Count of indices in Array having all prefix elements less than all in suffix
Last Updated :
17 Oct, 2023
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 'arr' and its size 'n' as input.
- Initialize a variable 'count' to 0
- 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'.
- Return the value of 'count'.
- Define the main function
- Initialize an integer array 'arr' with given elements and its size 'n' to the size of the array
- Call the function 'countIndices' with 'arr' and 'n' 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 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
# 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)
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);
}
}
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);
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,
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 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 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# 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.
<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>
Time Complexity: O(N)
Auxiliary Space: O(N)
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