Length of the longest subsequence such that XOR of adjacent elements is equal to K

19 June 2025

0

Given an array arr[] of N non-negative integers and an integer K, the idea is to find the length of the longest subsequence having Xor of adjacent elements equal to K.

Examples:

Input: N = 5, arr[] = {3, 2, 4, 3, 5}, K = 1
Output: 3
Explanation:
All the subsequences having Xor of adjacent element equal to K are {3, 2}, {2, 3}, {4, 5}, {3, 2, 3}.
Therefore, the length of the longest subsequence having xor of adjacent element as 1 is 3.

Input: N = 8, arr[] = {4, 5, 4, 7, 3, 5, 4, 6}, K = 2
Output: 3
Explanation:
All the subsequences having Xor of adjacent element equal to K are {4, 6}, {5, 7}, {7, 5}, {5, 7, 5}.
Therefore, the length of the longest subsequence having xor of adjacent element as 1 is 3

Naive Approach: The idea is to use Dynamic Programming. The given problem can be solved based on the following observations:

Suppose Dp(i) represent maximum length of subsequence ending at index i.

Then, transition of one state to another state will be as follows:

Find index j such that j < i and a[j] ^ a[i] = k.

Therefore, Dp(i) = max(Dp(j)+1, Dp(i))

Follow the steps below to solve the problem:

Initialize an integer, say ans, to store the length of the longest subsequence and an array, say dp[], to store the dp states.
Define base case as dp[0] = 1.
Iterate over the range [1, N – 1]:
- Iterate over the range [0, i-1] and update dp[i] as max(dp[i], dp[j] + 1) if a[i] ^ a[j] = K.
- Update ans as max(ans, dp[i]).
Finally, print the maximum length of the longest subsequence ans.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find maximum length
// of subsequence having XOR of
// adjacent elements equal to K
int xorSubsequence(int a[], int n, int k)
{
 
  // Store maximum length of subsequence
  int ans = 0;
 
  // Stores the dp-states
  int dp[n] = {0};
 
  // Base case
  dp[0] = 1;
 
  // Iterate over the range [1, N-1]
  for (int i = 1; i < n; i++) {
 
    // Iterate over the range [0, i - 1]
    for (int j = i - 1; j >= 0; j--) {
 
      // If arr[i]^arr[j] == K
      if ((a[i] ^ a[j]) == k)
 
        // Update the dp[i]
        dp[i] = max(dp[i], dp[j] + 1);
    }
 
    // Update the maximum subsequence length
    ans = max(ans, dp[i]);
    dp[i] = max(1, dp[i]);
  }
 
  // If length of longest subsequence
  // is less than 2 then return 0
  return ans >= 2 ? ans : 0;
}
 
// Driver Code
int main()
{
 
  // Input
  int arr[] = { 3, 2, 4, 3, 5 };
  int K = 1;
  int N = sizeof(arr) / sizeof(arr[0]);
 
  // Print the length of longest subsequence
  cout << xorSubsequence(arr, N, K);
  return 0;
}
 
// This code is contributed by Dharanendra L V

Java

// Java program for the above approach
 
import java.io.*;
import java.util.*;
class GFG {
 
    // Function to find maximum length
    // of subsequence having XOR of
    // adjacent elements equal to K
    public static int xorSubsequence(int a[],
                                     int n, int k)
    {
 
        // Store maximum length of subsequence
        int ans = 0;
 
        // Stores the dp-states
        int dp[] = new int[n];
 
        // Base case
        dp[0] = 1;
 
        // Iterate over the range [1, N-1]
        for (int i = 1; i < n; i++) {
 
            // Iterate over the range [0, i - 1]
            for (int j = i - 1; j >= 0; j--) {
 
                // If arr[i]^arr[j] == K
                if ((a[i] ^ a[j]) == k)
 
                    // Update the dp[i]
                    dp[i] = Math.max(dp[i], dp[j] + 1);
            }
            // Update the maximum subsequence length
            ans = Math.max(ans, dp[i]);
 
            dp[i] = Math.max(1, dp[i]);
        }
        // If length of longest subsequence
        // is less than 2 then return 0
        return ans >= 2 ? ans : 0;
    }
    // Driver Code
    public static void main(String[] args)
    {
        // Input
        int arr[] = { 3, 2, 4, 3, 5 };
        int K = 1;
        int N = arr.length;
        // Print the length of longest subsequence
        System.out.println(xorSubsequence(arr, N, K));
    }
}

Python3

# Python program for the above approach
 
# Function to find maximum length
# of subsequence having XOR of
# adjacent elements equal to K
def xorSubsequence(a, n, k):
   
    # Store maximum length of subsequence
    ans = 0;
 
    # Stores the dp-states
    dp = [0] * n;
 
    # Base case
    dp[0] = 1;
 
    # Iterate over the range [1, N-1]
    for i in range(1, n):
 
        # Iterate over the range [0, i - 1]
        for j in range(i - 1, -1, -1):
 
            # If arr[i]^arr[j] == K
            if ((a[i] ^ a[j]) == k):
 
                # Update the dp[i]
                dp[i] = max(dp[i], dp[j] + 1);
 
        # Update the maximum subsequence length
        ans = max(ans, dp[i]);
        dp[i] = max(1, dp[i]);
 
    # If length of longest subsequence
    # is less than 2 then return 0
    return ans if ans >= 2 else 0;
 
# Driver Code
if __name__ == '__main__':
   
    # Input
    arr = [3, 2, 4, 3, 5];
    K = 1;
    N = len(arr);
     
    # Print the length of longest subsequence
    print(xorSubsequence(arr, N, K));
 
    # This code contributed by shikhasingrajput

C#

// C# program of the above approach
using System;
 
class GFG{
     
// Function to find maximum length 
// of subsequence having XOR of 
// adjacent elements equal to K 
static int xorSubsequence(int[] a, int n,
                          int k) 
{ 
     
    // Store maximum length of subsequence 
    int ans = 0; 
 
    // Stores the dp-states 
    int[] dp = new int[n]; 
 
    // Base case 
    dp[0] = 1; 
 
    // Iterate over the range [1, N-1] 
    for(int i = 1; i < n; i++)
    { 
         
        // Iterate over the range [0, i - 1] 
        for(int j = i - 1; j >= 0; j--) 
        { 
             
            // If arr[i]^arr[j] == K 
            if ((a[i] ^ a[j]) == k) 
             
                // Update the dp[i] 
                dp[i] = Math.Max(dp[i], dp[j] + 1); 
        } 
         
        // Update the maximum subsequence length 
        ans = Math.Max(ans, dp[i]); 
 
        dp[i] = Math.Max(1, dp[i]); 
    } 
     
    // If length of longest subsequence 
    // is less than 2 then return 0 
    return ans >= 2 ? ans : 0; 
}  
 
// Driver code    
static void Main() 
{
     
    // Input 
    int[] arr = { 3, 2, 4, 3, 5 }; 
    int K = 1; 
    int N = arr.Length; 
     
    // Print the length of longest subsequence 
    Console.WriteLine(xorSubsequence(arr, N, K));  
}
}
 
// This code is contributed by divyeshrabadiya07

Javascript

<script>
 
    // Javascript program of the above approach
     
    // Function to find maximum length
    // of subsequence having XOR of
    // adjacent elements equal to K
    function xorSubsequence(a, n, k)
    {
 
        // Store maximum length of subsequence
        let ans = 0;
 
        // Stores the dp-states
        let dp = new Array(n);
        dp.fill(0);
 
        // Base case
        dp[0] = 1;
 
        // Iterate over the range [1, N-1]
        for(let i = 1; i < n; i++)
        {
 
            // Iterate over the range [0, i - 1]
            for(let j = i - 1; j >= 0; j--)
            {
 
                // If arr[i]^arr[j] == K
                if ((a[i] ^ a[j]) == k)
 
                    // Update the dp[i]
                    dp[i] = Math.max(dp[i], dp[j] + 1);
            }
 
            // Update the maximum subsequence length
            ans = Math.max(ans, dp[i]);
 
            dp[i] = Math.max(1, dp[i]);
        }
 
        // If length of longest subsequence
        // is less than 2 then return 0
        return ans >= 2 ? ans : 0;
    } 
     
    let arr = [ 3, 2, 4, 3, 5 ];
    let K = 1;
    let N = arr.length;
      
    // Print the length of longest subsequence
    document.write(xorSubsequence(arr, N, K)); 
     
</script>

Output:

Time Complexity: O(N²)
Auxiliary Space: O(N)

Efficient Approach: The above approach can be optimized by using the property of Xor and Hashmap with dynamic programming to store the maximum length of subsequence ending at an integer, resulting in constant time transition of states in DP.

Initialize an integer say ans =0 to store the length of the longest subsequence and an array say dp[] to store the state of DP.
Initialize a HashMap say mp to store the longest length of subsequence ending at an element.
Define base case as dp[0] = 1 and push the pair {arr[0], 1} in mp.
Iterate over the range [1, N-1]:
- Find the length of the longest subsequence say dpj ending at element arr[i] ^K from HashMap mp.
- Update dp[i] as max(dp[i], dpj+1) and update the longest length of subsequence ending at element arr[i] in HashMap mp.
- Update the ans = max(ans, dp[i]).
Finally, print the maximum length of the longest subsequence ans.

Below is the implementation of the above approach:

C++

// C++ program for above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find maximum length of subsequence
int xorSubsequence(int a[], int n, int k)
{
    
    // Stores maximum length of subsequence
    int ans = 0;
 
    // Dictionary to store the longest length of
    // subsequence ending at an integer, say X
    map<int, int> map;
 
    // Stores the maximum length of
    // subsequence ending at index i
    int dp[n] = {0};
 
    // Base case
    map[a[0]] = 1;
    dp[0] = 1;
 
    // Iterate over the range [1, N-1]
    for (int i = 1; i < n; i++) 
    {
         
        int dpj;
         
        // Retrieve the longest length of
        // subsequence ending at integer a[]^K
        if(map.find(a[i] ^ k) != map.end())
        {
            dpj = map[a[i] ^ k];
        }
        else{
            dpj = -1;
        }
 
        // If dpj is not NULL
        if (dpj != 0)
 
            // Update dp[i]
            dp[i] = max(dp[i], dpj + 1);
 
        // Update ans
        ans = max(ans, dp[i]);
 
        // Update the maximum length of subsequence
        // ending at element is a[i] in Dictionary
        if(map.find(a[i]) != map.end())
        {
            map[a[i]] = max(map[a[i]]+1, dp[i]);
        }
        else
        {
            map[a[i]] = max(1, dp[i]);
        }
    }
 
    // Return the ans if ans>=2.
    // Otherwise, return 0
    return ans >= 2 ? ans : 0;
}
     
int main()
{
    // Input
    int arr[] = { 3, 2, 4, 3, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 1;
 
    // Print the length of the longest subsequence
    cout << (xorSubsequence(arr, N, K));
 
    return 0;
}
 
// This code is contributed by divyesh072019.

Java

// Java program for above approach
import java.io.*;
import java.util.*;
class GFG
{
 
  // Function to find maximum length of subsequence
  public static int xorSubsequence(int a[], int n, int k)
  {
    // Stores maximum length of subsequence
    int ans = 0;
 
    // HashMap to store the longest length of
    // subsequence ending at an integer, say X
    HashMap<Integer, Integer> map = new HashMap<>();
 
    // Stores the maximum length of
    // subsequence ending at index i
    int dp[] = new int[n];
 
    // Base case
    map.put(a[0], 1);
    dp[0] = 1;
 
    // Iterate over the range [1, N-1]
    for (int i = 1; i < n; i++) {
 
      // Retrieve the longest length of
      // subsequence ending at integer a[]^K
      Integer dpj = map.get(a[i] ^ k);
 
      // If dpj is not NULL
      if (dpj != null)
 
        // Update dp[i]
        dp[i] = Math.max(dp[i], dpj + 1);
 
      // Update ans
      ans = Math.max(ans, dp[i]);
 
      // Update the maximum length of subsequence
      // ending at element is a[i] in HashMap
      map.put(
        a[i],
        Math.max(map.getOrDefault(a[i], 1), dp[i]));
    }
 
    // Return the ans if ans>=2.
    // Otherwise, return 0
    return ans >= 2 ? ans : 0;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    // Input
    int arr[] = { 3, 2, 4, 3, 5 };
    int N = arr.length;
    int K = 1;
 
    // Print the length of the longest subsequence
    System.out.println(xorSubsequence(arr, N, K));
  }
}

Python3

# Python 3 program for above approach
 
# Function to find maximum length of subsequence
def xorSubsequence( a, n, k):
 
    # Stores maximum length of subsequence
    ans = 0
 
    # HashMap to store the longest length of
    # subsequence ending at an integer, say X
    map = {}
     
    # Stores the maximum length of
    # subsequence ending at index i
    dp = [0]* n
 
    # Base case
    map[a[0]] =  1
    dp[0] = 1
 
    # Iterate over the range[1, N-1]
    for i in range(1, n):
 
        # Retrieve the longest length of
        # subsequence ending at integer a[] ^ K
         
        # If dpj is not NULL
        if (a[i] ^ k  in map):
 
            # Update dp[i]
               dp[i] = max(dp[i], map[a[i] ^ k] + 1)
 
        # Update ans
        ans = max(ans, dp[i])
 
        # Update the maximum length of subsequence
        # ending at element is a[i] in HashMap
        if a[i] in map:
              map[a[i]] = max(map[a[i]],dp[i])
        else:
            map[a[i]] = max(1, dp[i])
 
    # Return the ans if ans >= 2.
    # Otherwise, return 0
    if ans >= 2 :
      return ans
    return 0
 
# Driver Code
if __name__ == "__main__":
 
    # Input
    arr = [3, 2, 4, 3, 5]
    N = len(arr)
    K = 1
 
    # Print the length of the longest subsequence
    print(xorSubsequence(arr, N, K))
 
    # This code is contributed by chitranayal.

C#

// C# program for above approach
using System;
using System.Collections.Generic;
 
public class GFG 
{
 
    // Function to find maximum length of subsequence
    public static int xorSubsequence(int []a, int n, int k)
    {
       
        // Stores maximum length of subsequence
        int ans = 0;
 
        // Dictionary to store the longest length of
        // subsequence ending at an integer, say X
        Dictionary<int, int> map = new Dictionary<int, int>();
 
        // Stores the maximum length of
        // subsequence ending at index i
        int []dp = new int[n];
 
        // Base case
        map.Add(a[0], 1);
        dp[0] = 1;
 
        // Iterate over the range [1, N-1]
        for (int i = 1; i < n; i++) 
        {
 
            // Retrieve the longest length of
            // subsequence ending at integer []a^K
            int dpj = map.ContainsKey(a[i] ^ k)?map[a[i] ^ k]:-1;
 
            // If dpj is not NULL
            if (dpj != 0)
 
                // Update dp[i]
                dp[i] = Math.Max(dp[i], dpj + 1);
 
            // Update ans
            ans = Math.Max(ans, dp[i]);
 
            // Update the maximum length of subsequence
            // ending at element is a[i] in Dictionary
            if(map.ContainsKey(a[i]))
            {
                map[a[i]] = Math.Max(map[a[i]]+1, dp[i]); ;
            }
            else
            {
                map.Add(a[i], Math.Max(1, dp[i]));
            }
        }
 
        // Return the ans if ans>=2.
        // Otherwise, return 0
        return ans >= 2 ? ans : 0;
    }
   
    // Driver Code
    public static void Main(String[] args)
    {
        // Input
        int []arr = { 3, 2, 4, 3, 5 };
        int N = arr.Length;
        int K = 1;
 
        // Print the length of the longest subsequence
        Console.WriteLine(xorSubsequence(arr, N, K));
    }
}
 
// This code is contributed by shikhasingrajput

Javascript

<script>
 
// Javascript program for above approach
 
// Function to find maximum length of subsequence
function xorSubsequence(a, n, k)
{
    
    // Stores maximum length of subsequence
    var ans = 0;
 
    // Dictionary to store the longest length of
    // subsequence ending at an integer, say X
    var map = new Map();
 
    // Stores the maximum length of
    // subsequence ending at index i
    var dp = Array(n).fill(0);
 
    // Base case
    map.set(a[0], 1)
    dp[0] = 1;
 
    // Iterate over the range [1, N-1]
    for (var i = 1; i < n; i++) 
    {
         
        var dpj;
         
        // Retrieve the longest length of
        // subsequence ending at integer []a^K
        if(map.has(a[i] ^ k))
        {
            dpj = map.get(a[i] ^ k);
        }
        else{
            dpj = -1;
        }
 
        // If dpj is not NULL
        if (dpj != 0)
 
            // Update dp[i]
            dp[i] = Math.max(dp[i], dpj + 1);
 
        // Update ans
        ans = Math.max(ans, dp[i]);
 
        // Update the maximum length of subsequence
        // ending at element is a[i] in Dictionary
        if(map.has(a[i]))
        {
            map.set(a[i] , Math.max(map.get(a[i])+1, dp[i]));
        }
        else
        {
            map.set(a[i], Math.max(1, dp[i]));
        }
    }
 
    // Return the ans if ans>=2.
    // Otherwise, return 0
    return ans >= 2 ? ans : 0;
}
     
// Input
var arr = [3, 2, 4, 3, 5];
var N = arr.length;
var K = 1;
 
// Print the length of the longest subsequence
document.write(xorSubsequence(arr, N, K));
 
// This code is contributed by famously.
</script>

Output:

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

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Length of the longest subsequence such that XOR of adjacent elements is equal to K

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