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

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 satisifes the above property then the given range cannot contain a crest otherwise the crest is always possible.
  • To find which element satisifes 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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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Output:

Yes


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