1;
for (int i = 0; i < N; i++) {
arr[arr[i] % m - 1] += m;
if ((arr[arr[i] % m - 1] / m) == 2)
printf("%d ", arr[i] % m);
}
}
int main()
{
int arr[] = { 4, 2, 4, 5, 2, 3, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
printf("Repeating elements are ");
twoRepeated(arr, n);
return 0;
}
|
Java
import java.io.*;
class GFG {
public static void twoRepeated(int arr[], int N)
{
int m = N - 1;
for (int i = 0; i < N; i++) {
arr[arr[i] % m - 1] += m;
if ((arr[arr[i] % m - 1] / m) == 2)
System.out.print(arr[i] % m + " ");
}
}
public static void main(String[] args)
{
int arr[] = { 4, 2, 4, 5, 2, 3, 1 };
int n = 7;
System.out.print("Repeating elements are ");
twoRepeated(arr, n);
}
}
|
Python3
def twoRepeated(arr, N):
m = N - 1
for i in range(N):
arr[arr[i] % m - 1] += m
if ((arr[arr[i] % m - 1] // m) == 2):
print(arr[i] % m, end=" ")
arr = [4, 2, 4, 5, 2, 3, 1]
n = len(arr)
print("Repeating elements are ", end="")
twoRepeated(arr, n)
|
C#
using System;
public class GFG {
public static void twoRepeated(int[] arr, int N)
{
int m = N - 1;
for (int i = 0; i < N; i++) {
arr[arr[i] % m - 1] += m;
if ((arr[arr[i] % m - 1] / m) == 2)
Console.Write(arr[i] % m + " ");
}
}
public static void Main(String[] args)
{
int[] arr = { 4, 2, 4, 5, 2, 3, 1 };
int n = 7;
Console.Write("Repeating elements are ");
twoRepeated(arr, n);
}
}
|
Javascript
<script>
function twoRepeated(arr, N)
{
let m = N - 1;
for(let i = 0; i < N; i++)
{
arr[parseInt(arr[i] % m) - 1] += m;
if (parseInt(arr[parseInt(arr[i] % m) - 1] / m) == 2)
document.write(parseInt(arr[i] % m) + " ");
}
}
var arr = [ 4, 2, 4, 5, 2, 3, 1 ];
var n = 7;
document.write("Repeating elements are ");
twoRepeated(arr, n);
</script>
|
Output
Repeating elements are 4 2
Time Complexity: O(N), Traversing over the array of size N.
Auxiliary Space: O(1)
Find the two repeating elements in a given array using Hash Set:
The idea is to use a set, insert the elements in the set, and check simultaneously whether that is already present the set or not.
Follow the steps below to solve the problem:
- Traverse over the array and check whether the ith element is present in the set or not.
- If it present then print that element
- Otherwise that element into the set.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void printRepeating(int arr[], int size)
{
unordered_set<int> s;
cout << "Repeating elements are ";
for (int i = 0; i < size; i++) {
if (s.empty() == false && s.find(arr[i]) != s.end())
cout << arr[i] << " ";
s.insert(arr[i]);
}
}
int main()
{
int arr[] = { 4, 2, 4, 5, 2, 3, 1 };
int arr_size = sizeof(arr) / sizeof(arr[0]);
printRepeating(arr, arr_size);
return 0;
}
|
Java
import java.util.*;
class GFG {
static void printRepeating(int arr[], int size)
{
HashSet<Integer> s = new HashSet<>();
System.out.print("Repeating elements are ");
for (int i = 0; i < size; i++) {
if (!s.isEmpty() && s.contains(arr[i]))
System.out.print(arr[i] + " ");
s.add(arr[i]);
}
}
public static void main(String[] args)
{
int arr[] = { 4, 2, 4, 5, 2, 3, 1 };
int arr_size = arr.length;
printRepeating(arr, arr_size);
}
}
|
Python3
def printRepeating(arr, size):
s = set()
print("Repeating elements are ", end="")
for i in range(size):
if (len(s) and arr[i] in s):
print(arr[i], end=" ")
s.add(arr[i])
arr = [4, 2, 4, 5, 2, 3, 1]
arr_size = len(arr)
printRepeating(arr, arr_size)
|
C#
using System;
using System.Collections.Generic;
public class GFG {
static void printRepeating(int[] arr, int size)
{
HashSet<int> s = new HashSet<int>();
Console.Write("Repeating elements are ");
for (int i = 0; i < size; i++) {
if (s.Count != 0 && s.Contains(arr[i]))
Console.Write(arr[i] + " ");
s.Add(arr[i]);
}
}
public static void Main()
{
int[] arr = { 4, 2, 4, 5, 2, 3, 1 };
int arr_size = arr.Length;
printRepeating(arr, arr_size);
}
}
|
Javascript
<script>
function printRepeating(arr, size)
{
const s = new Set();
document.write("Repeating elements are ");
for(let i = 0; i < size; i++)
{
if(s.size != 0 && s.has(arr[i]))
document.write(arr[i] + " ");
s.add(arr[i]);
}
}
var arr = [ 4, 2, 4, 5, 2, 3, 1 ];
var arr_size = 7;
printRepeating(arr, arr_size);
</script>
|
Output
Repeating elements are 4 2
Time Complexity: O(N), Traversing over the array of size N.
Auxiliary Space: O(N), Space used to store the elements in the hash set
Find the two repeating elements in a given array using Sorting:
The idea is to sort the given array and then traverse it. While traversing find if two consecutive elements are equal then that is the repeating element.
Code-
C++
#include <bits/stdc++.h>
using namespace std;
void printRepeating(int arr[], int size)
{
sort(arr,arr+size);
cout << "Repeating elements are ";
for (int i = 0; i < size-1; i++) {
if (arr[i]==arr[i+1]){
cout << arr[i] << " ";
}
}
}
int main()
{
int arr[] = { 4, 2, 4, 5, 2, 3, 1 };
int arr_size = sizeof(arr) / sizeof(arr[0]);
printRepeating(arr, arr_size);
return 0;
}
|
Java
import java.util.Arrays;
public class Main {
static void printRepeating(int arr[], int size) {
Arrays.sort(arr);
System.out.print("Repeating elements are ");
for (int i = 0; i < size-1; i++) {
if (arr[i] == arr[i+1]) {
System.out.print(arr[i] + " ");
}
}
}
public static void main(String[] args) {
int arr[] = { 4, 2, 4, 5, 2, 3, 1 };
int arr_size = arr.length;
printRepeating(arr, arr_size);
}
}
|
Python3
def printRepeating(arr):
arr.sort()
print("Repeating elements are ", end="")
for i in range(len(arr) - 1):
if arr[i] == arr[i + 1]:
print(arr[i], end=" ")
if __name__ == "__main__":
arr = [4, 2, 4, 5, 2, 3, 1]
printRepeating(arr)
|
Javascript
function printRepeating(arr) {
arr.sort();
process.stdout.write("Repeating elements are ");
for (let i = 0; i < arr.length - 1; i++) {
if (arr[i] == arr[i + 1]) {
process.stdout.write(arr[i] + " ");
}
}
}
let arr = [4, 2, 4, 5, 2, 3, 1];
printRepeating(arr);
|
C#
using System;
using System.Linq;
class Program
{
static void PrintRepeating(int[] arr, int size)
{
Array.Sort(arr);
Console.Write("Repeating elements are ");
for (int i = 0; i < size - 1; i++)
{
if (arr[i] == arr[i + 1])
{
Console.Write(arr[i] + " ");
}
}
}
static void Main(string[] args)
{
int[] arr = { 4, 2, 4, 5, 2, 3, 1 };
int arrSize = arr.Length;
PrintRepeating(arr, arrSize);
}
}
|
Output
Repeating elements are 2 4
Time Complexity: O(NlogN), because of sorting
Auxiliary Space: O(1)
Please write comments if you find the above codes/algorithms incorrect, or find better ways to solve the same problem.
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