Find duplicates in constant array with elements 0 to N-1 in O(1) space
Given a constant array of n elements which contains elements from 1 to n-1, with any of these numbers appearing any number of times. Find any one of these repeating numbers in O(n) and using only constant memory space.
Examples:
Input : arr[] = {1, 2, 3, 4, 5, 6, 3}
Output : 3
As the given array is constant methods given in below articles will not work.
- Find duplicates in O(n) time and O(1) extra space | Set 1
- Duplicates in an array in O(n) and by using O(1) extra space | Set-2
So, here is an approach that is based on Floyd’s cycle finding algorithm. We use this to detect loop in a linked list.
The idea is to consider array items as linked list nodes. Any particular index is pointing to the value at that index. And you will see that there is loop as shown in the image below- In case of duplicate, two indexes will have same value and they will form a cycle just like in the image given below.
Linked list formed for above example would be :
1->2->3->4->5->6->3
So we can find the entry point of cycle in the linked list and that will be our duplicate element.
- We maintain two pointers fast and slow
- For each step fast will move to the index that is equal to arr[arr[fast]](two jumps at a time) and slow will move to the index arr[slow](one step at a time)
- When fast==slow that means now we are in a cycle.
- Fast and slow will meet in a circle and the entry point of that circle will be the duplicate element.
- Now we need to find entry point so we will start with fast=0 and visit one step at a time for both fast and slow.
- When fast==slow that will be entry point.
- Return the entry point.
C++
// CPP program to find a duplicate // element in an array with values in // range from 0 to n-1. #include <bits/stdc++.h> using namespace std; // function to find one duplicate int findduplicate(const int arr[], int n) { // return -1 because in these cases // there can not be any repeated element if (n <= 1) return -1; // initialize fast and slow int slow = arr[0]; int fast = arr[arr[0]]; // loop to enter in the cycle while (fast != slow) { // move one step for slow slow = arr[slow]; // move two step for fast fast = arr[arr[fast]]; } // loop to find entry point of the cycle fast = 0; while (slow != fast) { slow = arr[slow]; fast = arr[fast]; } return slow; } int main() { const int arr[] = { 1, 2, 3, 4, 5, 6, 3 }; int n = sizeof(arr) / sizeof(arr[0]); cout << findduplicate(arr, n); return 0; } |
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Java
// Java program to find a duplicate // element in an array with values in // range from 0 to n-1. import java.io.*; import java.util.*; public class GFG { // function to find one duplicate static int findduplicate(int []arr, int n) { // return -1 because in these cases // there can not be any repeated element if (n <= 1) return -1; // initialize fast and slow int slow = arr[0]; int fast = arr[arr[0]]; // loop to enter in the cycle while (fast != slow) { // move one step for slow slow = arr[slow]; // move two step for fast fast = arr[arr[fast]]; } // loop to find entry // point of the cycle fast = 0; while (slow != fast) { slow = arr[slow]; fast = arr[fast]; } return slow; } // Driver Code public static void main(String args[]) { int []arr = {1, 2, 3, 4, 5, 6, 3}; int n = arr.length; System.out.print(findduplicate(arr, n)); } } // This code is contributed by // Manish Shaw (manishshaw1) |
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Python 3
# Python 3 program to find a duplicate # element in an array with values in # range from 0 to n-1. # function to find one duplicate def findduplicate(arr, n): # return -1 because in these cases # there can not be any repeated element if (n <= 1): return -1 # initialize fast and slow slow = arr[0] fast = arr[arr[0]] # loop to enter in the cycle while (fast != slow) : # move one step for slow slow = arr[slow] # move two step for fast fast = arr[arr[fast]] # loop to find entry point of the cycle fast = 0 while (slow != fast): slow = arr[slow] fast = arr[fast] return slow # Driver Code if __name__ == "__main__": arr = [1, 2, 3, 4, 5, 6, 3 ] n = len(arr) print(findduplicate(arr, n)) # This code is contributed by ita_c |
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C#
// C# program to find a duplicate // element in an array with values in // range from 0 to n-1. using System; using System.Collections.Generic; class GFG { // function to find one duplicate static int findduplicate(int []arr, int n) { // return -1 because in these cases // there can not be any repeated element if (n <= 1) return -1; // initialize fast and slow int slow = arr[0]; int fast = arr[arr[0]]; // loop to enter in the cycle while (fast != slow) { // move one step for slow slow = arr[slow]; // move two step for fast fast = arr[arr[fast]]; } // loop to find entry // point of the cycle fast = 0; while (slow != fast) { slow = arr[slow]; fast = arr[fast]; } return slow; } // Driver Code public static void Main() { int []arr = {1, 2, 3, 4, 5, 6, 3}; int n = arr.Length; Console.Write(findduplicate(arr, n)); } } // This code is contributed by // Manish Shaw (manishshaw1) |
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PHP
<?php // PHP program to find a duplicate // element in an array with values in // range from 0 to n-1. // function to find one duplicate function findduplicate($arr, $n) { // return -1 because in these cases // there can not be any repeated element if ($n <= 1) return -1; // initialize fast and slow $slow = $arr[0]; $fast = $arr[$arr[0]]; // loop to enter in the cycle while ($fast != $slow) { // move one step for slow $slow = $arr[$slow]; // move two step for fast $fast = $arr[$arr[$fast]]; } // loop to find entry point of the cycle $fast = 0; while ($slow != $fast) { $slow = $arr[$slow]; $fast = $arr[$fast]; } return $slow; } // Driver Code $arr = array( 1, 2, 3, 4, 5, 6, 3 ); $n = sizeof($arr); echo findduplicate($arr, $n); // This code is contributed by Tushil ?> |
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Output:
3
Time Complexity : O(n)
Auxiliary Space : O(1)
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