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Number of buildings facing the sun

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Given an array representing heights of buildings. The array has buildings from left to right as shown in below diagram, count number of buildings facing the sunset. It is assumed that heights of all buildings are distinct.

Examples: 

Input : arr[] = {7, 4, 8, 2, 9}
Output: 3
Explanation: As 7 is the first element, it can 
see the sunset.
4 can't see the sunset as 7 is hiding it. 
8 can see.
2 can't see the sunset.
9 also can see the sunset.

Input : arr[] = {2, 3, 4, 5}
Output : 4

Asked in : Amazon Interview

It can be easily observed that only the maximum element found so far will see the sunlight 
i.e. curr_max will see the sunlight and then only the element greater than curr_max will see the sunlight. We traverse given array from left to right. We keep track of maximum element seen so far. Whenever an element becomes more than current max, increment result and update current max.

Implementation:

C++

// C++ program to count buildings that can
// see sunlight.
#include <iostream>
using namespace std;
  
// Returns count buildings 
// that can see sunlight
int countBuildings(int arr[], int n)
{
    // Initialize result (Note that first building
    // always sees sunlight)
    int count = 1;
  
    // Start traversing element
    int curr_max = arr[0];
    for (int i = 1; i < n; i++) {
        
        // If curr_element is maximum
        // or current element is
        // equal, update maximum and increment count
        if (arr[i] > curr_max || arr[i] == curr_max) {
            count++;
            curr_max = arr[i];
        }
    }
  
    return count;
}
  
// Driver code
int main()
{
    int arr[] = { 7, 4, 8, 2, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countBuildings(arr, n);
    return 0;
}

                    

Java

// Java program to count buildings that can
// see sunlight.
  
class Test {
    // Returns count buildings that can see sunlight
    static int countBuildings(int arr[], int n)
    {
        // Initialize result  (Note that first building
        // always sees sunlight)
        int count = 1;
  
        // Start traversing element
        int curr_max = arr[0];
        for (int i = 1; i < n; i++) {
            
            // If curr_element is maximum 
            // or current element
            // is equal, update maximum and increment count
            if (arr[i] > curr_max || arr[i] == curr_max) {
                count++;
                curr_max = arr[i];
            }
        }
  
        return count;
    }
  
    // Driver method
    public static void main(String[] args)
    {
        int arr[] = { 7, 4, 8, 2, 9 };
  
        System.out.println(countBuildings(arr, arr.length));
    }
}

                    

Python3

# Python3 program to count buildings
# that can see sunlight.
  
# Returns count buildings that
# can see sunlight
  
  
def countBuildings(arr, n):
  
    # Initialize result (Note that first
    # building always sees sunlight)
    count = 1
  
    # Start traversing element
    curr_max = arr[0]
    for i in range(1, n):
  
        # If curr_element is maximum or 
        # current element is equal,
        # update maximum and increment count
        if (arr[i] > curr_max or arr[i] == curr_max):
  
            count += 1
            curr_max = arr[i]
  
    return count
  
  
# Driver code
arr = [7, 4, 8, 2, 9]
n = len(arr)
print(countBuildings(arr, n))
  
# This code is contributed by Rohit.

                    

C#

// C# program to count buildings that can
// see sunlight.
using System;
  
class GFG {
    
    // Returns count buildings 
    // that can see sunlight
    static int countBuildings(int[] arr, int n)
    {
          
        // Initialize result (Note that first building
        // always sees sunlight)
        int count = 1;
  
        // Start traversing element
        int curr_max = arr[0];
        for (int i = 1; i < n; i++) {
            
            // If curr_element is maximum 
            // or current element
            // is equal, update maximum 
            // and increment count
            if (arr[i] > curr_max || arr[i] == curr_max) {
                count++;
                curr_max = arr[i];
            }
        }
  
        return count;
    }
  
    // Driver method
    public static void Main()
    {
        int[] arr = { 7, 4, 8, 2, 9 };
  
        Console.Write(countBuildings(arr, arr.Length));
    }
}
  
// This code is contributed by Rohit.

                    

PHP

<?php
// php program to count buildings
// that can see sunlight.
  
// Returns count buildings that 
// can see sunlight
function countBuildings($arr, $n)
{
    // Initialize result (Note that
    // first building always sees 
    // sunlight)
    $count = 1;
  
    // Start traversing element
    $curr_max = $arr[0];
    for ( $i = 1; $i < $n; $i++)
    {
          
        // If curr_element is maximum or
        // current element is equal,
        // update maximum and 
        // increment count
        if ($arr[$i] > $curr_max || $arr[$i] == $curr_max)
        {
            $count++;
            $curr_max=$arr[$i];
        }
    }
  
    return $count;
}
  
// Driver code
$arr = array(7, 4, 8, 2, 9);
$n = sizeof($arr) / sizeof($arr[0]);
echo countBuildings($arr, $n);
  
// This code is contributed by 
// Rohit
?>

                    

Javascript

<script>
  
// Javascript program to count buildings that can
// see sunlight.
  
// Returns count buildings 
// that can see sunlight
function countBuildings( arr, n)
{
    // Initialize result (Note that first building
    // always sees sunlight)
    let count = 1;
  
    // Start traversing element
    let curr_max = arr[0];
    for (let i = 1; i < n; i++) {
        
        // If curr_element is maximum
        // or current element is
        // equal, update maximum and increment count
        if (arr[i] > curr_max || arr[i] == curr_max) {
            count++;
            curr_max = arr[i];
        }
    }
  
    return count;
}
  
      
    // Driver program
      
    let arr = [ 7, 4, 8, 2, 9 ];
    let n = arr.length;
    document.write(countBuildings(arr, n));
         
</script>

                    

Output
3

Time Complexity: O(n) 
Auxiliary Space: O(1)



Last Updated : 14 Sep, 2023
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