Dada una array arr[] , la tarea es encontrar la subsecuencia más larga con un valor OR dado M . Si no existe tal subsecuencia, imprima 0 .
Ejemplos:
Entrada: arr[] = {3, 7, 2, 3}, M = 3
Salida: 3
{3, 2, 3} es la subsecuencia requerida
3 | 2 | 3 = 3
Entrada: arr[] = {2, 2}, M = 3
Salida: 0
Enfoque: una solución simple es generar todas las subsecuencias posibles y luego encontrar la más grande entre ellas con el valor OR requerido. Sin embargo, para valores más pequeños de M , se puede utilizar un enfoque de programación dinámica .
Veamos primero la relación de recurrencia.
dp[i][curr_or] = max(dp[i + 1][curr_or], dp[i + 1][curr_or | arr[i]] + 1)
Comprendamos los estados de DP ahora. Aquí, dp[i][curr_or] almacena la subsecuencia más larga del subarreglo arr[i…N-1] tal que curr_or da M cuando obtiene OR con esta subsecuencia. En cada paso, elija el índice i y actualice curr_or o rechace el índice i y continúe.
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, 0, n, m); if (ans == -1) cout << 0; else cout << ans; return 0; }
Java
// Java implementation of the approach class GFG { static int maxN = 20; static int maxM = 64; // 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, 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 np maxN = 20 maxM = 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 approach using System; class GFG { static int maxN = 20; static int maxM = 64; // 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, 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 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, 0, n, m); if (ans == -1) document.write( 0); else document.write( ans); </script>
3
Complejidad de tiempo: O(N * maxArr) donde maxArr es el elemento máximo de la array.
Publicación traducida automáticamente
Artículo escrito por DivyanshuShekhar1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA