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

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:
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:
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 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 > > 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; }

 # 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

Output
5

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

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Improved By : Sanjit_Prasad, ujjwalgoel1103