Divide Matrix into K groups of adjacent cells having minimum difference between maximum and minimum sized groups
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 <bits/stdc++.h>using namespace std;// Function to fill the matrix with// the given conditionsvoid 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 filledint** 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 matrixvoid 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 Codeint 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 approachimport 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 conditionsstatic 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 filledstatic 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 matrixstatic 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 Codepublic 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 |
Python3
# Python3 program for the above approach# Function to fill the matrix with# the given conditionsdef fillMatrix( sizeOfpart,noOfPart, m): global start, flag, size, rem, row, col, mat # Count of parts with size sizeOfPart for i in range(noOfPart): count = 0; while (count < sizeOfpart): # Assigning the cell # with no of groups mat[row][col] = start; # Update row if (col == m - 1 and flag == 1) : row+= 1; col = m; flag = 0; elif (col == 0 and flag == 0): row+= 1; col = -1; flag = 1; # Update col if (flag == 1): col += 1; else: col -= 1; # Increment count count += 1; # For new group increment start start += 1; # Function to return the reference of# the matrix to be filleddef findMatrix(k): global start, flag, size, rem, row, col, mat # Create matrix of size N*M mat = [ [0] * N for _ in range(M) ] # Starting index of the matrix row = 0; col = 0; # Size of one group size = int((N * M) / k); 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 matrixdef printMatrix(): for row in mat: print(*row) # Driver Code# Given N, M, KN = 5;M = 5;K = 6;# Function CallfindMatrix(K);# Function Call to prmatrixprintMatrix();# This code is contributed by phasing17 |
C#
// C# program for the // above approachusing System;class GFG{ static int N, M;static int [,]mat;static int row, col, start, flag;// Function to fill the // matrix with the given // conditionsstatic 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 filledstatic 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 // matrixstatic 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 Codepublic 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 |
Javascript
// JS program for the above approachlet N, M;let mat;let row, col, start, flag;// Function to fill the matrix with// the given conditionsfunction fillMatrix( sizeOfpart, noOfPart, m){ // Count of parts with size sizeOfPart for(let i = 0; i < noOfPart; ++i) { let 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 filledfunction findMatrix(k){ // Create matrix of size N*M mat = new Array(M); for (var i = 0; i < M; i++) mat[i] = new Array(N).fill(0); // Starting index of the matrix row = 0; col = 0; // Size of one group let size = Math.floor((N * M) / k); let 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 matrixfunction printMatrix(){ // Traverse the rows for(let i = 0; i < N; ++i) { // Traverse the columns for(let j = 0; j < M; ++j) { process.stdout.write(mat[i][j] + " "); } console.log(); }}// Driver Code // Given N, M, K N = 5; M = 5; let K = 6; // Function Call findMatrix(K); // Function Call to print matrix printMatrix();// This code is contributed by phasing17 |
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)
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