Longest subsequence with a given OR value : Dynamic Programming Approach

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Longest subsequence with a given OR value : Dynamic Programming Approach

Last Updated : 13 May, 2021

Given an array arr[], the task is to find the longest subsequence with a given OR value M. If there is no such sub-sequence then print 0.

Examples:

Input: arr[] = {3, 7, 2, 3}, M = 3
Output: 3
{3, 2, 3} is the required subsequence
3 | 2 | 3 = 3
Input: arr[] = {2, 2}, M = 3
Output: 0

Approach: A simple solution is to generate all the possible sub-sequences and then find the largest among them with the required OR value. However, for smaller values of M, a dynamic programming approach can be used.
Let’s look at the recurrence relation first.

dp[i][curr_or] = max(dp[i + 1][curr_or], dp[i + 1][curr_or | arr[i]] + 1)

Let’s understand the states of DP now. Here, dp[i][curr_or] stores the longest subsequence of the subarray arr[i…N-1] such the curr_or gives M when gets ORed with this subsequence. At each step, either choose the index i and update curr_or or reject index i and continue.

Below is the implementation of the above approach:

Python3
Javascript

// C++ implementation of the approach

#include <bits/stdc++.h>

using namespace std;

#define maxN 20

#define maxM 64

// To store the states of DP

int dp[maxN][maxM];

bool v[maxN][maxM];

// Function to return the required length

int findLen(int* arr, int i, int curr,

int n, int m)

{

// Base case

if (i == n) {

if (curr == m)

return 0;

else

return -1;

}

// If the state has been solved before

// return the value of the state

if (v[i][curr])

return dp[i][curr];

// Setting the state as solved

v[i][curr] = 1;

// Recurrence relation

int l = findLen(arr, i + 1, curr, n, m);

int r = findLen(arr, i + 1, curr | arr[i], n, m);

dp[i][curr] = l;

if (r != -1)

dp[i][curr] = max(dp[i][curr], r + 1);

return dp[i][curr];

}

// Driver code

int main()

{

int arr[] = { 3, 7, 2, 3 };

int n = sizeof(arr) / sizeof(int);

int m = 3;

int ans = findLen(arr, 0, 0, n, m);

if (ans == -1)

cout << 0;

else

cout << ans;

return 0;

}

Output:

Time Complexity: O(N * maxArr) where maxArr is the maximum element from the array.

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