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# 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, 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

Naive Approach:

The naive approach is that we traverse each element of the array and for each element find maximum element say max in the left side which is including itself also and find minimum element say min in the right side which is after ith element. Finding maximum and minimum will be requiring another loop traversal. Now, check if max is less than min or not. If lesser then increase the count of elements, else continue.

Algorithm:

1. Define function countIndices(arr, n) that takes an integer array ‘arr‘ and its size ‘n’ as input.
2. Initialize a variable ‘count‘ to 0
3. Traverse the array ‘arr‘ from index 0 to n-1:
•  Initialize a variable ‘max_left‘ to INT_MIN
•  Traverse the left side of ‘arr‘ including the current element and find the maximum element among them, store it in ‘max_left
•  Initialize a variable ‘min_right‘ to INT_MAX
•  Traverse the right side of ‘arr‘ after the current element and find the minimum element among them, store it in ‘min_right
•  If the value of ‘min_right‘ is not INT_MAX and ‘max_left‘ is less than ‘min_right‘, then increment the value of ‘count‘.
4. Return the value of ‘count‘.
5. Define the main function
6. Initialize an integer array ‘arr‘ with given elements and its size ‘n’ to the size of the array
7. Call the function ‘countIndices‘ with ‘arr‘ and ‘n‘ as input
8. Print the returned value from the function

Below is the implementation of the approach:

## C++

 `// C++ code for the approach` `#include ``using` `namespace` `std;` `// Function to return total count``// of sorted points in the array``int` `countIndices(``int` `arr[], ``int` `n) {``    ``int` `count = 0;``  ` `    ``// Traverse through each element of the array``    ``for` `(``int` `i = 0; i < n; i++) {` `        ``int` `max_left = INT_MIN;``        ``int` `j;``        ``// Find maximum element in left side``        ``for` `(j = 0; j <= i; j++) {``            ``max_left = max(max_left, arr[j]);``        ``}` `        ``int` `min_right = INT_MAX;``        ``// Find minimum element in right side``        ``for` `(``int` `k = i + 1; k < n; k++) {``            ``min_right = min(min_right, arr[k]);``        ``}` `        ``// Check if max is less than min or not``        ``if` `(min_right != INT_MAX && max_left < min_right) {``            ``count++;``        ``}``    ``}` `    ``return` `count;``}` `// Driver Code``int` `main() {``      ``// Input array``    ``int` `arr[] = { 1, 5, 4, 2, 3, 8, 7, 9 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(``int``);``      ` `      ``// Function Call``    ``int` `result = countIndices(arr, n);``    ``cout << result << endl;` `    ``return` `0;``}`

## Java

 `// Java code for the approach``import` `java.io.*;``import` `java.util.*;` `public` `class` `GFG {``    ` `// Function to return total count``// of sorted points in the array``static` `int` `countIndices(``int` `arr[], ``int` `n) {``    ``int` `count = ``0``;` `    ``// Traverse through each element of the array``    ``for` `(``int` `i = ``0``; i < n; i++) {` `        ``int` `max_left = Integer.MIN_VALUE;``        ``int` `j;``        ``// Find maximum element in left side``        ``for` `(j = ``0``; j <= i; j++) {``            ``max_left = Math.max(max_left, arr[j]);``        ``}` `        ``int` `min_right = Integer.MAX_VALUE;``        ``// Find minimum element in right side``        ``for` `(``int` `k = i + ``1``; k < n; k++) {``            ``min_right = Math.min(min_right, arr[k]);``        ``}` `        ``// Check if max is less than min or not``        ``if` `(min_right != Integer.MAX_VALUE && max_left < min_right) {``            ``count++;``        ``}``    ``}` `    ``return` `count;``}` `// Driver Code``public` `static` `void` `main(String[] args) {``    ``// Input array``    ``int` `arr[] = { ``1``, ``5``, ``4``, ``2``, ``3``, ``8``, ``7``, ``9` `};``    ``int` `n = arr.length;``    ` `    ``// Function Call``    ``int` `result = countIndices(arr, n);``    ``System.out.println(result);` `}``}` `// This code has been contributed by Pushpesh Raj`

## Javascript

 `function` `GFG(arr) {``    ``let count = 0;``    ``// Traverse through each element of``    ``// the array``    ``for` `(let i = 0; i < arr.length; i++) {``        ``let max_left = -Infinity;``        ``for` `(let j = 0; j <= i; j++) {``            ``max_left = Math.max(max_left, arr[j]);``        ``}``        ``let min_right = Infinity;``        ``// Find minimum element in the``        ``// right side``        ``for` `(let k = i + 1; k < arr.length; k++) {``            ``min_right = Math.min(min_right, arr[k]);``        ``}``        ``// Check if max is less than min or not``        ``if` `(min_right !== Infinity && max_left < min_right) {``            ``count++;``        ``}``    ``}``    ``return` `count;``}``// Driver Code``const arr = [1, 5, 4, 2, 3, 8, 7, 9];``// Function Call``const result = GFG(arr);``console.log(result);`

Output

```3

```

Time Complexity: O(n*n) where n is size of input array. This is because two nested for loops are being executed in function countIndices.
Auxiliary Space: O(1)  as no extra space has been used.

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)