Dada una array arr[] y un entero M , la tarea es encontrar la subsecuencia más larga con un valor AND dado M . Si no existe tal subsecuencia, imprima 0 .
Ejemplos:
Entrada: arr[] = {3, 7, 2, 3}, M = 3
Salida: 3
La subsecuencia más larga con valor AND 3 es {3, 7, 3}.
Entrada: arr[] = {2, 2}, M = 3
Salida: 0
Enfoque: una forma simple de resolver esto será generar todas las subsecuencias posibles y luego encontrar la más grande entre ellas con el valor AND requerido.
Sin embargo, para valores más pequeños de M , se puede utilizar un enfoque basado en la programación dinámica .
Veamos primero la relación de recurrencia.
dp[i][curr_and] = max(dp[i + 1][curr_and], dp[i + 1][curr_and & arr[i]] + 1)
Comprendamos los estados de DP ahora. Aquí, dp[i][curr_and] almacena la subsecuencia más larga del subarreglo arr[i…N-1] tal que curr_and & AND de esta subsecuencia es igual a M . En cada paso, se puede elegir el índice i actualizando curr_and o se puede rechazar.
A continuación se muestra la implementación del enfoque anterior:
C++
// 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; }
Java
// 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
Python3
# 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
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 approach var maxN = 20 var maxM = 64 // To store the states of DP var dp = Array.from(Array(maxN), ()=> Array(maxM)); var v = Array.from(Array(maxN), ()=> Array(maxM)); // Function to return the required length function 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 code var 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>
3
Complejidad de tiempo: O (N * maxVal) donde maxVal es el elemento máximo de la array dada.
Espacio Auxiliar: O(maxN * maxM)
Publicación traducida automáticamente
Artículo escrito por DivyanshuShekhar1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA