Find if a crest is present in the index range [L, R] of the given array

Last Updated : 08 Mar, 2022

Given an array arr[] of N distinct elements and an index range [L, R]. The task is to find whether a crest is present in that index range in the array or not. Any element arr[i] in the subarray arr[L…R] is called a crest if all the elements of the subarray arr[L…i] are strictly increasing and all elements of the subarray arr[i…R] are strictly decreasing.
Examples:

Input: arr[] = {2, 1, 3, 5, 12, 11, 7, 9}, L = 2, R = 6
Output: Yes
Element 12 is a crest in the subarray {3, 5, 12, 11, 7}.
Input: arr[] = {2, 1, 3, 5, 12, 11, 7, 9}, L = 0, R = 2
Output: No

Approach:

Check whether in the given index range [L, R] there exists an element which satisfies the Property where arr[i – 1] ? arr[i] ? arr[i + 1].
If any element in the given range satisfies the above property then the given range cannot contain a crest otherwise the crest is always possible.
To find which element satisfies the above property, maintain an array present[] where present[i] will be 1 if arr[i – 1] ? arr[i] ? arr[i + 1] else present[i] will be 0.
Now convert the present[] array to its cumulative sum where present[i] will now represent number of elements in the index range [0, i] that satisfy the property.
For an index range [L, R] to contain a crest, present[L] must be equal to present[R – 1].

Below is the implementation of the above approach:

C++

   // C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns true if the
// array contains a crest in
// the index range [L, R]
bool hasCrest(int arr[], int n, int L, int R)
{
    // To keep track of elements
    // which satisfy the Property
    int present[n] = { 0 };
 
    for (int i = 1; i <= n - 2; i++) {
 
        // Property is satisfied for
        // the current element
        if ((arr[i] <= arr[i + 1])
            && (arr[i] <= arr[i - 1])) {
            present[i] = 1;
        }
    }
 
    // Cumulative Sum
    for (int i = 1; i < n; i++) {
        present[i] += present[i - 1];
    }
 
    // If a crest is present in
    // the given index range
    if (present[L] == present[R - 1])
        return true;
 
    return false;
}
 
// Driver code
int main()
{
    int arr[] = { 2, 1, 3, 5, 12, 11, 7, 9 };
    int N = sizeof(arr) / sizeof(arr[0]);
    int L = 2;
    int R = 6;
 
    if (hasCrest(arr, N, L, R))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}
 
  

Java

   // Java implementation of the approach
import java.util.*;
 
class GFG
{
     
// Function that returns true if the
// array contains a crest in
// the index range [L, R]
static boolean hasCrest(int arr[], int n, 
                        int L, int R)
{
    // To keep track of elements
    // which satisfy the Property
    int []present = new int[n];
    for(int i = 0; i < n; i++)
    {
        present[i] = 0;
    }
 
    for (int i = 1; i <= n - 2; i++) 
    {
 
        // Property is satisfied for
        // the current element
        if ((arr[i] <= arr[i + 1]) && 
            (arr[i] <= arr[i - 1])) 
        {
            present[i] = 1;
        }
    }
 
    // Cumulative Sum
    for (int i = 1; i < n; i++) 
    {
        present[i] += present[i - 1];
    }
 
    // If a crest is present in
    // the given index range
    if (present[L] == present[R - 1])
        return true;
 
    return false;
}
 
// Driver code
public static void main(String args[])
{
    int arr[] = { 2, 1, 3, 5, 12, 11, 7, 9 };
    int N = arr.length;
    int L = 2;
    int R = 6;
 
    if (hasCrest(arr, N, L, R))
        System.out.println("Yes");
    else
        System.out.println("No");
 
}
}
 
// This code is contributed by Surendra_Gangwar
 
  

Python3

   # Python3 implementation of the approach 
 
# Function that returns true if the 
# array contains a crest in 
# the index range [L, R] 
def hasCrest(arr, n, L, R) : 
 
    # To keep track of elements 
    # which satisfy the Property 
    present = [0] * n ; 
 
    for i in range(1, n - 2 + 1) :
 
        # Property is satisfied for 
        # the current element 
        if ((arr[i] <= arr[i + 1]) and
            (arr[i] <= arr[i - 1])) :
            present[i] = 1; 
 
    # Cumulative Sum 
    for i in range(1, n) :
        present[i] += present[i - 1]; 
 
    # If a crest is present in 
    # the given index range 
    if (present[L] == present[R - 1]) :
        return True; 
 
    return False; 
 
# Driver code 
if __name__ == "__main__" : 
 
    arr = [ 2, 1, 3, 5, 12, 11, 7, 9 ]; 
    N = len(arr); 
    L = 2; 
    R = 6; 
 
    if (hasCrest(arr, N, L, R)) :
        print("Yes"); 
    else :
        print("No"); 
 
# This code is contributed by AnkitRai01
 
  

C#

        
// C# implementation of the approach
using System;
 
class GFG
{
     
// Function that returns true if the
// array contains a crest in
// the index range [L, R]
static bool hasCrest(int []arr, int n, 
                        int L, int R)
{
    // To keep track of elements
    // which satisfy the Property
    int []present = new int[n];
    for(int i = 0; i < n; i++)
    {
        present[i] = 0;
    }
 
    for (int i = 1; i <= n - 2; i++) 
    {
 
        // Property is satisfied for
        // the current element
        if ((arr[i] <= arr[i + 1]) && 
            (arr[i] <= arr[i - 1])) 
        {
            present[i] = 1;
        }
    }
 
    // Cumulative Sum
    for (int i = 1; i < n; i++) 
    {
        present[i] += present[i - 1];
    }
 
    // If a crest is present in
    // the given index range
    if (present[L] == present[R - 1])
        return true;
 
    return false;
}
 
// Driver code
public static void Main(String []args)
{
    int []arr = { 2, 1, 3, 5, 12, 11, 7, 9 };
    int N = arr.Length;
    int L = 2;
    int R = 6;
 
    if (hasCrest(arr, N, L, R))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed by PrinciRaj1992
 
  

Javascript

   <script>
 
// JavaScript implementation of the approach 
 
// Function that returns true if the 
// array contains a crest in 
// the index range [L, R] 
function hasCrest(arr, n, L, R) 
{ 
    // To keep track of elements 
    // which satisfy the Property 
    let present = new Uint8Array(n); 
 
    for (let i = 1; i <= n - 2; i++) { 
 
        // Property is satisfied for 
        // the current element 
        if ((arr[i] <= arr[i + 1]) 
            && (arr[i] <= arr[i - 1])) { 
            present[i] = 1; 
        } 
    } 
 
    // Cumulative Sum 
    for (let i = 1; i < n; i++) { 
        present[i] += present[i - 1]; 
    } 
 
    // If a crest is present in 
    // the given index range 
    if (present[L] == present[R - 1]) 
        return true; 
 
    return false; 
} 
 
// Driver code 
 
    let arr = [ 2, 1, 3, 5, 12, 11, 7, 9 ]; 
    let N = arr.length; 
    let L = 2; 
    let R = 6; 
 
    if (hasCrest(arr, N, L, R)) 
        document.write("Yes"); 
    else
        document.write("No"); 
 
 
// This code is contributed by Surbhi Tyagi.
 
</script>
 
  

Output:

Yes

Time Complexity: O(n)

Auxiliary Space: O(n)

Find the peak index of a given array

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Find if a crest is present in the index range [L, R] of the given array

C++

Java

Python3

C#

Javascript

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