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:
- Define function countIndices(arr, n) that takes an integer array βarrβ and its size βnβ as input.
- Initialize a variable βcountβ to 0
- Traverse the array βarrβ from index 0 to n-1:
- Initialize a variable βmax_leftβ to INT_MIN
- Traverse the left side of βarrβ including the current element and find the maximum element among them, store it in βmax_leftβ
- Initialize a variable βmin_rightβ to INT_MAX
- Traverse the right side of βarrβ after the current element and find the minimum element among them, store it in βmin_rightβ
- If the value of βmin_rightβ is not INT_MAX and βmax_leftβ is less than βmin_rightβ, then increment the value of βcountβ.
- Return the value of βcountβ.
- Define the main function
- Initialize an integer array βarrβ with given elements and its size βnβ to the size of the array
- Call the function βcountIndicesβ with βarrβ and βnβ as input
- Print the returned value from the function
Below is the implementation of the approach:
C++
// C++ code for the approach #include <bits/stdc++.h> 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 |
Python3
# Function to return the total count of sorted points in the array def countIndices(arr, n): count = 0 # Traverse through each element of the array for i in range (n): max_left = float ( "-inf" ) # Find the maximum element on the left side for j in range (i + 1 ): max_left = max (max_left, arr[j]) min_right = float ( "inf" ) # Find the minimum element on the right side for k in range (i + 1 , n): min_right = min (min_right, arr[k]) # Check if the maximum on the left is less than the minimum on the right if min_right ! = float ( "inf" ) and max_left < min_right: count + = 1 return count # Driver Code if __name__ = = "__main__" : # Input array arr = [ 1 , 5 , 4 , 2 , 3 , 8 , 7 , 9 ] n = len (arr) # Function Call result = countIndices(arr, n) print (result) |
C#
using System; class GFG { // Function to return total count // of sorted points in the array static int CountIndices( int [] arr) { int count = 0; int n = arr.Length; // Traverse through each element of the array for ( int i = 0; i < n; i++) { int maxLeft = int .MinValue; int j; // Find maximum element in left side for (j = 0; j <= i; j++) { maxLeft = Math.Max(maxLeft, arr[j]); } int minRight = int .MaxValue; // Find minimum element in right side for ( int k = i + 1; k < n; k++) { minRight = Math.Min(minRight, arr[k]); } // Check if max is less than min or not if (minRight != int .MaxValue && maxLeft < minRight) { count++; } } return count; } // Driver Code static void Main() { // Input array int [] arr = { 1, 5, 4, 2, 3, 8, 7, 9 }; // Function Call int result = CountIndices(arr); Console.WriteLine(result); } } |
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); |
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 <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> |
3
Time Complexity: O(N)
Auxiliary Space: O(N)
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