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Count cells in a grid from which maximum number of cells can be reached by K vertical or horizontal jumps

  • Last Updated : 13 Aug, 2021

Given a matrix mat[][] of dimensions N*M and a positive integer K, the task is to find the number of cells in a grid from which maximum cells can be reached by K jumps in the vertical or horizontal direction.

Examples:

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Input: N = 3, M = 3, K = 2
Output: 4
Explanation:
The cells represented as X are the cells from which maximum cells (= 2) can be reached by K (= 2) jumps:
X O X
O O O
X O X

Input: N = 5, M = 5, K = 2
Output: 9



Approach: The given problem can be solved by counting the number of possible cells for the rows and columns separately based on the following observations:

  • Consider any row, say i such that i <= K, then the number of rows from i that can be reached using jumps of K is (N – i) / K + 1.
  • If the rows say i is i > K, then these cells are connected to some row X where X <= K, hence they are already considered.

Therefore, from the above observations, the idea is to find the count of rows that are connected to the maximum number of rows in a variable, say count_R. Similarly, for the columns, find the count of columns that are connected to maximum columns in a variable, say count_C.

Now, the number of cells in the grid such that the maximum cell is reachable from that cell with a jump of K in the vertical or horizontal direction is given by (count_R)*(count_C).

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the number of cells
// in the grid such that maximum cell is
// reachable with a jump of K
long long countCells(int n, int m, int s)
{
    // Maximum reachable rows
    // from the current row
    int mx1 = -1;
 
    // Stores the count of cell that are
    // reachable from the current row
    int cont1 = 0;
 
    for (int i = 0; i < s && i < n; ++i) {
 
        // Count of reachable rows
        int aux = (n - (i + 1)) / s + 1;
 
        // Update the maximum value
        if (aux > mx1) {
            mx1 = cont1 = aux;
        }
 
        // Add it to the count
        else if (aux == mx1)
            cont1 += aux;
    }
 
    // Maximum reachable columns from
    // the current column
    int mx2 = -1;
 
    // Stores the count of cell that are
    // reachable from the current column
    int cont2 = 0;
 
    for (int i = 0; i < s && i < m; ++i) {
 
        // Count of rechable columns
        int aux = (m - (i + 1)) / s + 1;
 
        // Update the maximum value
        if (aux > mx2)
            mx2 = cont2 = aux;
 
        // Add it to the count
        else if (aux == mx2)
            cont2 += aux;
    }
 
    // Return the total count of cells
    return (long long)(cont1 * cont2);
}
 
// Driver Code
int main()
{
    int N = 5, M = 5, K = 2;
    cout << countCells(N, M, K);
 
    return 0;
}

Java




// Java program for the above approach
class GFG
{
 
  // Function to count the number of cells
  // in the grid such that maximum cell is
  // reachable with a jump of K
  public static long countCells(int n, int m, int s)
  {
     
    // Maximum reachable rows
    // from the current row
    int mx1 = -1;
 
    // Stores the count of cell that are
    // reachable from the current row
    int cont1 = 0;
 
    for (int i = 0; i < s && i < n; ++i) {
 
      // Count of reachable rows
      int aux = (n - (i + 1)) / s + 1;
 
      // Update the maximum value
      if (aux > mx1) {
        mx1 = cont1 = aux;
      }
 
      // Add it to the count
      else if (aux == mx1)
        cont1 += aux;
    }
 
    // Maximum reachable columns from
    // the current column
    int mx2 = -1;
 
    // Stores the count of cell that are
    // reachable from the current column
    int cont2 = 0;
 
    for (int i = 0; i < s && i < m; ++i) {
 
      // Count of rechable columns
      int aux = (m - (i + 1)) / s + 1;
 
      // Update the maximum value
      if (aux > mx2)
        mx2 = cont2 = aux;
 
      // Add it to the count
      else if (aux == mx2)
        cont2 += aux;
    }
 
    // Return the total count of cells
    return (long) (cont1 * cont2);
  }
 
  // Driver Code
  public static void main(String args[]) {
    int N = 5, M = 5, K = 2;
    System.out.println(countCells(N, M, K));
 
  }
 
}
 
// This code is contributed by gfgking.

Python3




# Python 3 program for the above approach
 
# Function to count the number of cells
# in the grid such that maximum cell is
# reachable with a jump of K
def countCells(n, m, s):
    # Maximum reachable rows
    # from the current row
    mx1 = -1
 
    # Stores the count of cell that are
    # reachable from the current row
    cont1 = 0
    i = 0
    while(i < s and i < n):
        # Count of reachable rows
        aux = (n - (i + 1)) // s + 1
 
        # Update the maximum value
        if (aux > mx1):
            mx1 = cont1 = aux
 
        # Add it to the count
        elif(aux == mx1):
            cont1 += aux
        i += 1
 
    # Maximum reachable columns from
    # the current column
    mx2 = -1
 
    # Stores the count of cell that are
    # reachable from the current column
    cont2 = 0
     
    i = 0
    while(i < s and i < m):
        # Count of rechable columns
        aux = (m - (i + 1)) // s + 1
 
        # Update the maximum value
        if (aux > mx2):
            mx2 = cont2 = aux
 
        # Add it to the count
        elif(aux == mx2):
            cont2 += aux
        i += 1
 
    # Return the total count of cells
    return cont1 * cont2
 
# Driver Code
if __name__ == '__main__':
    N = 5
    M = 5
    K = 2
    print(countCells(N, M, K))
     
    # This code is contributed by ipg2016107

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to count the number of cells
// in the grid such that maximum cell is
// reachable with a jump of K
static int countCells(int n, int m, int s)
{
   
    // Maximum reachable rows
    // from the current row
    int mx1 = -1;
 
    // Stores the count of cell that are
    // reachable from the current row
    int cont1 = 0;
 
    for (int i = 0; i < s && i < n; ++i) {
 
        // Count of reachable rows
        int aux = (n - (i + 1)) / s + 1;
 
        // Update the maximum value
        if (aux > mx1) {
            mx1 = cont1 = aux;
        }
 
        // Add it to the count
        else if (aux == mx1)
            cont1 += aux;
    }
 
    // Maximum reachable columns from
    // the current column
    int mx2 = -1;
 
    // Stores the count of cell that are
    // reachable from the current column
    int cont2 = 0;
 
    for (int i = 0; i < s && i < m; ++i) {
 
        // Count of rechable columns
        int aux = (m - (i + 1)) / s + 1;
 
        // Update the maximum value
        if (aux > mx2)
            mx2 = cont2 = aux;
 
        // Add it to the count
        else if (aux == mx2)
            cont2 += aux;
    }
 
    // Return the total count of cells
    return (cont1 * cont2);
}
 
// Driver Code
public static void Main()
{
    int N = 5, M = 5, K = 2;
    Console.Write(countCells(N, M, K));
}
}
 
// This code is contributed by ipg2016107.

Javascript




<script>
// Javascript program for the above approach
 
// Function to count the number of cells
// in the grid such that maximum cell is
// reachable with a jump of K
function countCells(n, m, s)
{
 
  // Maximum reachable rows
  // from the current row
  let mx1 = -1;
 
  // Stores the count of cell that are
  // reachable from the current row
  let cont1 = 0;
 
  for (let i = 0; i < s && i < n; ++i) {
    // Count of reachable rows
    let aux = Math.floor((n - (i + 1)) / s + 1);
 
    // Update the maximum value
    if (aux > mx1) {
      mx1 = cont1 = aux;
    }
 
    // Add it to the count
    else if (aux == mx1) cont1 += aux;
  }
 
  // Maximum reachable columns from
  // the current column
  let mx2 = -1;
 
  // Stores the count of cell that are
  // reachable from the current column
  let cont2 = 0;
 
  for (let i = 0; i < s && i < m; ++i) {
    // Count of rechable columns
    let aux = Math.floor((m - (i + 1)) / s + 1);
 
    // Update the maximum value
    if (aux > mx2) mx2 = cont2 = aux;
    // Add it to the count
    else if (aux == mx2) cont2 += aux;
  }
 
  // Return the total count of cells
  return cont1 * cont2;
}
 
// Driver Code
 
let N = 5,
  M = 5,
  K = 2;
document.write(countCells(N, M, K));
 
// This code is contributed by gfgking.
</script>
Output: 
9

 

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




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