Longest chain of arr[i], arr[arr[i]], .. without repetition

Given an array of size n such that elements in array are distinct and in range from 0 to n-1. We need to find out length of the longest chain {arr[i], arr[arr[i]], arr[arr[arr[i]]]……} such that no set element repeats.

Examples:

Input : arr[] = [5, 4, 0, 3, 1, 6, 2]
Output :4
Explanation:
The longest chain without repetition is
{arr[0], arr[5], arr[6], arr[2]} = {5, 6, 2, 0}


Input : arr[] = {1, 0, 4, 2, 3}
Output :3
Explanation:
The longest chain without repetition is
{arr[2], arr[4], arr[3]} = {4, 2, 3}

A naive solution is to find length of longest chain beginning from every element. To keep track of visited nodes, keep a visited array and reset this array after every finding longest chain from an element.

An efficient solution is to traverse each index and set them as -1. Once we see an index that we have set as -1, we know we have come across the same index so we stop and compare if the current count is greater than the max we have seen so far. We can do this setting to -1 in place so that we don’t keep revisiting the same indexes again and again. Why this works is because since each number is distinct, there is always just one way to reach a particular index. So no matter which index you start at in the particular cycle, you will always see the same cycle and hence the same count. So once a cycle is completely visited we can just skip checking for all the indexes in this cycle.

Implementation:

C++

// C++ program to find longest chain of
// arr[i], arr[arr[i]], arr[arr[arr[i]]]
// without repetition.
#include <iostream>
using namespace std;

void aNesting(int arr[], int start, int& max)
{
    // Local maximum
    int c_max = 0;

    // if current element is not visited then
    // increase c_max by one, set arr[start]
    // and change the start to current value.
    while (arr[start] != -1) {
        c_max++;
        int temp = arr[start];
        arr[start] = -1;
        start = temp;
    }

   // if local max is greater than global max
   // then update global maximum
   if (c_max > max) max = c_max;
}

int maxLength(int arr[], int n)
{
    int max = 0;
    for (int i = 0; i < n; i++) {
        aNesting(arr, i, max);
    }
    return max;
}

// Driver code
int main()
{
    int arr[] = { 5, 4, 0, 3, 1, 6, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << maxLength(arr, n);
    return 0;
}

Java

// Java program to find longest chain of
// arr[i], arr[arr[i]], arr[arr[arr[i]]]
// without repetition.
import java.util.*;

class solution
{

static int aNesting(int arr[], int start, int max)
{
    // Local maximum
    int c_max = 0;

    // if current element is not visited then
    // increase c_max by one, set arr[start]
    // and change the start to current value.
    while (arr[start] != -1)
    {
        c_max++;
        int temp = arr[start];
        arr[start] = -1;
        start = temp;
    }

// if local max is greater than global max
// then update global maximum
     if (c_max > max)
      max = c_max;

     return max;
  }

static int maxLength(int[] arr, int n)
{
    int max = 0,max1;
    for (int i = 0; i < n; i++)
    {
        max1 = aNesting(arr, i, max);
        if(max1>max)
         max = max1;
    }
    return max;
 }

// Driver code
public static void main(String args[])
{
    int arr[] = { 5, 4, 0, 3, 1, 6, 2 };
    int n = arr.length;
    System.out.println(maxLength(arr, n));
}
}

// This code is contributed by
// Surendra_Gangwar

Python3

# Python 3 program to find longest chain
# of arr[i], arr[arr[i]], arr[arr[arr[i]]]
# without repetition.
def aNesting(arr,start, max):

    # Local maximum
    c_max = 0

    # if current element is not visited then
    # increase c_max by one, set arr[start]
    # and change the start to current value.
    while (arr[start] != -1):
        c_max += 1
        temp = arr[start]
        arr[start] = -1
        start = temp

    # if local max is greater than global
    # max then update global maximum
    if (c_max > max):
        max = c_max

    return max

def maxLength(arr, n):
    max = 0
    for i in range(0, n, 1):
        max__ = aNesting(arr, i, max)
        if (max__>max):
            max = max__
    return max__

# Driver code
if __name__ =='__main__':
    arr = [5, 4, 0, 3, 1, 6, 2]
    n = len(arr)
    print(maxLength(arr, n))

# This code is contributed by
# Shashank_Sharma

C#

// C# program to find longest chain of
// arr[i], arr[arr[i]], arr[arr[arr[i]]]
// without repetition.
using System;

class GFG
{

static int aNesting(int[] arr, int start, int max)
{
    // Local maximum
    int c_max = 0;

    // if current element is not visited then
    // increase c_max by one, set arr[start]
    // and change the start to current value.
    while (arr[start] != -1)
    {
        c_max++;
        int temp = arr[start];
        arr[start] = -1;
        start = temp;
    }

    // if local max is greater than global max
    // then update global maximum
    if (c_max > max)
    max = c_max;

    return max;
}

static int maxLength(int[] arr, int n)
{
    int max = 0, max1;
    for (int i = 0; i < n; i++)
    {
        max1 = aNesting(arr, i, max);
        if(max1>max)
        max = max1;
    }
    return max;
}

// Driver code
public static void Main()
{
    int[] arr = { 5, 4, 0, 3, 1, 6, 2 };
    int n = arr.Length;
    Console.WriteLine(maxLength(arr, n));
}
}

// This code is contributed by
// Akanksha Rai

PHP

<?php
// PHP program to find longest chain of
// arr[i], arr[arr[i]], arr[arr[arr[i]]]
// without repetition.

function aNesting($arr, $start, &$max)
{
    // Local maximum
    $c_max = 0;

    // if current element is not visited then
    // increase c_max by one, set arr[start]
    // and change the start to current value.
    while ($arr[$start] != -1)
    {
        $c_max++;
        $temp = $arr[$start];
        $arr[$start] = -1;
        $start = $temp;
    }

// if local max is greater than global
// max then update global maximum
if ($c_max > $max)
    $max = $c_max;
}

function maxLength($arr, $n)
{
    $max = 0;
    for ($i = 0; $i < $n; $i++)
    {
        aNesting($arr, $i, $max);
    }
    return $max;
}

// Driver code
$arr = array(5, 4, 0, 3, 1, 6, 2 );
$n = sizeof($arr);
echo maxLength($arr, $n);

// This code is contributed
// by Akanksha Rai
?>

Javascript

<script>

// Javascript program to find longest chain of
// arr[i], arr[arr[i]], arr[arr[arr[i]]]
// without repetition.
function aNesting(arr, start, max)
{

    // Local maximum
    var c_max = 0;

    // If current element is not visited then
    // increase c_max by one, set arr[start]
    // and change the start to current value.
    while (arr[start] != -1)
    {
        c_max++;
        var temp = arr[start];
        arr[start] = -1;
        start = temp;
    }

    // If local max is greater than global
    // max then update global maximum
    if (c_max > max)
        max = c_max;

    return max;
}

function maxLength(arr, n)
{
    var max = 0, max1;
    for(var i = 0; i < n; i++)
    {
        max1 = aNesting(arr, i, max);

        if (max1 > max)
            max = max1;
    }
    return max;
}

// Driver code
var arr = [ 5, 4, 0, 3, 1, 6, 2 ];
var n = arr.length;

document.write(maxLength(arr, n));

// This code is contributed by bunnyram19

</script>

Output

4

Complexity Analysis:

• Time Complexity: O(n) 
• Auxiliary Space: O(1)