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Maximize number of circular buildings that can be covered by L length wire

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  • Difficulty Level : Medium
  • Last Updated : 30 Jun, 2021

Given an array arr[] of size N, representing the diameter of N circular buildings and a straight wire of length L. The task is to find the maximum number of continuous buildings the wire can cover if it is round it once around the building.

Note: Distance between each building is 1 unit length, so it takes 1 unit length extra to reach one build to the next building.

Examples:

Input: arr[] = {4, 1, 6}, L = 24
Output: 2
Explanation: 1 round of building will require  pi * d length of wire, where pi is 3.14159 and d is the diameter of circle.
For 1st building → 12.566 length of wire required, remaining wire→ 24-12.566=11.434 -1( to reach next building)=10.434
Similarly, for second building 3.141 length of wire required, remaining wire→ 10.434-3.141= 7.292 -1= 6.292 
For third building 18.849, which is > remaining wire i.e, 18.849>6.292
Therefore, Maximum of 2 building can be covered.

Input: arr[] = {2, 5, 3, 4}, L = 36
Output: 3

 

Approach: The idea is to use the greedy approach. Follow the steps below to solve the problem:

  • Initialize variables curr_sum, start, curr_count, max_count to calculate the current sum of elements, current count, starting index of the current subarray, and maximum count of covered buildings.
  • Traverse the array for i in range [0, N – 1],
    • Update current sum of wire length required, curr_sum += arr[i] * 3.14
    • If i is greater than 0, Increment curr_sum by 1.
    • If curr_sum ≤ L. Increment curr_count by 1
    • Otherwise, exclude arr[start] from the current group
      • Update curr_sum = curr_sum – ((double)arr[start] * 3.14)
      • Decrement curr_sum by 1
      • Increment start pointer by 1
      • Decrement curr_count by 1
  • Update max_count i.e, max_count = max(curr_count, max_count).
  • After completing the above steps, print the value of max_count as the result.

Below is the implementation of the above approach.

C++14




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
const double Pi = 3.141592;
 
// Function to find the maximum
// number of buildings covered
int MaxBuildingsCovered(int arr[], int N, int L)
{
    // Store the current sum
    double curr_sum = 0;
 
    int start = 0, curr_count = 0, max_count = 0;
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Add the length of wire required for
        // current building to cur_sum
        curr_sum = curr_sum + ((double)arr[i] * Pi);
 
        // Add extra unit distance 1
        if (i != 0)
            curr_sum += 1;
 
        // If curr_sum <= length of wire
        // increment count by 1
        if (curr_sum <= L) {
            curr_count++;
        }
 
        // If curr_sum > length of wire
        // increment start by 1 and
        // decrement count by 1 and
        // update the new curr_sum
        else if (curr_sum > L) {
            curr_sum = curr_sum - ((double)arr[start] * Pi);
            curr_sum -= 1;
            start++;
            curr_count--;
        }
 
        // Update the max_count
        max_count = max(curr_count, max_count);
    }
 
    // Return the max_count
    return max_count;
}
 
// Driver Code
int main()
{
    // Given Input
    int arr[] = { 4, 1, 6, 2 };
    int L = 24;
 
    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << MaxBuildingsCovered(arr, N, L);
 
    return 0;
}


Java




// Java program for the above approach
import java.io.*;
 
class GFG{
 
static final double Pi = 3.141592;
 
// Function to find the maximum
// number of buildings covered
static int MaxBuildingsCovered(int arr[], int N,
                               int L)
{
     
    // Store the current sum
    double curr_sum = 0;
 
    int start = 0, curr_count = 0, max_count = 0;
 
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
         
        // Add the length of wire required for
        // current building to cur_sum
        curr_sum = curr_sum + ((double)arr[i] * Pi);
 
        // Add extra unit distance 1
        if (i != 0)
            curr_sum += 1;
 
        // If curr_sum <= length of wire
        // increment count by 1
        if (curr_sum <= L)
        {
            curr_count++;
        }
 
        // If curr_sum > length of wire
        // increment start by 1 and
        // decrement count by 1 and
        // update the new curr_sum
        else if (curr_sum > L)
        {
            curr_sum = curr_sum - ((double)arr[start] * Pi);
            curr_sum -= 1;
            start++;
            curr_count--;
        }
 
        // Update the max_count
        max_count = Math.max(curr_count, max_count);
    }
 
    // Return the max_count
    return max_count;
}
 
// Driver Code
public static void main (String[] args)
{
     
    // Given Input
    int arr[] = { 4, 1, 6, 2 };
    int L = 24;
 
    // Size of the array
    int N = arr.length;
 
    // Function Call
    System.out.println(MaxBuildingsCovered(arr, N, L));
}
}
 
// This code is contributed by Dharanendra L V.


Python3




# Python3 program for the above approach
Pi = 3.141592
 
# Function to find the maximum
# number of buildings covered
def MaxBuildingsCovered(arr, N,  L):
    # Store the current sum
    curr_sum = 0
 
    start = 0
    curr_count = 0
    max_count = 0
 
    # Traverse the array
    for i in range(N):
        # Add the length of wire required for
        # current building to cur_sum
        curr_sum = curr_sum + (arr[i] * Pi)
 
        # Add extra unit distance 1
        if (i != 0):
            curr_sum += 1
 
        # If curr_sum <= length of wire
        # increment count by 1
        if (curr_sum <= L):
            curr_count += 1
 
        # If curr_sum > length of wire
        # increment start by 1 and
        # decrement count by 1 and
        # update the new curr_sum
        elif(curr_sum > L):
            curr_sum = curr_sum - (arr[start] * Pi)
            curr_sum -= 1
            start += 1
            curr_count -= 1
 
        # Update the max_count
        max_count = max(curr_count, max_count)
 
    # Return the max_count
    return max_count
 
# Driver Code
if __name__ == '__main__':
    # Given Input
    arr = [4, 1, 6, 2]
    L = 24
 
    # Size of the array
    N = len(arr)
 
    # Function Call
    print(MaxBuildingsCovered(arr, N, L))
     
    # This code is contributed by SURENDRA_GANGWAR.


C#




// C# program for the above approach
using System;
 
class GFG{
 
static double Pi = 3.141592;
 
// Function to find the maximum
// number of buildings covered
static int MaxBuildingsCovered(int[] arr, int N,
                               int L)
{
     
    // Store the current sum
    double curr_sum = 0;
 
    int start = 0, curr_count = 0, max_count = 0;
 
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
         
        // Add the length of wire required for
        // current building to cur_sum
        curr_sum = curr_sum + ((double)arr[i] * Pi);
 
        // Add extra unit distance 1
        if (i != 0)
            curr_sum += 1;
 
        // If curr_sum <= length of wire
        // increment count by 1
        if (curr_sum <= L)
        {
            curr_count++;
        }
 
        // If curr_sum > length of wire
        // increment start by 1 and
        // decrement count by 1 and
        // update the new curr_sum
        else if (curr_sum > L)
        {
            curr_sum = curr_sum - ((double)arr[start] * Pi);
            curr_sum -= 1;
            start++;
            curr_count--;
        }
 
        // Update the max_count
        max_count = Math.Max(curr_count, max_count);
    }
 
    // Return the max_count
    return max_count;
}
 
// Driver code
static void Main()
{
     
    // Given Input
    int[] arr = { 4, 1, 6, 2 };
    int L = 24;
 
    // Size of the array
    int N = arr.Length;
 
    // Function Call
    Console.Write(MaxBuildingsCovered(arr, N, L));
}
}
 
// This code is contributed by code_hunt


Javascript




<script>
      // JavaScript program for the above approach
      var Pi = 3.141592;
 
      // Function to find the maximum
      // number of buildings covered
      function MaxBuildingsCovered(arr, N, L) {
        // Store the current sum
        var curr_sum = 0;
 
        var start = 0,
          curr_count = 0,
          max_count = 0;
 
        // Traverse the array
        for (var i = 0; i < N; i++) {
          // Add the length of wire required for
          // current building to cur_sum
          curr_sum = curr_sum + parseFloat(arr[i]) * Pi;
 
          // Add extra unit distance 1
          if (i != 0) curr_sum += 1;
 
          // If curr_sum <= length of wire
          // increment count by 1
          if (curr_sum <= L) {
            curr_count++;
          }
 
          // If curr_sum > length of wire
          // increment start by 1 and
          // decrement count by 1 and
          // update the new curr_sum
          else if (curr_sum > L) {
            curr_sum = curr_sum - parseFloat(arr[start]) * Pi;
            curr_sum -= 1;
            start++;
            curr_count--;
          }
 
          // Update the max_count
          max_count = Math.max(curr_count, max_count);
        }
 
        // Return the max_count
        return max_count;
      }
 
      // Driver code
      // Given Input
      var arr = [4, 1, 6, 2];
      var L = 24;
 
      // Size of the array
      var N = arr.length;
 
      // Function Call
      document.write(MaxBuildingsCovered(arr, N, L));
       
      // This code is contributed by rdtank.
    </script>


Output

2

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

 


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