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📜  计算前缀数组中最大值小于后缀数组中的最大值的索引

📅  最后修改于: 2021-10-26 05:46:41             🧑  作者: Mango

给定一个大小为N的数组arr[] ,任务是找到最大值小于剩余后缀数组中最大元素的前缀数组的数量。

例子:

朴素的方法:最简单的方法是找到给定数组的所有可能的前缀和后缀,并计算前缀数组中最大元素小于后缀数组中最大元素的索引数量。

时间复杂度: O(N 2 )
辅助空间: O(1)

高效方法:对上述方法进行优化,其思想是存储数组中每个前缀和后缀的最大值,然后计算最大值小于其对应后缀数组的前缀数组的数量。请按照以下步骤解决问题:

  • 初始化一个变量,比如ans和两个大小为N 的数组prefix[]suffix[]
  • [0, N – 1]范围内遍历数组arr[]并且对于prefix[] 中的每个i索引,将最大元素存储为prefix[i] = max(prefix[i – 1], arr[i] )
  • [N – 1, 0]范围内以相反的顺序遍历给定数组,对于suffix[] 中的每个i索引,将最大元素存储为suffix[i] = max(suffix[i + 1], arr[我])
  • 现在,在[0, N – 2]范围内遍历数组arr[] ,如果prefix[i]小于suffix[i] ,则将ans增加1
  • 完成以上步骤后,打印ans的值作为结果。

下面是上述方法的实现:

C++
// C++ program for the above approach
#include 
using namespace std;
 
// Function to print the count of indices
// in which the maximum in prefix arrays
// is less than that in the suffix array
void count(int a[], int n)
{
    // If size of array is 1
    if (n == 1) {
        cout << 0;
        return;
    }
 
    // pre[]: Prefix array
    // suf[]: Suffix array
    int pre[n - 1], suf[n - 1];
    int max = a[0];
 
    // Stores the required count
    int ans = 0, i;
 
    pre[0] = a[0];
 
    // Find the maximum in prefix array
    for (i = 1; i < n - 1; i++) {
 
        if (a[i] > max)
            max = a[i];
 
        pre[i] = max;
    }
 
    max = a[n - 1];
    suf[n - 2] = a[n - 1];
 
    // Find the maximum in suffix array
    for (i = n - 2; i >= 1; i--) {
 
        if (a[i] > max)
            max = a[i];
 
        suf[i - 1] = max;
    }
 
    // Traverse the array
    for (i = 0; i < n - 1; i++) {
 
        // If maximum in prefix array
        // is less than maximum in
        // the suffix array
        if (pre[i] < suf[i])
            ans++;
    }
 
    // Print the answer
    cout << ans;
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 3, 4, 8, 1, 4 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    count(arr, N);
 
    return 0;
}


Java
// Java program for the above approach
public class gfg
{
  // Function to print the count of indices
  // in which the maximum in prefix arrays
  // is less than that in the suffix array
  static void count(int a[], int n)
  {
 
    // If size of array is 1
    if (n == 1)
    {
      System.out.print(0);
      return;
    }
 
    // pre[]: Prefix array
    // suf[]: Suffix array
    int[] pre = new int[n - 1];
    int[] suf = new int[n - 1];
    int max = a[0];
 
    // Stores the required count
    int ans = 0, i;         
    pre[0] = a[0];
 
    // Find the maximum in prefix array
    for(i = 1; i < n - 1; i++)
    {
      if (a[i] > max)
        max = a[i];
 
      pre[i] = max;
    }       
    max = a[n - 1];
    suf[n - 2] = a[n - 1];
 
    // Find the maximum in suffix array
    for(i = n - 2; i >= 1; i--)
    {
      if (a[i] > max)
        max = a[i];                 
      suf[i - 1] = max;
    }
 
    // Traverse the array
    for(i = 0; i < n - 1; i++)
    {
 
      // If maximum in prefix array
      // is less than maximum in
      // the suffix array
      if (pre[i] < suf[i])
        ans++;
    }
 
    // Print the answer
    System.out.print(ans);
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int arr[] = { 2, 3, 4, 8, 1, 4 };
    int N = arr.length;
 
    // Function Call
    count(arr, N);
  }
}
 
// This code is contributed by divyesh072019.


Python3
# Python program for the above approach
 
# Function to print the count of indices
# in which the maximum in prefix arrays
# is less than that in the suffix array
def count(a, n) :
 
    # If size of array is 1
    if (n == 1) :
        print(0)
        return
 
    # pre[]: Prefix array
    # suf[]: Suffix array
    pre = [0] * (n - 1)
    suf = [0] * (n - 1)
    max = a[0]
 
    # Stores the required count
    ans = 0
 
    pre[0] = a[0]
 
    # Find the maximum in prefix array
    for i in range(n-1):
 
        if (a[i] > max):
            max = a[i]
 
        pre[i] = max
 
    max = a[n - 1]
    suf[n - 2] = a[n - 1]
 
    # Find the maximum in suffix array
    for i in range(n-2, 0, -1):
 
        if (a[i] > max):
            max = a[i]
 
        suf[i - 1] = max
 
    # Traverse the array
    for i in range(n - 1):
 
        # If maximum in prefix array
        # is less than maximum in
        # the suffix array
        if (pre[i] < suf[i]) :
            ans += 1
 
    # Print the answer
    print(ans)
     
 
# Driver Code
 
arr = [ 2, 3, 4, 8, 1, 4 ]
N = len(arr)
 
# Function Call
count(arr, N)
 
# This code is contributed by code_hunt.


C#
// C# program for the above approach
using System;
 
class GFG{
     
// Function to print the count of indices
// in which the maximum in prefix arrays
// is less than that in the suffix array
static void count(int[] a, int n)
{
     
    // If size of array is 1
    if (n == 1)
    {
        Console.Write(0);
        return;
    }
   
    // pre[]: Prefix array
    // suf[]: Suffix array
    int[] pre = new int[n - 1];
    int[] suf = new int[n - 1];
    int max = a[0];
   
    // Stores the required count
    int ans = 0, i;
     
    pre[0] = a[0];
     
    // Find the maximum in prefix array
    for(i = 1; i < n - 1; i++)
    {
        if (a[i] > max)
            max = a[i];
             
        pre[i] = max;
    }
   
    max = a[n - 1];
    suf[n - 2] = a[n - 1];
     
    // Find the maximum in suffix array
    for(i = n - 2; i >= 1; i--)
    {
        if (a[i] > max)
            max = a[i];
             
        suf[i - 1] = max;
    }
   
    // Traverse the array
    for(i = 0; i < n - 1; i++)
    {
         
        // If maximum in prefix array
        // is less than maximum in
        // the suffix array
        if (pre[i] < suf[i])
            ans++;
    }
     
    // Print the answer
    Console.Write(ans);
}
 
// Driver code
static void Main()
{
    int[] arr = { 2, 3, 4, 8, 1, 4 };
    int N = arr.Length;
     
    // Function Call
    count(arr, N);
}
}
 
// This code is contributed by divyeshrabadiya07


Javascript


输出:
3

时间复杂度: O(N)
辅助空间: O(N)

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