Find any one of the multiple repeating elements in read only array | Set 2
Given a read-only array arr[] of size N + 1, find one of the multiple repeating elements in the array where the array contains integers only between 1 and N.
Note: Read-only array means that the contents of the array can’t be modified.
Examples:
Input: N = 5, arr[] = {1, 1, 2, 3, 5, 4}
Output: 1
Explanation:
1 is the only number repeated in the array.
Input: N = 10, arr[] = {10, 1, 2, 3, 5, 4, 9, 8, 5, 6, 4}
Output: 5
Explanation:
5 is the one of the number repeated in the array.
In the previous post, we have discussed the same article with a space complexity O(N) and O(sqrt(N)).
Approach: This approach is based on Floyd’s Tortoise and Hare Algorithm (Cycle Detection Algorithm).
- Use the function f(x) = arr[x] to construct the sequence:
arr[0], arr[arr[0]], arr[arr[arr[0]]], arr[arr[arr[arr[0]]]] …….
- Each new element in the sequence is an element in arr[] at the index of the previous element.
- Starting from x = arr[0], it will produce a linked list with a cycle.
- The cycle appears because arr[] contains duplicate elements(at least one). The duplicate value is an entrance to the cycle. Given below is an example to show how cycle exists:
For Example: Let the array arr[] = {2, 6, 4, 1, 3, 1, 5}
index |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
arr |
2 |
6 |
4 |
1 |
3 |
1 |
5 |
Starting from index 0, the traversal looks as follows:
arr[0] = 2 –> arr[2] = 4 –> arr[4] = 3 –> arr[3] = 1 –> arr[1] = 6 –> arr[6] = 5 –> arr[5] = 1.
The sequence forms cycle as shown below:
- Algorithm consists of two parts and uses two pointers, usually called tortoise and hare.
- hare = arr[arr[hare]] is twice as fast as tortoise = arr[tortoise].
- Since the hare goes fast, it would be the first one who enters the cycle and starts to run around the cycle.
- At some point, the tortoise enters the cycle as well, and since it’s moving slower the hare catches the tortoise up at some intersection point.
- Note that the intersection point is not the cycle entrance in the general case, but the two intersect at somewhere middle in cycle.
- Move tortoise to the starting point of sequence and hare remains within cycle and both move with the same speed i.e. tortoise = arr[tortoise] and hare = arr[hare]. Now they intersect at duplicate element.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void findDuplicate( int arr[])
{
int tortoise = arr[0];
int hare = arr[0];
while (1) {
tortoise = arr[tortoise];
hare = arr[arr[hare]];
if (tortoise == hare)
break ;
}
tortoise = arr[0];
while (tortoise != hare) {
tortoise = arr[tortoise];
hare = arr[hare];
}
cout << tortoise;
}
int main()
{
int arr[] = { 2, 6, 4, 1, 3, 1, 5 };
findDuplicate(arr);
return 0;
}
|
Java
class GFG{
static void findDuplicate( int arr[])
{
int tortoise = arr[ 0 ];
int hare = arr[ 0 ];
while ( true )
{
tortoise = arr[tortoise];
hare = arr[arr[hare]];
if (tortoise == hare)
break ;
}
tortoise = arr[ 0 ];
while (tortoise != hare)
{
tortoise = arr[tortoise];
hare = arr[hare];
}
System.out.print(tortoise);
}
public static void main (String []args)
{
int arr[] = { 2 , 6 , 4 , 1 , 3 , 1 , 5 };
findDuplicate(arr);
}
}
|
Python3
def findDuplicate(arr):
tortoise = arr[ 0 ]
hare = arr[ 0 ]
while ( 1 ):
tortoise = arr[tortoise]
hare = arr[arr[hare]]
if (tortoise = = hare):
break
tortoise = arr[ 0 ]
while (tortoise ! = hare):
tortoise = arr[tortoise]
hare = arr[hare]
print (tortoise)
arr = [ 2 , 6 , 4 , 1 , 3 , 1 , 5 ]
findDuplicate(arr)
|
C#
using System;
class GFG{
static void findDuplicate( int []arr)
{
int tortoise = arr[0];
int hare = arr[0];
while ( true )
{
tortoise = arr[tortoise];
hare = arr[arr[hare]];
if (tortoise == hare)
break ;
}
tortoise = arr[0];
while (tortoise != hare)
{
tortoise = arr[tortoise];
hare = arr[hare];
}
Console.Write(tortoise);
}
public static void Main(String []args)
{
int []arr = { 2, 6, 4, 1, 3, 1, 5 };
findDuplicate(arr);
}
}
|
Javascript
<script>
function findDuplicate(arr)
{
let tortoise = arr[0];
let hare = arr[0];
while ( true )
{
tortoise = arr[tortoise];
hare = arr[arr[hare]];
if (tortoise == hare)
break ;
}
tortoise = arr[0];
while (tortoise != hare)
{
tortoise = arr[tortoise];
hare = arr[hare];
}
document.write(tortoise);
}
let arr = [ 2, 6, 4, 1, 3, 1, 5 ];
findDuplicate(arr);
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(1)
Last Updated :
22 Oct, 2021
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