Count of indices in Array having all prefix elements less than all in suffix

Given an array arr[], the task is to calculate the total number of indices where all elements in the left part is less than all elements in the right part of the array.

Examples:

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

Approach: The approach is based on the following idea:

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.

Follow the steps below to solve the given problem:

• Initialize Max = INT_MIN, Min = INT_MAX and Count = 0
• Now, create two arrays left and right of size N.
• Run one loop from start to end.
  • In each iteration update Max as Max = max(Max, arr[i]) and also assign left[i] = Max
• Run another loop from end to start.
  • In each iteration update Min as Min = min(Min, arr[i]) and also assign right[i] = Min
• Traverse the array from start to end.
• If, left[i] <= right[i+1], then a sorted point is achieved,
  • Increment Count by 1

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
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

<script>
    // 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

</script>

Output

3

Time Complexity: O(N)
Auxiliary Space: O(N)