Length of the longest subsequence such that xor of adjacent elements is non-decreasing

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Given a sequence arr of N positive integers, the task is to find the length of the longest subsequence such that xor of adjacent integers in the subsequence must be non-decreasing.

Examples:

Input: N = 8, arr = {1, 100, 3, 64, 0, 5, 2, 15}
Output: 6
The subsequence of maximum length is {1, 3, 0, 5, 2, 15}
with XOR of adjacent elements as {2, 3, 5, 7, 13}
Input: N = 3, arr = {1, 7, 10}
Output: 3
The subsequence of maximum length is {1, 3, 7}
with XOR of adjacent elements as {2, 4}.

Approach:

This problem can be solved using dynamic programming where dp[i] will store the length of the longest valid subsequence that ends at index i.
First, store the xor of all the pairs of elements i.e. arr[i] ^ arr[j] and the pair (i, j) also and then sort them according to the value of xor as they need to be non-decreasing.
Now if the pair (i, j) is considered then the length of the longest subsequence that ends at j will be max(dp[j], 1 + dp[i]). In this way, calculate the maximum possible value of dp[] array for each position and then take the maximum of them.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the length of the longest
// subsequence such that the XOR of adjacent
// elements in the subsequence must
// be non-decreasing
int LongestXorSubsequence(int arr[], int n)
{
 
    vector<pair<int, pair<int, int> > > v;
 
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
 
            // Computing xor of all the pairs
            // of elements and store them
            // along with the pair (i, j)
            v.push_back(make_pair(arr[i] ^ arr[j],
                                  make_pair(i, j)));
        }
    }
 
    // Sort all possible xor values
    sort(v.begin(), v.end());
 
    int dp[n];
 
    // Initialize the dp array
    for (int i = 0; i < n; i++) {
        dp[i] = 1;
    }
 
    // Calculating the dp array
    // for each possible position
    // and calculating the max length
    // that ends at a particular index
    for (auto i : v) {
        dp[i.second.second]
            = max(dp[i.second.second],
                  1 + dp[i.second.first]);
    }
 
    int ans = 1;
 
    // Taking maximum of all position
    for (int i = 0; i < n; i++)
        ans = max(ans, dp[i]);
 
    return ans;
}
 
// Driver code
int main()
{
 
    int arr[] = { 2, 12, 6, 7, 13, 14, 8, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << LongestXorSubsequence(arr, n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
import java.util.*;
import java.util.stream.Collectors;
 
class GFG {
    // Function to find the length of the longest
    // subsequence such that the XOR of adjacent
    // elements in the subsequence must
    // be non-decreasing
    static int LongestXorSubsequence(int[] arr, int n)
    {
 
        List<int[]> v = new ArrayList<>();
 
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
 
                // Computing xor of all the pairs
                // of elements and store them
                // along with the pair (i, j)
                int[] l1 = { arr[i] ^ arr[j], i, j };
                v.add(l1);
            }
        }
 
        // Sort all possible xor values
        Comparator<int[]> byFirstElement
            = (int[] a, int[] b) -> a[0] - b[0];
 
        List<int[]> v1 = v.stream()
                             .sorted(byFirstElement)
                             .collect(Collectors.toList());
 
        int[] dp = new int[n];
 
        // Initialize the dp array
        for (int i = 0; i < n; i++) {
            dp[i] = 1;
        }
 
        // Calculating the dp array
        // for each possible position
        // and calculating the max length
        // that ends at a particular index
        Iterator<int[]> iter = v1.iterator();
        while (iter.hasNext()) {
            int[] list = iter.next();
            dp[list[2]]
                = Math.max(dp[list[2]], 1 + dp[list[1]]);
        }
 
        int ans = 1;
 
        // Taking maximum of all position
        for (int i = 0; i < n; i++)
            ans = Math.max(ans, dp[i]);
 
        return ans;
    }
    public static void main(String[] args)
    {
        // Driver code
        int[] arr = { 2, 12, 6, 7, 13, 14, 8, 6 };
        int n = arr.length;
 
        System.out.println(LongestXorSubsequence(arr, n));
    }
}
 
// this code is contributed by phasing17

Python3

# Python3 implementation of the approach
 
# Function to find the length of the longest
# subsequence such that the XOR of adjacent
# elements in the subsequence must
# be non-decreasing
def LongestXorSubsequence(arr, n):
 
    v = []
 
    for i in range(0, n): 
        for j in range(i + 1, n): 
 
             # Computing xor of all the pairs
            # of elements and store them
            # along with the pair (i, j)
            v.append([(arr[i] ^ arr[j]), (i, j)])
 
        # v.push_back(make_pair(arr[i] ^ arr[j], make_pair(i, j)))
         
    # Sort all possible xor values
    v.sort()
     
    # Initialize the dp array
    dp = [1 for x in range(88)]
 
    # Calculating the dp array
    # for each possible position
    # and calculating the max length
    # that ends at a particular index
    for a, b in v:
        dp[b[1]] = max(dp[b[1]], 1 + dp[b[0]])
     
    ans = 1
 
    # Taking maximum of all position
    for i in range(0, n):
        ans = max(ans, dp[i])
 
    return ans
 
# Driver code
arr = [ 2, 12, 6, 7, 13, 14, 8, 6 ]
n = len(arr)
print(LongestXorSubsequence(arr, n))
 
# This code is contributed by Sanjit Prasad

C#

// C# implementation of the approach
using System;
using System.Linq;
using System.Collections.Generic;
 
class GFG
{
 
  // Function to find the length of the longest
  // subsequence such that the XOR of adjacent
  // elements in the subsequence must
  // be non-decreasing
  static int LongestXorSubsequence(int[] arr, int n)
  {
 
    List<int[]> v = new List<int[]>();
 
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
 
        // Computing xor of all the pairs
        // of elements and store them
        // along with the pair (i, j)
        int[] l1 = { arr[i] ^ arr[j], i, j };
        v.Add(l1);
      }
    }
 
    // Sorting the array by First Value
    List<int[]> v1 = v.OrderBy(a => a[0])
      .ThenBy(a => a[1])
      .ToList();
 
    int[] dp = new int[n];
 
    // Initialize the dp array
    for (int i = 0; i < n; i++) {
      dp[i] = 1;
    }
 
    // Calculating the dp array
    // for each possible position
    // and calculating the max length
    // that ends at a particular index
    foreach(var list in v1) dp[list[2]]
      = Math.Max(dp[list[2]], 1 + dp[list[1]]);
 
    int ans = 1;
 
    // Taking maximum of all position
    for (int i = 0; i < n; i++)
      ans = Math.Max(ans, dp[i]);
 
    return ans;
  }
  public static void Main(string[] args)
  {
    // Driver code
    int[] arr = { 2, 12, 6, 7, 13, 14, 8, 6 };
    int n = arr.Length;
 
    Console.WriteLine(LongestXorSubsequence(arr, n));
  }
}
 
// This code is contributed by phasing17

Javascript

<script>
// Javascript implementation of the approach
 
// Function to find the length of the longest
// subsequence such that the XOR of adjacent
// elements in the subsequence must
// be non-decreasing
function LongestXorSubsequence(arr, n) {
 
    let v = [];
 
    for (let i = 0; i < n; i++) {
        for (let j = i + 1; j < n; j++) {
 
            // Computing xor of all the pairs
            // of elements and store them
            // along with the pair (i, j)
            v.push([arr[i] ^ arr[j], [i, j]]);
        }
    }
 
    // Sort all possible xor values
    v.sort((a, b) => a[0] - b[0]);
 
    let dp = new Array(n);
 
    // Initialize the dp array
    for (let i = 0; i < n; i++) {
        dp[i] = 1;
    }
 
    // Calculating the dp array
    // for each possible position
    // and calculating the max length
    // that ends at a particular index
    for (let i of v) {
        dp[i[1][1]]
            = Math.max(dp[i[1][1]],
                1 + dp[i[1][0]]);
    }
 
    let ans = 1;
 
    // Taking maximum of all position
    for (let i = 0; i < n; i++)
        ans = Math.max(ans, dp[i]);
 
    return ans;
}
 
// Driver code
let arr = [2, 12, 6, 7, 13, 14, 8, 6];
let n = arr.length;
 
document.write(LongestXorSubsequence(arr, n));
 
// This code is contributed by _saurabh_jaiswal.
</script>

Output

Time Complexity: O(N* N)
Auxiliary Space: O(N)

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