Given a sorted array of n elements containing elements in range from 1 to n-1 i.e. one element occurs twice, the task is to find the repeating element in an array.
Examples :
Input : arr[] = { 1, 2 , 3 , 4 , 4} Output : 4 Input : arr[] = { 1 , 1 , 2 , 3 , 4} Output : 1
Brute Force:
- Traverse the input array using a for a loop.
- For each element in the array, traverse the remaining part of the array using another for loop.
- For each subsequent element, check if it is equal to the current element.
- If a match is found, return the current element.
- If no match is found, continue with the next element in the outer loop.
- If the outer loop completes without finding a match, return -1 to indicate that there is no repeating element in the array.
Below is the implementation of the above approach:
C++
// C++ program to find the only repeating element in an // array of size n and elements from range 1 to n-1. #include <bits/stdc++.h> using namespace std;
// Returns index of second appearance of a repeating element // The function assumes that array elements are in range from // 1 to n-1. int FindRepeatingElement( int arr[], int size){
for ( int i=0; i<size; i++){
for ( int j=i+1; j<size; j++){
if (arr[i] == arr[j])
return i;
}
}
return -1;
} // Driver code int main()
{ int arr[] = {1, 2 , 3 , 4 , 4};
int n = sizeof (arr) / sizeof (arr[0]);
int index = FindRepeatingElement(arr, n);
if (index != -1)
cout << arr[index];
return 0;
} |
Java
import java.util.*;
public class Main {
// Returns index of second appearance of a repeating element
// The function assumes that array elements are in range from
// 1 to n-1.
public static int findRepeatingElement( int arr[], int size){
for ( int i= 0 ; i<size; i++){
for ( int j=i+ 1 ; j<size; j++){
if (arr[i] == arr[j])
return i;
}
}
return - 1 ;
}
// Driver code
public static void main(String[] args) {
int arr[] = { 1 , 2 , 3 , 4 , 4 };
int n = arr.length;
int index = findRepeatingElement(arr, n);
if (index != - 1 )
System.out.println(arr[index]);
}
} |
Python3
# Python3 program to find the only repeating element in an # array of size n and elements from range 1 to n-1. # Returns second appearance of a repeating element # The function assumes that array elements are in range from # 1 to n-1. def FindRepeatingElement(arr, size):
for i in range (size):
for j in range (i + 1 , size):
if arr[i] = = arr[j]:
return arr[i]
return - 1
# Driver code if __name__ = = '__main__' :
arr = [ 1 , 2 , 3 , 4 , 4 ]
n = len (arr)
element = FindRepeatingElement(arr, n)
if element ! = - 1 :
print (element)
|
C#
using System;
public class Program
{ // Returns index of second appearance of a repeating element
// The function assumes that array elements are in range from
// 1 to n-1.
public static int FindRepeatingElement( int [] arr, int size)
{
for ( int i = 0; i < size; i++)
{
for ( int j = i + 1; j < size; j++)
{
if (arr[i] == arr[j])
{
return i;
}
}
}
return -1;
}
// Driver code
public static void Main()
{
int [] arr = { 1, 2, 3, 4, 4 };
int n = arr.Length;
int index = FindRepeatingElement(arr, n);
if (index != -1)
{
Console.WriteLine(arr[index]);
}
}
} |
Javascript
// JavaScript program to find the only repeating element in an // array of size n and elements from range 1 to n-1. // Returns index of second appearance of a repeating element // The function assumes that array elements are in range from // 1 to n-1. function FindRepeatingElement(arr, size)
{ for (let i=0; i<size; i++)
{
for (let j=i+1; j<size; j++)
{
if (arr[i] == arr[j])
return i;
}
}
return -1;
} // Driver code let arr = [1, 2 , 3 , 4 , 4]; let n = arr.length; let index = FindRepeatingElement(arr, n); if (index != -1)
console.log(arr[index]); // This code is contributed by akashish__ |
Output
4
Time Complexity: O(N^2)
Auxiliary Space: O(1)
An efficient method is to use Binary Search.
Observation: If an element ‘X’ is repeating, then it must be at index ‘X’ in the array. So the problem reduces to find any element whose value is same as its index.
C++
// C++ program to find the only repeating element in an // array of size n and elements from range 1 to n-1. #include <bits/stdc++.h> using namespace std;
// Returns index of second appearance of a repeating element // The function assumes that array elements are in range from // 1 to n-1. int FindRepeatingElement( int arr[], int size){
int lo = 0;
int hi = size - 1;
int mid;
while (lo <= hi){
mid = (lo+hi)/2;
if (arr[mid] <= mid){
hi = mid-1;
}
else {
lo = mid + 1;
}
}
return lo;
} // Driver code int main()
{ int arr[] = {1, 2, 3, 3, 4, 5};
int n = sizeof (arr) / sizeof (arr[0]);
int index = FindRepeatingElement(arr, n);
if (index != -1)
cout << arr[index];
return 0;
} |
Java
// Java program to find the only repeating element in an // array of size n and elements from range 1 to n-1. class Test
{ // Returns index of second appearance of a repeating element
// The function assumes that array elements are in range from
// 1 to n-1.
static int findRepeatingElement( int arr[], int low, int high)
{
// low = 0 , high = n-1;
if (low > high)
return - 1 ;
int mid = (low + high) / 2 ;
// Check if the mid element is the repeating one
if (arr[mid] != mid + 1 )
{
if (mid > 0 && arr[mid]==arr[mid- 1 ])
return mid;
// If mid element is not at its position that means
// the repeated element is in left
return findRepeatingElement(arr, low, mid- 1 );
}
// If mid is at proper position then repeated one is in
// right.
return findRepeatingElement(arr, mid+ 1 , high);
}
// Driver method
public static void main(String[] args)
{
int arr[] = { 1 , 2 , 3 , 3 , 4 , 5 };
int index = findRepeatingElement(arr, 0 , arr.length- 1 );
if (index != - 1 )
System.out.println(arr[index]);
}
} |
Python3
# Python program to find the only repeating element in an # array of size n and elements from range 1 to n-1 # Returns index of second appearance of a repeating element # The function assumes that array elements are in range from # 1 to n-1. def findRepeatingElement(arr, low, high):
# low = 0 , high = n-1
if low > high:
return - 1
mid = (low + high) / / 2
# Check if the mid element is the repeating one
if (arr[mid] ! = mid + 1 ):
if (mid > 0 and arr[mid] = = arr[mid - 1 ]):
return mid
# If mid element is not at its position that means
# the repeated element is in left
return findRepeatingElement(arr, low, mid - 1 )
# If mid is at proper position then repeated one is in
# right.
return findRepeatingElement(arr, mid + 1 , high)
# Driver code arr = [ 1 , 2 , 3 , 3 , 4 , 5 ]
n = len (arr)
index = findRepeatingElement(arr, 0 , n - 1 )
if (index is not - 1 ):
print (arr[index])
#This code is contributed by Afzal Ansari |
Javascript
<script> // JavaScript program to find the only repeating element in an
// array of size n and elements from range 1 to n-1.
// Returns index of second appearance of a repeating element
// The function assumes that array elements are in range from
// 1 to n-1.
function findRepeatingElement(arr, low, high)
{
// low = 0 , high = n-1;
if (low > high) return -1;
var mid = parseInt((low + high) / 2);
// Check if the mid element is the repeating one
if (arr[mid] != mid + 1)
{
if (mid > 0 && arr[mid] == arr[mid - 1]) return mid;
// If mid element is not at its position that means
// the repeated element is in left
return findRepeatingElement(arr, low, mid - 1);
}
// If mid is at proper position then repeated one is in
// right.
return findRepeatingElement(arr, mid + 1, high);
}
// Driver code
var arr = [1, 2, 3, 3, 4, 5];
var n = arr.length;
var index = findRepeatingElement(arr, 0, n - 1);
if (index != -1) document.write(arr[index]);
// This code is contributed by rdtank.
</script>
|
C#
// C# program to find the only repeating // element in an array of size n and // elements from range 1 to n-1. using System;
class Test
{ // Returns index of second appearance of a
// repeating element. The function assumes that
// array elements are in range from 1 to n-1.
static int findRepeatingElement( int []arr, int low,
int high)
{
// low = 0 , high = n-1;
if (low > high)
return -1;
int mid = (low + high) / 2;
// Check if the mid element
// is the repeating one
if (arr[mid] != mid + 1)
{
if (mid > 0 && arr[mid]==arr[mid-1])
return mid;
// If mid element is not at its position
// that means the repeated element is in left
return findRepeatingElement(arr, low, mid-1);
}
// If mid is at proper position
// then repeated one is in right.
return findRepeatingElement(arr, mid+1, high);
}
// Driver method
public static void Main()
{
int []arr = {1, 2, 3, 3, 4, 5};
int index = findRepeatingElement(arr, 0, arr.Length-1);
if (index != -1)
Console.Write(arr[index]);
}
} // This code is contributed by Nitin Mittal. |
PHP
<?php // PHP program to find the only // repeating element in an array // of size n and elements from // range 1 to n-1. // Returns index of second // appearance of a repeating // element. The function assumes // that array elements are in // range from 1 to n-1. function findRepeatingElement( $arr ,
$low ,
$high )
{ // low = 0 , high = n-1;
if ( $low > $high )
return -1;
$mid = floor (( $low + $high ) / 2);
// Check if the mid element
// is the repeating one
if ( $arr [ $mid ] != $mid + 1)
{
if ( $mid > 0 && $arr [ $mid ] ==
$arr [ $mid - 1])
return $mid ;
// If mid element is not at
// its position that means
// the repeated element is in left
return findRepeatingElement( $arr , $low ,
$mid - 1);
}
// If mid is at proper position
// then repeated one is in right.
return findRepeatingElement( $arr , $mid + 1,
$high );
} // Driver code $arr = array (1, 2, 3, 3, 4, 5);
$n = sizeof( $arr );
$index = findRepeatingElement( $arr , 0,
$n - 1);
if ( $index != -1)
echo $arr [ $index ];
// This code is contributed // by nitin mittal. ?> |
Output
3
Time Complexity : O(log n)
Space Complexity: O(1)