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++ 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 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 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# 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
<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
5
Time Complexity: O(N* N)
Auxiliary Space: O(N)
