📜  具有给定 OR 值的最长子序列:动态规划方法

📅  最后修改于: 2021-09-17 06:58:31             🧑  作者: Mango

给定一个数组arr[] ,任务是找到具有给定 OR 值M的最长子序列。如果没有这样的子序列,则打印0

例子:

方法:一个简单的解决方案是生成所有可能的子序列,然后找到其中具有所需 OR 值的最大子序列。然而,对于较小的M值,可以使用动态规划方法。
我们先来看递归关系。

现在让我们了解DP的状态。这里, dp[i][curr_or]存储子数组arr[i…N-1]的最长子序列,因此curr_or在与该子序列进行 OR 运算时给出M。在每一步,要么选择索引i并更新curr_or,要么拒绝索引i并继续。

下面是上述方法的实现:

C++
// C++ implementation of the approach
#include 
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


输出:
3

时间复杂度: O(N * maxArr) 其中 maxArr 是数组中的最大元素。

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