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Minimum sprinklers required to water a rectangular park

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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:
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:
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: 

  1. 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.
  2. If M is even, then clearly N* (M / 2) sprinklers are required.
  3. 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++




// 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




// 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




# 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#




// 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


Javascript




<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)



Last Updated : 25 Sep, 2022
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