Array 中所有前缀元素少于后缀中的所有元素的索引计数

给定一个数组arr[] ，任务是计算索引的总数，其中左侧的所有元素都小于数组右侧的所有元素。

例子：

Input: arr[] = {1, 5, 4, 2, 3, 8, 7, 9}
Output: 3
Explanation: 

• Lets consider left part = [1], right part = [5, 4, 2, 3, 8, 7, 9]
  Here, left_Max (1) < right_Min (2). So, it can be considered as sorted point.
• Again, If we consider left part = [1, 5, 4, 2, 3], right part = [8, 7, 9]
  Here also, left_Max < right_Min, So, it can also be considered as sorted point.
• Similarly, If we consider left part = [1, 5, 4, 2, 3, 8, 7], right part = [9]
  Here, left_Max < right_Min, So, it can also be considered as sorted point.

Hence, total 3 sorted points are found.

Input: arr[] = {5, 2, 3, 4, 1}
Output: 0

方法：该方法基于以下思想：

The idea to solve the problem is by traversing the array and initialize two arrays to store the left part of the array and the right part of the array. 
Then check if the maximum element of the left part of the array is less than the minimum element of the right part of the array. 
If this condition is satisfied it is the sorted point and hence, increment the count by one and so on.

请按照以下步骤解决给定的问题：

• 初始化Max = INT_MIN、 Min = INT_MAX 和Count = 0
• 现在，创建大小为N的左右两个数组。
• 从头到尾运行一个循环。
  • 在每次迭代中将 Max 更新为Max = max(Max, arr[i]) 并分配 left[i] = Max
• 从头到尾运行另一个循环。
  • 在每次迭代中将 Min 更新为Min = min(Min, arr[i]) 并分配 right[i] = Min
• 从头到尾遍历数组。
• 如果left[i] <= right[i+1] ，则达到排序点，
  • 将计数增加 1

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to return total count
// of sorted points in the array
int countSortedPoints(int* arr, int N)
{
 
    int left[N];
    int right[N];
 
    // Initialize the variables
    int Min = INT_MAX;
    int Max = INT_MIN;
    int Count = 0;
 
    // Make Maximum array
    for (int i = 0; i < N; i++) {
 
        Max = max(arr[i], Max);
        left[i] = Max;
    }
 
    // Make Minimum array
    for (int i = N - 1; i >= 0; i--) {
 
        Min = min(arr[i], Min);
        right[i] = Min;
    }
 
    // Count of sorted points
    for (int i = 0; i < N - 1; i++) {
        if (left[i] <= right[i + 1])
            Count++;
    }
 
    // Return count of sorted points
    return Count;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 5, 4, 2, 3, 8, 7, 9 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    cout << countSortedPoints(arr, N);
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
class GFG {
 
  // Function to return total count
  // of sorted points in the array
  static int countSortedPoints(int []arr, int N)
  {
 
    int []left = new int[N];
    int []right = new int[N];
 
    // Initialize the variables
    int Min = Integer.MAX_VALUE;
    int Max = Integer.MIN_VALUE;
    int Count = 0;
 
    // Make Maximum array
    for (int i = 0; i < N; i++) {
 
      Max = Math.max(arr[i], Max);
      left[i] = Max;
    }
 
    // Make Minimum array
    for (int i = N - 1; i >= 0; i--) {
 
      Min = Math.min(arr[i], Min);
      right[i] = Min;
    }
 
    // Count of sorted points
    for (int i = 0; i < N - 1; i++) {
      if (left[i] <= right[i + 1])
        Count++;
    }
 
    // Return count of sorted points
    return Count;
  }
 
  // Driver Code
  public static void main (String[] args) {
    int arr[] = { 1, 5, 4, 2, 3, 8, 7, 9 };
    int N = arr.length;
 
    // Function call
    System.out.print(countSortedPoints(arr, N));
  }
}
 
// This code is contributed by hrithikgarg03188.

Python3

# Python3 implementation of the approach
INT_MIN = -2147483648
INT_MAX = 2147483647
 
# Function to return total count
# of sorted points in the array
def countSortedPoints(arr, N):
 
    left = [0 for i in range(N)]
    right = [0 for i in range(N)]
 
    # Initialize the variables
    Min = INT_MAX
    Max = INT_MIN
    Count = 0
 
    # Make Maximum array
    for i in range(N):
 
        Max = max(arr[i], Max)
        left[i] = Max
 
            # Make Minimum array
    for i in range(N - 1, -1, -1):
 
        Min = min(arr[i], Min)
        right[i] = Min
 
    # Count of sorted points
    for i in range(0, N - 1):
        if (left[i] <= right[i + 1]):
            Count += 1
 
            # Return count of sorted points
    return Count
 
# Driver Code
arr = [1, 5, 4, 2, 3, 8, 7, 9]
N = len(arr)
 
# Function call
print(countSortedPoints(arr, N))
 
# This code is contributed by shinjanpatra

C#

// C# program for the above approach
using System;
class GFG
{
 
// Function to return total count
// of sorted points in the array
static int countSortedPoints(int []arr, int N)
{
 
    int []left = new int[N];
    int []right = new int[N];
 
    // Initialize the variables
    int Min = Int32.MaxValue;
    int Max = Int32.MinValue;
    int Count = 0;
 
    // Make Maximum array
    for (int i = 0; i < N; i++) {
 
        Max = Math.Max(arr[i], Max);
        left[i] = Max;
    }
 
    // Make Minimum array
    for (int i = N - 1; i >= 0; i--) {
 
        Min = Math.Min(arr[i], Min);
        right[i] = Min;
    }
 
    // Count of sorted points
    for (int i = 0; i < N - 1; i++) {
        if (left[i] <= right[i + 1])
            Count++;
    }
 
    // Return count of sorted points
    return Count;
}
 
// Driver Code
public static void Main()
{
    int []arr = { 1, 5, 4, 2, 3, 8, 7, 9 };
    int N = arr.Length;
 
    // Function call
    Console.Write(countSortedPoints(arr, N));
}
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

    // JavaScript program for the above approach
 
    const INT_MIN = -2147483647 - 1;
    const INT_MAX = 2147483647;
 
    // Function to return total count
    // of sorted points in the array
    const countSortedPoints = (arr, N) => {
 
        let left = new Array(N).fill(0);
        let right = new Array(N).fill(0);
 
        // Initialize the variables
        let Min = INT_MAX;
        let Max = INT_MIN;
        let Count = 0;
 
        // Make Maximum array
        for (let i = 0; i < N; i++) {
 
            Max = Math.max(arr[i], Max);
            left[i] = Max;
        }
 
        // Make Minimum array
        for (let i = N - 1; i >= 0; i--) {
 
            Min = Math.min(arr[i], Min);
            right[i] = Min;
        }
 
        // Count of sorted points
        for (let i = 0; i < N - 1; i++) {
            if (left[i] <= right[i + 1])
                Count++;
        }
 
        // Return count of sorted points
        return Count;
    }
 
    // Driver Code
 
    let arr = [1, 5, 4, 2, 3, 8, 7, 9];
    let N = arr.length;
 
    // Function call
    document.write(countSortedPoints(arr, N));
 
// This code is contributed by rakeshsahni
 

输出

3

时间复杂度：O(N)
辅助空间：O(N)