# Divide Matrix into K groups of adjacent cells having minimum difference between maximum and minimum sized groups

• Last Updated : 03 Nov, 2020

Given N rows and M columns of a matrix, the task is to fill the matrix cells using first K integers such that:

• Each group of cells is denoted by a single integer.
• The difference between the group containing the maximum and the minimum number of cells should be minimum.
• All the cells of the same group should be contiguous i.e., for any group, two adjacent cells should follow the rule |xi+1 – xi| + |yi+1 – yi| = 1.

Examples:

Input: N = 5, M = 5, K = 6
Output:
1 1 1 1 1
3 2 2 2 2
3 3 3 4 4
5 5 5 4 4
5 6 6 6 6
Explanation:
The above matrix follows all the conditions above and dividing the matrix into K different groups.

Input: N = 2, M = 3, K = 3
Output:
1 1 2
3 3 2
Explanation:
For making three group of the matrix each should have the group of size two.
So, to reduce the difference between the group containing maximum and minimum no of cells and all the matrix cells are used to make the K different groups having all the adjacent elements of the same group follow the |xi + 1 – xi| + |yi + 1 – yi| = 1 as well.

Approach: Below are the steps:

• Create the matrix of size N * M.
• To reduce the difference between group containing max and min no of cells, fill all the part with at least (N*M)/ K cells.
• The remaining part will contain (N * M)/ K + 1 no of cells.
• To follow the given rules, traverse the matrix and fill the matrix with the different parts accordingly.
• Print the matrix after the above steps.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to fill the matrix with``// the given conditions``void` `fillMatrix(``int``** mat, ``int``& row,``                ``int``& col, ``int` `sizeOfpart,``                ``int` `noOfPart, ``int``& start,``                ``int` `m, ``int``& flag)``{` `    ``// Count of parts with size sizeOfPart``    ``for` `(``int` `i = 0; i < noOfPart; ++i) {` `        ``int` `count = 0;` `        ``while` `(count < sizeOfpart) {` `            ``// Assigning the cell``            ``// with no of groups``            ``mat[row][col] = start;` `            ``// Update row``            ``if` `(col == m - 1``                ``&& flag == 1) {``                ``row++;``                ``col = m;``                ``flag = 0;``            ``}``            ``else` `if` `(col == 0``                     ``&& flag == 0) {``                ``row++;``                ``col = -1;``                ``flag = 1;``            ``}` `            ``// Update col``            ``if` `(flag == 1) {``                ``col++;``            ``}``            ``else` `{``                ``col--;``            ``}` `            ``// Increment count``            ``count++;``        ``}` `        ``// For new group increment start``        ``start++;``    ``}``}` `// Function to return the reference of``// the matrix to be filled``int``** findMatrix(``int` `N, ``int` `M, ``int` `k)``{` `    ``// Create matrix of size N*M``    ``int``** mat = (``int``**)``malloc``(``        ``N * ``sizeof``(``int``*));` `    ``for` `(``int` `i = 0; i < N; ++i) {``        ``mat[i] = (``int``*)``malloc``(``            ``M * ``sizeof``(``int``));``    ``}` `    ``// Starting index of the matrix``    ``int` `row = 0, col = 0;` `    ``// Size of one group``    ``int` `size = (N * M) / k;``    ``int` `rem = (N * M) % k;` `    ``// Element to assigned to matrix``    ``int` `start = 1, flag = 1;` `    ``// Fill the matrix that have rem``    ``// no of parts with size size + 1``    ``fillMatrix(mat, row, col, size + 1,``               ``rem, start, M, flag);` `    ``// Fill the remaining number of parts``    ``// with each part size is 'size'``    ``fillMatrix(mat, row, col, size,``               ``k - rem, start, M, flag);` `    ``// Return the matrix``    ``return` `mat;``}` `// Function to print the matrix``void` `printMatrix(``int``** mat, ``int` `N,``                 ``int` `M)``{``    ``// Traverse the rows``    ``for` `(``int` `i = 0; i < N; ++i) {` `        ``// Traverse the columns``        ``for` `(``int` `j = 0; j < M; ++j) {``            ``cout << mat[i][j] << ``" "``;``        ``}``        ``cout << endl;``    ``}``}` `// Driver Code``int` `main()``{``    ``// Given N, M, K``    ``int` `N = 5, M = 5, K = 6;` `    ``// Function Call``    ``int``** mat = findMatrix(N, M, K);` `    ``// Function Call to print matrix``    ``printMatrix(mat, N, M);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;` `class` `GFG{``    ` `static` `int` `N, M;``static` `int` `[][]mat;``static` `int` `row, col, start, flag;` `// Function to fill the matrix with``// the given conditions``static` `void` `fillMatrix(``int` `sizeOfpart,``                       ``int` `noOfPart,``                       ``int` `m)``{``    ` `    ``// Count of parts with size sizeOfPart``    ``for``(``int` `i = ``0``; i < noOfPart; ++i)``    ``{``        ``int` `count = ``0``;` `        ``while` `(count < sizeOfpart)``        ``{``            ` `            ``// Assigning the cell``            ``// with no of groups``            ``mat[row][col] = start;` `            ``// Update row``            ``if` `(col == m - ``1` `&& flag == ``1``)``            ``{``                ``row++;``                ``col = m;``                ``flag = ``0``;``            ``}``            ``else` `if` `(col == ``0` `&& flag == ``0``)``            ``{``                ``row++;``                ``col = -``1``;``                ``flag = ``1``;``            ``}` `            ``// Update col``            ``if` `(flag == ``1``)``            ``{``                ``col++;``            ``}``            ``else``            ``{``                ``col--;``            ``}` `            ``// Increment count``            ``count++;``        ``}` `        ``// For new group increment start``        ``start++;``    ``}``}` `// Function to return the reference of``// the matrix to be filled``static` `void` `findMatrix(``int` `k)``{``    ` `    ``// Create matrix of size N*M``    ``mat = ``new` `int``[M][N];` `    ``// Starting index of the matrix``    ``row = ``0``;``    ``col = ``0``;` `    ``// Size of one group``    ``int` `size = (N * M) / k;``    ``int` `rem = (N * M) % k;` `    ``// Element to assigned to matrix``    ``start = ``1``;``    ``flag = ``1``;` `    ``// Fill the matrix that have rem``    ``// no of parts with size size + 1``    ``fillMatrix(size + ``1``, rem, M);` `    ``// Fill the remaining number of parts``    ``// with each part size is 'size'``    ``fillMatrix(size, k - rem, M);``}` `// Function to print the matrix``static` `void` `printMatrix()``{``    ` `    ``// Traverse the rows``    ``for``(``int` `i = ``0``; i < N; ++i)``    ``{``        ` `        ``// Traverse the columns``        ``for``(``int` `j = ``0``; j < M; ++j)``        ``{``            ``System.out.print(mat[i][j] + ``" "``);``        ``}``        ``System.out.println();``    ``}``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ` `    ``// Given N, M, K``    ``N = ``5``;``    ``M = ``5``;``    ``int` `K = ``6``;` `    ``// Function Call``    ``findMatrix(K);` `    ``// Function Call to print matrix``    ``printMatrix();``}``}` `// This code is contributed by Amit Katiyar`

## C#

 `// C# program for the``// above approach``using` `System;``class` `GFG{``    ` `static` `int` `N, M;``static` `int` `[,]mat;``static` `int` `row, col,``           ``start, flag;` `// Function to fill the``// matrix with the given``// conditions``static` `void` `fillMatrix(``int` `sizeOfpart,``                       ``int` `noOfPart,``                       ``int` `m)``{   ``  ``// Count of parts with size``  ``// sizeOfPart``  ``for``(``int` `i = 0;``          ``i < noOfPart; ++i)``  ``{``    ``int` `count = 0;` `    ``while` `(count < sizeOfpart)``    ``{``      ``// Assigning the cell``      ``// with no of groups``      ``mat[row, col] = start;` `      ``// Update row``      ``if` `(col == m - 1 &&``          ``flag == 1)``      ``{``        ``row++;``        ``col = m;``        ``flag = 0;``      ``}``      ``else` `if` `(col == 0 &&``               ``flag == 0)``      ``{``        ``row++;``        ``col = -1;``        ``flag = 1;``      ``}` `      ``// Update col``      ``if` `(flag == 1)``      ``{``        ``col++;``      ``}``      ``else``      ``{``        ``col--;``      ``}` `      ``// Increment count``      ``count++;``    ``}` `    ``// For new group increment``    ``// start``    ``start++;``  ``}``}` `// Function to return the``// reference of the matrix``// to be filled``static` `void` `findMatrix(``int` `k)``{   ``  ``// Create matrix of``  ``// size N*M``  ``mat = ``new` `int``[M, N];` `  ``// Starting index of the``  ``// matrix``  ``row = 0;``  ``col = 0;` `  ``// Size of one group``  ``int` `size = (N * M) / k;``  ``int` `rem = (N * M) % k;` `  ``// Element to assigned to``  ``// matrix``  ``start = 1;``  ``flag = 1;` `  ``// Fill the matrix that have``  ``// rem no of parts with size``  ``// size + 1``  ``fillMatrix(size + 1,``             ``rem, M);` `  ``// Fill the remaining number``  ``// of parts with each part``  ``// size is 'size'``  ``fillMatrix(size, k - rem, M);``}` `// Function to print the``// matrix``static` `void` `printMatrix()``{   ``  ``// Traverse the rows``  ``for``(``int` `i = 0; i < N; ++i)``  ``{``    ``// Traverse the columns``    ``for``(``int` `j = 0; j < M; ++j)``    ``{``      ``Console.Write(mat[i, j] +``                    ``" "``);``    ``}``    ``Console.WriteLine();``  ``}``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{   ``  ``// Given N, M, K``  ``N = 5;``  ``M = 5;``  ``int` `K = 6;` `  ``// Function Call``  ``findMatrix(K);` `  ``// Function Call to``  ``// print matrix``  ``printMatrix();``}``}` `// This code is contributed by 29AjayKumar`

Output:

```1 1 1 1 1
3 2 2 2 2
3 3 3 4 4
5 5 5 4 4
5 6 6 6 6

```

Time Complexity: O(N * M)
Auxiliary Space: O(N * M)

My Personal Notes arrow_drop_up