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:
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, 0, n, m); if (ans == -1) cout << 0; else cout << ans; return 0;} |
Java
// Java implementation of the approachclass GFG {static int maxN = 20;static int maxM = 64;// 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, 0, n, m); if (ans == -1) System.out.println(0); else System.out.println(ans);}}// This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the approach import numpy as npmaxN = 20maxM = 64# 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, 0, n, m); if (ans == -1) : print(0); else : print(ans); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System; class GFG{static int maxN = 20;static int maxM = 64;// To store the states of DPstatic int [,]dp = new int[maxN,maxM];static bool [,]v = new bool[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, 0, n, m); if (ans == -1) Console.WriteLine(0); else Console.WriteLine(ans);}}// This code is contributed by PrinciRaj1992 |
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, 0, n, m);if (ans == -1) document.write( 0);else document.write( ans);</script> |
Output:
3
Time Complexity: O(N * maxArr) where maxArr is the maximum element from the array.
Auxiliary Space: O(maxN * maxM)
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