Longest subsequence with a given AND value

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

Examples:

Input: arr[] = {3, 7, 2, 3}, M = 3
Output: 3
Longest sub-sequence with AND value 3 is {3, 7, 3}.



Input: arr[] = {2, 2}, M = 3
Output: 0

Approach: A simple way to solve this will be to generate all the possible sub-sequences and then find the largest among them with required AND value.
However, for smaller values of M, an approach based on dynamic programming can be used.

Let’s look at the recurrence relation first.

dp[i][curr_and] = max(dp[i + 1][curr_and], dp[i + 1][curr_and & arr[i]] + 1)

Let’s understand the states of DP now. Here, dp[i][curr_and] stores the longest subsequence of subarray arr[i…N-1] such that curr_and & the AND of this subsequence is equal to M. At each step, either the index i can be chosen updating curr_and or it can be rejected.

Below is the implementation of the above approach:

C++

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// 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, ((1 << 8) - 1),
                      n, m);
    if (ans == -1)
        cout << 0;
    else
        cout << ans;
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*;
  
class GFG
{
static int maxN = 300;
static int maxM = 300;
  
// To store the states of DP
static int dp[][] = new int[maxN][maxM];
static boolean v[][] = new boolean[maxN][maxM];
  
// Function to return the required length
static 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] = true;
  
    // 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] = Math.max(dp[i][curr], r + 1);
    return dp[i][curr];
}
  
// Driver code
public static void main(String args[])
{
    int arr[] = { 3, 7, 2, 3 };
    int n = arr.length;
    int m = 3;
  
    int ans = findLen(arr, 0, ((1 << 8) - 1), n, m);
    if (ans == -1)
        System.out.print( 0);
    else
        System.out.print( ans);
}
}
  
// This code is contributed by Arnab Kundu

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Python3

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# Python3 implementation of the approach 
import numpy as np
  
maxN = 20
maxM = 256
  
# To store the states of DP 
dp = np.zeros((maxN, maxM)); 
v = np.zeros((maxN, maxM)); 
  
# Function to return the required length 
def findLen(arr, i, curr, n, 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 
    l = findLen(arr, i + 1, curr, n, m);
    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 
if __name__ == "__main__" :
  
    arr = [ 3, 7, 2, 3 ]; 
    n = len(arr); 
    m = 3
  
    ans = findLen(arr, 0, ((1 << 8) - 1), n, m); 
      
    if (ans == -1) :
        print(0); 
    else :
        print(ans); 
  
# This code is contributed by AnkitRai01

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C#

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// C# implementation of the approach 
using System;
  
class GFG 
static int maxN = 300; 
static int maxM = 300; 
  
// To store the states of DP 
static int[,] dp = new int[maxN, maxM]; 
static bool[,] v = new bool[maxN, maxM]; 
  
// Function to return the required length 
static 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] = true
  
    // 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] = Math.Max(dp[i, curr], r + 1); 
    return dp[i, curr]; 
  
// Driver code 
public static void Main(String[] args) 
    int[] arr = { 3, 7, 2, 3 }; 
    int n = arr.Length; 
    int m = 3; 
  
    int ans = findLen(arr, 0, ((1 << 8) - 1), n, m); 
    if (ans == -1) 
        Console.WriteLine(0); 
    else
        Console.WriteLine(ans); 
  
// This code is contributed by
// sanjeev2552

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Output:

3

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




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