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Count of squares reachable by a Bishop initially placed at top left on a given NxM chessboard

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  • Last Updated : 29 Nov, 2021

Given two integers N and M representing a N x M chessboard, the task is to find the maximum number of squares that the bishop can reach using any number of moves if initially it is placed in the top left corner of the chessboard.

Examples:

Input: N = 8, M = 8
Output: 32
Explanation: The bishop is initially standing on (1, 1) which is either a white or a black colour tile. Therefore, either all the black or the white tiles can be visited from (1, 1) using a sequence of moves depending on the color of the (1, 1) tile. 

Input: N = 7, M = 3
Output: 11

 

Approach: The given problem can be solved by observing the fact that the number of reachable tiles from (1, 1) are the tiles with the same color as that of (1, 1). The count of such tiles can be calculated by the formula ceil((N*M)/2). The case where the above-mentioned statement proves wrong is the case where either N = 1 or M = 1. In such cases, no cells are reachable from (1, 1) and hence the required answer is 1.

Below is the implementation of the above approach:

C++




// C++ program of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum number of
// reachable squares by a bishop in an
// N x M chessboard
int maximumSquares(int N, int M)
{
    // If either of N or M is equal to 1
    if (N == 1 || M == 1) {
        return 1;
    }
 
    // Otherwise the required
    // answer is Ceil(N*M/2)
    // Return Answer
    return (N * M + 1) / 2;
}
 
// Driver Code
int main()
{
    int N = 7;
    int M = 3;
 
    cout << maximumSquares(N, M);
 
    return 0;
}

Java




// Java program for the above approach
 
class GFG {
 
// Function to find the maximum number of
// reachable squares by a bishop in an
// N x M chessboard
static int maximumSquares(int N, int M)
{
   
    // If either of N or M is equal to 1
    if (N == 1 || M == 1) {
        return 1;
    }
 
    // Otherwise the required
    // answer is Ceil(N*M/2)
    // Return Answer
    return (N * M + 1) / 2;
}
 
// Driver Code
public static void main (String[] args) {
         
    int N = 7;
    int M = 3;
 
    System.out.println(maximumSquares(N, M));
}
}
 
// This code is contributed by target_2.

Python




# Python program of the above approach
 
# Function to find the maximum number of
# reachable squares by a bishop in an
# N x M chessboard
def maximumSquares(N, M) :
 
    # If either of N or M is equal to 1
    if (N == 1 or M == 1) :
        return 1
 
    # Otherwise the required
    # answer is Ceil(N*M/2)
    # Return Answer
    return (N * M + 1) // 2
 
# Driver Code
if __name__ == "__main__" :
    N = 7
    M = 3
 
    print(maximumSquares(N, M))
     
    # This code is contributed by Samim Hossain Mondal.

C#




// C# program for the above approach
using System;
 
class GFG {
 
// Function to find the maximum number of
// reachable squares by a bishop in an
// N x M chessboard
static int maximumSquares(int N, int M)
{
   
    // If either of N or M is equal to 1
    if (N == 1 || M == 1) {
        return 1;
    }
 
    // Otherwise the required
    // answer is Ceil(N*M/2)
    // Return Answer
    return (N * M + 1) / 2;
}
 
// Driver Code
public static void Main () {
         
    int N = 7;
    int M = 3;
 
    Console.Write(maximumSquares(N, M));
}
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript




<script>
// Javascript program of the above approach
 
// Function to find the maximum number of
// reachable squares by a bishop in an
// N x M chessboard
function maximumSquares(N, M)
{
    // If either of N or M is equal to 1
    if (N == 1 || M == 1) {
        return 1;
    }
 
    // Otherwise the required
    // answer is Ceil(N*M/2)
    // Return Answer
    return (N * M + 1) / 2;
}
 
// Driver Code
let N = 7;
let M = 3;
 
document.write(maximumSquares(N, M));
 
// This code is contributed by Samim Hossain Mondal.
</script>

 
 

Output
11

 

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

 


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