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

• Last Updated : 24 Mar, 2022

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, leftMax (1) < rightMin (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, leftMax < rightMin, 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, leftMax < rightMin, 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 ` `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

 ``

Output

`3`

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

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