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++ 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 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 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# 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 |
<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>
|
2
Time Complexity: O(N)
Auxiliary Space: O(1)