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📜  最大子序列,以使所有索引和所有值分别为倍数

📅  最后修改于: 2021-05-17 06:29:55             🧑  作者: Mango

给定N个正整数的数组arr [] ,任务是找到arr []的最大严格增加的子序列,以使arr []中所选元素的索引以及所选元素分别是彼此的倍数。
注意:考虑对数组arr []基于1的索引。
例子:

天真的方法:
天真的方法是简单地生成所有可能的子序列,并对每个子序列检查两个条件:

  • 首先检查元素是否按照严格的顺序排列,以及
  • 其次,检查arr []中所选元素的索引是否彼此倍数。

在满足给定两个条件的所有可能子序列中,选择最大的子序列。
时间复杂度: O(N * 2 N )
辅助空间: O(N)
高效方法:
我们可以通过使用动态编程来优化代码,方法是通过缓存其结果来避免重复子问题的冗余计算。

  1. 创建一个大小等于arr []大小的数组dp [] ,其中dp [i]表示满足给定条件的第i个索引之前最大子序列的大小。
  2. 用0初始化数组dp []
  3. 现在,从末尾迭代数组arr []
  4. 对于每个索引,找到将当前索引除以索引j ,然后检查当前索引处的值是否大于索引j处的元素。
  5. 如果是,则将dp [j]更新为:
  1. 最后,遍历数组dp []并打印最大值。

下面是有效方法的实现:

C++
// C++ program for the above approach
#include 
using namespace std;
 
// Function that print maximum length
// of array
void maxLength(int arr[], int n)
{
    // dp[] array to store the
    // maximum length
    vector dp(n, 1);
 
    for (int i = n - 1; i > 1; i--) {
 
        // Find all divisors of i
        for (int j = 1;
             j <= sqrt(i); j++) {
 
            if (i % j == 0) {
                int s = i / j;
 
                if (s == j) {
 
                    // If the current value
                    // is greater than the
                    // divisor's value
                    if (arr[i] > arr[s]) {
 
                        dp[s] = max(dp[i] + 1,
                                    dp[s]);
                    }
                }
                else {
 
                    // If current value
                    // is greater
                    // than the divisor's value
                    // and s is not equal
                    // to current index
                    if (s != i
                        && arr[i] > arr[s])
                        dp[s] = max(dp[i] + 1,
                                    dp[s]);
 
                    // Condition if current
                    // value is greater
                    // than the divisor's value
                    if (arr[i] > arr[j]) {
                        dp[j] = max(dp[i] + 1,
                                    dp[j]);
                    }
                }
            }
        }
    }
 
    int max = 0;
 
    // Computing the greatest value
    for (int i = 1; i < n; i++) {
        if (dp[i] > max)
            max = dp[i];
    }
 
    // Printing maximum length of array
    cout << max << "\n";
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 0, 1, 4, 2, 3, 6, 4, 9 };
    int size = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    maxLength(arr, size);
    return 0;
}


Java
// Java program for the above approach
import java.util.*;
import java.io.*;
 
class GFG{
     
// Function that print maximum length
// of array
static void maxLength(int arr[], int n)
{
     
    // dp[] array to store the
    // maximum length
    int dp[] = new int[n];
    for(int i = 1; i < n; i++)
    {
        dp[i] = 1;
    }
 
    for(int i = n - 1; i > 1; i--)
    {
         
        // Find all divisors of i
        for(int j = 1;
                j <= Math.sqrt(i); j++)
        {
            if (i % j == 0)
            {
                int s = i / j;
 
                if (s == j)
                {
                     
                    // If the current value
                    // is greater than the
                    // divisor's value
                    if (arr[i] > arr[s])
                    {
                        dp[s] = Math.max(dp[i] + 1,
                                         dp[s]);
                    }
                }
                else
                {
                     
                    // If current value is greater
                    // than the divisor's value
                    // and s is not equal
                    // to current index
                    if (s != i && arr[i] > arr[s])
                        dp[s] = Math.max(dp[i] + 1,
                                         dp[s]);
     
                    // Condition if current
                    // value is greater
                    // than the divisor's value
                    if (arr[i] > arr[j])
                    {
                        dp[j] = Math.max(dp[i] + 1,
                                         dp[j]);
                    }
                }
            }
        }
    }
    int max = 0;
 
    // Computing the greatest value
    for(int i = 1; i < n; i++)
    {
        if (dp[i] > max)
            max = dp[i];
    }
     
    // Printing maximum length of array
    System.out.println(max);
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array arr[]
    int arr[] = { 0, 1, 4, 2, 3, 6, 4, 9 };
    int size = arr.length;
 
    // Function call
    maxLength(arr, size);
}
}
 
// This code is contributed by sanjoy_62


Python3
# Python3 program for the above approach
from math import *
 
# Function that print maximum length
# of array
def maxLength (arr, n):
 
    # dp[] array to store the
    # maximum length
    dp = [1] * n
 
    for i in range(n - 1, 1, -1):
 
        # Find all divisors of i
        for j in range(1, int(sqrt(i)) + 1):
            if (i % j == 0):
                s = i // j
 
                if (s == j):
 
                    # If the current value
                    # is greater than the
                    # divisor's value
                    if (arr[i] > arr[s]):
                        dp[s] = max(dp[i] + 1, dp[s])
 
                else:
                    # If current value
                    # is greater
                    # than the divisor's value
                    # and s is not equal
                    # to current index
                    if (s != i and arr[i] > arr[s]):
                        dp[s] = max(dp[i] + 1, dp[s])
 
                    # Condition if current
                    # value is greater
                    # than the divisor's value
                    if (arr[i] > arr[j]):
                        dp[j] = max(dp[i] + 1, dp[j])
 
    Max = 0
 
    # Computing the greatest value
    for i in range(1, n):
        if (dp[i] > Max):
            Max = dp[i]
 
    # Printing maximum length of array
    print(Max)
 
# Driver Code
if __name__ == '__main__':
 
    # Given array arr[]
    arr = [ 0, 1, 4, 2, 3, 6, 4, 9]
    size = len(arr)
 
    # Function call
    maxLength(arr, size)
 
# This code is contributed by himanshu77


C#
// C# program for the above approach
using System;
 
class GFG{
 
// Function that print maximum length
// of array
static void maxLength(int[] arr, int n)
{
     
    // dp[] array to store the
    // maximum length
    int[] dp = new int[n];
    for(int i = 1; i < n; i++)
    {
        dp[i] = 1;
    }
 
    for(int i = n - 1; i > 1; i--)
    {
         
        // Find all divisors of i
        for(int j = 1;
                j <= Math.Sqrt(i); j++)
        {
            if (i % j == 0)
            {
                int s = i / j;
                 
                if (s == j)
                {
                     
                    // If the current value
                    // is greater than the
                    // divisor's value
                    if (arr[i] > arr[s])
                    {
                        dp[s] = Math.Max(dp[i] + 1,
                                         dp[s]);
                    }
                }
                else
                {
                     
                    // If current value is greater
                    // than the divisor's value
                    // and s is not equal
                    // to current index
                    if (s != i && arr[i] > arr[s])
                        dp[s] = Math.Max(dp[i] + 1,
                                         dp[s]);
     
                    // Condition if current
                    // value is greater
                    // than the divisor's value
                    if (arr[i] > arr[j])
                    {
                        dp[j] = Math.Max(dp[i] + 1,
                                         dp[j]);
                    }
                }
            }
        }
    }
    int max = 0;
 
    // Computing the greatest value
    for(int i = 1; i < n; i++)
    {
        if (dp[i] > max)
            max = dp[i];
    }
 
    // Printing maximum length of array
    Console.WriteLine(max);
}
 
// Driver Code
public static void Main()
{
     
    // Given array arr[]
    int[] arr = new int[] { 0, 1, 4, 2,
                            3, 6, 4, 9 };
    int size = arr.Length;
 
    // Function call
    maxLength(arr, size);
}
}
 
// This code is contributed by sanjoy_62


Javascript


输出:
3

时间复杂度: O(N *(sqrt(N))由于既然对于数组的每个索引,我们都将计算其所有的除数,所以需要O(sqrt(N))
辅助空间: O(N)