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

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 ``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);` `    ``// 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

 ``

Output

`2`

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

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