Given an odd number **N** which represents a grid of size **N x N** which is initially filled by coins, the task is to find the minimum number of moves required for all the coins to move to any cell of the grid such that in each step, some arbitrary coin in the middle of the grid can move to any of the surrounding eight cells.

**Examples:**

Input:N = 3

Output:8

Explanation:

There are a total of 9 cells in a 3 x 3 grid. On assuming that all the coins have to be brought to the central cell, 8 coins should be moved and the number of steps is 8.

Input:N = 5

Output:40

**Approach:** The minimum number of steps is obtained only when all the coins are moved to the centre of the grid. Therefore, the idea is to divide the entire grid into multiple layers.

- Each layer of the grid K takes K moves to reach to the centre. That is:
- The coins at layer 1 take one step to move to the centre.
- The coins at layer 2 take two steps to move to the centre and so on.

- For example, let N = 5. Then, the grid is divided into the layers as follows:
- In the above illustration, the coins which are marked in the red is at layer 1 and the coins which are marked blue is at layer 2.
- Similarly, for a grid of size N x N, we can divide the coins into N//2 layers.
- In each layer K, the number of coins present is (8 * K). And, the number of steps required is K. Therefore, iterate through all the layers and find the total number of steps as
**8 * K**^{2}

Below is the implementation of the above approach:

## C++

`// C++ program to find the minimum number ` `// of moves taken to move the element of ` `// each cell to any one cell of the ` `// square matrix of odd length ` `#include <iostream> ` `using` `namespace` `std; ` ` ` `// Function to find the minimum number ` `// of moves taken to move the element ` `// of each cell to any one cell of the ` `// square matrix of odd length ` `int` `calculateMoves(` `int` `n) ` `{ ` ` ` ` ` `// Initializing count to 0 ` ` ` `int` `count = 0; ` ` ` ` ` `// Number of layers that are ` ` ` `// around the centre element ` ` ` `int` `layers = n / 2; ` ` ` ` ` `// Iterating over ranger of layers ` ` ` `for` `(` `int` `k = 1; k < layers + 1; k++) ` ` ` `{ ` ` ` ` ` `// Increase the value of count ` ` ` `// by 8 * k * k ` ` ` `count += 8 * k * k; ` ` ` `} ` ` ` `return` `count; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `N = 5; ` ` ` ` ` `cout << calculateMoves(N); ` `} ` ` ` `// This code is contributed by coder001 ` |

*chevron_right*

*filter_none*

## Java

`// Java program to find the minimum number ` `// of moves taken to move the element of ` `// each cell to any one cell of the ` `// square matrix of odd length ` `class` `GFG{ ` ` ` `// Function to find the minimum number ` `// of moves taken to move the element ` `// of each cell to any one cell of the ` `// square matrix of odd length ` `public` `static` `int` `calculateMoves(` `int` `n) ` `{ ` ` ` ` ` `// Initializing count to 0 ` ` ` `int` `count = ` `0` `; ` ` ` ` ` `// Number of layers that are ` ` ` `// around the centre element ` ` ` `int` `layers = n / ` `2` `; ` ` ` ` ` `// Iterating over ranger of layers ` ` ` `for` `(` `int` `k = ` `1` `; k < layers + ` `1` `; k++) ` ` ` `{ ` ` ` ` ` `// Increase the value of count ` ` ` `// by 8 * k * k ` ` ` `count += ` `8` `* k * k; ` ` ` `} ` ` ` `return` `count; ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` ` ` `int` `N = ` `5` `; ` ` ` ` ` `System.out.println(calculateMoves(N)); ` `} ` `} ` ` ` `// This code is contributed by divyeshrabadiya07 ` |

*chevron_right*

*filter_none*

## Python3

`# Python3 program to find the minimum number ` `# of moves taken to move the element of ` `# each cell to any one cell of the ` `# square matrix of odd length ` ` ` `# Function to find the minimum number ` `# of moves taken to move the element of ` `# each cell to any one cell of the ` `# square matrix of odd length ` `def` `calculateMoves(n): ` ` ` ` ` `# Initializing count to 0 ` ` ` `count ` `=` `0` ` ` ` ` `# Number of layers that are ` ` ` `# around the centre element ` ` ` `layers ` `=` `n` `/` `/` `2` ` ` ` ` `# Iterating over ranger of layers ` ` ` `for` `k ` `in` `range` `(` `1` `, layers ` `+` `1` `): ` ` ` ` ` `# Increase the value of count by ` ` ` `# 8 * k * k ` ` ` `count` `+` `=` `8` `*` `k` `*` `k ` ` ` ` ` `return` `count ` ` ` `# Driver code ` `if` `__name__ ` `=` `=` `"__main__"` `: ` ` ` ` ` `N ` `=` `5` ` ` ` ` `print` `(calculateMoves(N)) ` |

*chevron_right*

*filter_none*

## C#

`// C# program to find the minimum number ` `// of moves taken to move the element of ` `// each cell to any one cell of the ` `// square matrix of odd length ` `using` `System; ` `class` `GFG{ ` ` ` `// Function to find the minimum number ` `// of moves taken to move the element ` `// of each cell to any one cell of the ` `// square matrix of odd length ` `public` `static` `int` `calculateMoves(` `int` `n) ` `{ ` ` ` ` ` `// Initializing count to 0 ` ` ` `int` `count = 0; ` ` ` ` ` `// Number of layers that are ` ` ` `// around the centre element ` ` ` `int` `layers = n / 2; ` ` ` ` ` `// Iterating over ranger of layers ` ` ` `for` `(` `int` `k = 1; k < layers + 1; k++) ` ` ` `{ ` ` ` ` ` `// Increase the value of count ` ` ` `// by 8 * k * k ` ` ` `count += 8 * k * k; ` ` ` `} ` ` ` `return` `count; ` `} ` ` ` `// Driver code ` `public` `static` `void` `Main() ` `{ ` ` ` `int` `N = 5; ` ` ` ` ` `Console.Write(calculateMoves(N)); ` `} ` `} ` ` ` `// This code is contributed by Code_Mech ` |

*chevron_right*

*filter_none*

**Output:**

40

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.

## Recommended Posts:

- Find minimum moves to bring all elements in one cell of a matrix
- Maximum of all distances to the nearest 1 cell from any 0 cell in a Binary matrix
- Number of shortest paths to reach every cell from bottom-left cell in the grid
- Maximum path sum that starting with any cell of 0-th row and ending with any cell of (N-1)-th row
- Final cell position in the matrix
- Distance of nearest cell having 1 in a binary matrix
- Sum of cost of all paths to reach a given cell in a Matrix
- Final direction after visiting every cell of Matrix starting from (0, 0)
- Number of ways of cutting a Matrix such that atleast one cell is filled in each part
- Minimum cells traversed to reach corner where every cell represents jumps
- Minimum cells required to reach destination with jumps equal to cell values
- Find sum of all elements in a matrix except the elements in row and/or column of given cell?
- Transportation Problem | Set 3 (Least Cost Cell Method)
- Find if a 2-D array is completely traversed or not by following the cell values
- Check if cells numbered 1 to K in a grid can be connected after removal of atmost one blocked cell
- Make a fair coin from a biased coin
- Print all paths from top left to bottom right in a matrix with four moves allowed
- Find maximum path sum in a 2D matrix when exactly two left moves are allowed
- Expected number of moves to reach the end of a board | Matrix Exponentiation
- Minimize steps required to move all 1's in a matrix to a given index

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.