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

04 Jun, 2021
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++
Python3
Javascript
// 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-decreasingint 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 codeint main(){     int arr[] = { 2, 12, 6, 7, 13, 14, 8, 6 };    int n = sizeof(arr) / sizeof(arr[0]);     cout << LongestXorSubsequence(arr, n);     return 0;}
# 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-decreasingdef 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 codearr = [ 2, 12, 6, 7, 13, 14, 8, 6 ]n = len(arr)print(LongestXorSubsequence(arr, n)) # This code is contributed by Sanjit Prasad
<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-decreasingfunction 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 codelet 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>
5
Time Complexity: O(N* N)Auxiliary Space: O(N)
Sanjit_Prasad
ujjwalgoel1103
gfgking
Bitwise-XOR
subsequence
Arrays
C++ Programs
Dynamic Programming
Greedy
Arrays
Dynamic Programming
Greedy
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