Given an array arr[], the task is to find the maximum length sub-sequence from the array which satisfy the following condition:
Any element can be chosen as the first element of the sub-sequence but the index of the next element will be determined by arr[arr[i] + i] where i is the index of the previous element in the sequence.
Examples:
Input: arr[] = {1, 2, 3, 4, 5}
Output: 1 2 4
arr[0] = 1, arr[1 + 0] = arr[1] = 2, arr[2 + 1] = arr[3] = 4
Other possible sub-sequences are {2, 4}, {3}, {4} and {5}Input: arr[] = {1, 6, 3, 1, 12, 1, 4}
Output: 3 1 4
Approach:
- Make use of two arrays temp and print.
- The temp array will store the array elements that are currently under consideration and the print array will store the array elements that are to be printed as the final output.
- Iterate from 0 to n – 1 and consider the current element as the first element of the sequence.
- Store all the elements of the current sequence into temp array.
- If the size of the temp array becomes greater than print array then copy all the contents of the temp array to the print array.
- When all the sequences have been considered, print the contents of the print array.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to print the maximum length sub-sequencevoid maxLengthSubSeq(int a[], int n){ // Arrays to store the values to be printed int temp[n], print[n]; int y = 0; for (int i = 0; i < n; i++) { int j = 0; int x = 0; // Store the first value into the temp array temp[j++] = a[x]; // Index of the next element x = a[x] + x; // Iterate till index is in range of the array while (x < n) { temp[j++] = a[x]; x = a[x] + x; } // If the length (temp) > the length (print) then // copy the contents of the temp array into // the print array if (y < j) { for (int k = 0; k < j; k++) { print[k] = temp[k]; y = j; } } } // Print the contents of the array for (int i = 0; i < y; i++) cout << print[i] << " ";}// Driver codeint main(){ int a[] = { 1, 2, 3, 4, 5 }; int n = sizeof(a) / sizeof(a[0]); maxLengthSubSeq(a, n); return 0;} |
Java
//Java implementation of the approach/import java.io.*;class GFG { // Function to print the maximum length sub-sequencestatic void maxLengthSubSeq(int a[], int n){ // Arrays to store the values to be printed int temp[]=new int[n]; int print[]=new int[n]; int y = 0; for (int i = 0; i < n; i++) { int j = 0; int x = 0; // Store the first value into the temp array temp[j++] = a[x]; // Index of the next element x = a[x] + x; // Iterate till index is in range of the array while (x < n) { temp[j++] = a[x]; x = a[x] + x; } // If the length (temp) > the length (print) then // copy the contents of the temp array into // the print array if (y < j) { for (int k = 0; k < j; k++) { print[k] = temp[k]; y = j; } } } // Print the contents of the array for (int i = 0; i < y; i++) System.out.print(print[i] + " ");}// Driver code public static void main (String[] args) { int a[] = { 1, 2, 3, 4, 5 }; int n = a.length; maxLengthSubSeq(a, n); }//This code is contributed by @Tushil. } |
Python3
# Python 3 implementation of the approach# Function to print the maximum length # sub-sequencedef maxLengthSubSeq(a, n): # Arrays to store the values to be printed temp = [0 for i in range(n)] print1 = [0 for i in range(n)] y = 0 for i in range(0, n, 1): j = 0 x = 0 # Store the first value into # the temp array temp[j] = a[x] j += 1 # Index of the next element x = a[x] + x # Iterate till index is in range # of the array while (x < n): temp[j] = a[x] j += 1 x = a[x] + x # If the length (temp) > the length # (print) then copy the contents of # the temp array into the print array if (y < j): for k in range(0, j, 1): print1[k] = temp[k] y = j # Print the contents of the array for i in range(0, y, 1): print(print1[i], end = " ")# Driver codeif __name__ == '__main__': a = [1, 2, 3, 4, 5] n = len(a) maxLengthSubSeq(a, n)# This code is contributed by # Surendra_Gangwar |
C#
//C# implementation of the approach/using System;public class GFG{ // Function to print the maximum length sub-sequencestatic void maxLengthSubSeq(int []a, int n){ // Arrays to store the values to be printed int []temp=new int[n]; int []print=new int[n]; int y = 0; for (int i = 0; i < n; i++) { int j = 0; int x = 0; // Store the first value into the temp array temp[j++] = a[x]; // Index of the next element x = a[x] + x; // Iterate till index is in range of the array while (x < n) { temp[j++] = a[x]; x = a[x] + x; } // If the length (temp) > the length (print) then // copy the contents of the temp array into // the print array if (y < j) { for (int k = 0; k < j; k++) { print[k] = temp[k]; y = j; } } } // Print the contents of the array for (int i = 0; i < y; i++) Console.Write(print[i] + " ");}// Driver code static public void Main (){ int []a = { 1, 2, 3, 4, 5 }; int n = a.Length; maxLengthSubSeq(a, n); }//This code is contributed by ajit. } |
PHP
<?php// PHP implementation of the approach// Function to print the maximum// length sub-sequencefunction maxLengthSubSeq($a, $n){ $y = 0; for ($i = 0; $i < $n; $i++) { $j = 0; $x = 0; // Store the first value into // the temp array $temp[$j++] = $a[$x]; // Index of the next element $x = $a[$x] + $x; // Iterate till index is in // range of the array while ($x < $n) { $temp[$j++] = $a[$x]; $x = $a[$x] + $x; } // If the length (temp) > the length // (print) then copy the contents of // the temp array into the print array if ($y < $j) { for ($k = 0; $k < $j; $k++) { $print[$k] = $temp[$k]; $y = $j; } } } // Print the contents of the array for ($i = 0; $i < $y; $i++) echo $print[$i] . " ";}// Driver code$a = array(1, 2, 3, 4, 5);$n = sizeof($a);maxLengthSubSeq($a, $n);// This code is contributed// by Akanksha Rai |
Javascript
<script> // Javascript implementation of the approach // Function to print the maximum length sub-sequence function maxLengthSubSeq(a, n) { // Arrays to store the values to be printed let temp=new Array(n); temp.fill(0); let print=new Array(n); print.fill(0); let y = 0; for (let i = 0; i < n; i++) { let j = 0; let x = 0; // Store the first value into the temp array temp[j++] = a[x]; // Index of the next element x = a[x] + x; // Iterate till index is in range of the array while (x < n) { temp[j++] = a[x]; x = a[x] + x; } // If the length (temp) > the length (print) then // copy the contents of the temp array into // the print array if (y < j) { for (let k = 0; k < j; k++) { print[k] = temp[k]; y = j; } } } // Print the contents of the array for (let i = 0; i < y; i++) document.write(print[i] + " "); } let a = [ 1, 2, 3, 4, 5 ]; let n = a.length; maxLengthSubSeq(a, n); </script> |
Output
1 2 4
Complexity Analysis:
- Time Complexity: O(n*n)
- Auxiliary Space: O(n)
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