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++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define maxN 20#define maxM 64// To store the states of DPint dp[maxN][maxM];bool v[maxN][maxM];// Function to return the required lengthint 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 codeint 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;} |
Java
// Java implementation of the approachimport java.util.*;class GFG{static int maxN = 300;static int maxM = 300;// To store the states of DPstatic int dp[][] = new int[maxN][maxM];static boolean v[][] = new boolean[maxN][maxM];// Function to return the required lengthstatic 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 codepublic 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 |
Python3
# Python3 implementation of the approach import numpy as npmaxN = 20maxM = 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 |
C#
// 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 |
Javascript
<script>// Javascript implementation of the approachvar maxN = 20var maxM = 64// To store the states of DPvar dp = Array.from(Array(maxN), ()=> Array(maxM));var v = Array.from(Array(maxN), ()=> Array(maxM));// Function to return the required lengthfunction 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 var l = findLen(arr, i + 1, curr, n, m); var 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 codevar arr = [3, 7, 2, 3];var n = arr.length;var m = 3;var ans = findLen(arr, 0, ((1 << 8) - 1), n, m);if (ans == -1) document.write( 0);else document.write( ans);</script> |
Output:
3
Time Complexity: O(N * maxVal) where maxVal is the maximum element from the given array.
Auxiliary Space: O(maxN * maxM)
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