Given N * M rectangular park having N rows and M columns, each cell of the park is a square of unit area and boundaries between the cells is called hex and a sprinkler can be placed in the middle of the hex. The task is to find the minimum number of sprinklers required to water the entire park.
Examples:
Input: N = 3 M = 3
Output: 5
Explanation:
For the first two columns 3 sprinklers are required and for last column we are bound to use 2 sprinklers to water the last column.Input: N = 5 M = 3
Output: 8
Explanation:
For the first two columns 5 sprinklers are required and for last column we are bound to use 3 sprinklers to water the last column.
Approach:
- After making some observation one thing can be point out i.e for every two column, N sprinkler are required because we can placed them in between of two columns.
- If M is even, then clearly N* (M / 2) sprinklers are required.
- But if M is odd then solution for M – 1 column can be computed using even column formula, and for last column add ( N + 1) / 2 sprinkler to water the last column irrespective of N is odd or even.
// C++ program to find the // minimum number sprinklers // required to water the park. #include <iostream> using namespace std;
typedef long long int ll;
// Function to find the // minimum number sprinklers // required to water the park. void solve( int N, int M)
{ // General requirements of
// sprinklers
ll ans = (N) * (M / 2);
// if M is odd then add
// one additional sprinklers
if (M % 2 == 1) {
ans += (N + 1) / 2;
}
cout << ans << endl;
} // Driver code int main()
{ int N, M;
N = 5;
M = 3;
solve(N, M);
} |
// Java program to find minimum // number sprinklers required // to cover the park class GFG{
// Function to find the minimum // number sprinklers required // to water the park. public static int solve( int n, int m)
{ // General requirements of sprinklers
int ans = n * (m / 2 );
// If M is odd then add one
// additional sprinklers
if (m % 2 == 1 )
{
ans += (n + 1 ) / 2 ;
}
return ans;
} // Driver code public static void main(String args[])
{ int N = 5 ;
int M = 3 ;
System.out.println(solve(N, M));
} } // This code is contributed by grand_master |
# Python3 program to find the # minimum number sprinklers # required to water the park. # Function to find the # minimum number sprinklers # required to water the park. def solve(N, M) :
# General requirements of
# sprinklers
ans = int ((N) * int (M / 2 ))
# if M is odd then add
# one additional sprinklers
if (M % 2 = = 1 ):
ans + = int ((N + 1 ) / 2 )
print (ans)
# Driver code N = 5
M = 3
solve(N, M) # This code is contributed by yatinagg |
// C# program to find minimum // number sprinklers required // to cover the park using System;
class GFG{
// Function to find the minimum // number sprinklers required // to water the park. public static int solve( int n, int m)
{ // General requirements of sprinklers
int ans = n * (m / 2);
// If M is odd then add one
// additional sprinklers
if (m % 2 == 1)
{
ans += (n + 1) / 2;
}
return ans;
} // Driver code public static void Main(String []args)
{ int N = 5;
int M = 3;
Console.WriteLine(solve(N, M));
} } // This code is contributed by 29AjayKumar |
<script> // javascript program to find the // minimum number sprinklers // required to water the park. // Function to find the // minimum number sprinklers // required to water the park. function solve(N, M)
{ // General requirements of
// sprinklers
var ans = (N) * parseInt((M / 2));
// if M is odd then add
// one additional sprinklers
if (M % 2 == 1) {
ans += parseInt((N + 1) / 2);
}
document.write(ans);
} // Driver code var N, M;
N = 5;
M = 3;
solve(N, M);
// This code is contributed by SURENDRA_GANGWAR. </script> |
Output:
8
Time complexity: O(1)
Auxiliary space: O(1)