Given three integers R, C, N, and an array arr[] of size N. The task is to color all cells of a grid of R rows and C columns such that all same color cells are connected either horizontally or vertically. N represents the colors numbered from 1 to N and arr[] denotes the quantity of each color. The total quantity of color is exactly equal to the total number of cells of the grid.
Approach:
Input: R = 3, C = 5, N = 5, arr[] = {1, 2, 3, 4, 5}
Output:
1 4 4 4 3
2 5 4 5 3
2 5 5 5 3
Explanation: Available colors are 1(count = 1), 2(count = 2), 3(count = 3) etc.
For color 5: we can reach all color 5s by going horizontally or vertically through the same color 5.
Similarly for color 3, the rightmost row contains all 3 etc.
Similarly, for the rest of the colors 1, 2, 4.
Below is an invalid grid:
1 4 3 4 4
2 5 4 5 3
2 5 5 5 3
This is because the connection for the colors 3 and 4 has been broken by the invalid position of 3
in the position(0, 2).We can no longer traverse through all the 4s or all the 3s, horizontally or vertically, by passing through the respective 3s and 4s only.
Input: R = 2, C = 2, N = 3, arr[] = {2, 1, 1}
Output:
1 1
2 3
Approach:
At first glance, it might seem that graph algorithms are required. However, we are going to follow an optimized greedy algorithm.
- Create a new 2D array which will be our final grid. Let us call it dp[][].
- Traverse the color array A[]
- For each color, i having A[i] quantities
- If the row is an odd-numbered row, fill the dp array from left to right
- Else if it is an even row, fill it from right to left
- If the quantity of color is used up, move on to the next color greedily
Below is the implementation of the above approach:
C++
// C++ Program to Color a grid// such that all same color cells// are connected either// horizontally or vertically#include <bits/stdc++.h>using namespace std;void solve(vector<int>& arr, int r, int c){ // Current color int idx = 1; // final grid int dp[r]; for (int i = 0; i < r; i++) { // if even row if (i % 2 == 0) { // traverse from left to // right for (int j = 0; j < c; j++) { // if color has been exhausted //, move to the next color if (arr[idx - 1] == 0) idx++; // color the grid at // this position dp[i][j] = idx; // reduce the color count arr[idx - 1]--; } } else { // traverse from right to // left for odd rows for (int j = c - 1; j >= 0; j--) { if (arr[idx - 1] == 0) idx++; dp[i][j] = idx; arr[idx - 1]--; } } } // print the grid for (int i = 0; i < r; ++i) { for (int j = 0; j < c; ++j) { cout << dp[i][j] << " "; } cout << endl; }}// Driver codeint main(){ int r = 3, c = 5; int n = 5; vector<int> arr = { 1, 2, 3, 4, 5 }; solve(arr, r, c); return 0;} |
Java
// Java program to color a grid // such that all same color cells // are connected either // horizontally or vertically import java.util.*; class GFG{ static void solve(List<Integer> arr, int r, int c) { // Current color int idx = 1; // Final grid int[][] dp = new int[r]; for(int i = 0; i < r; i++) { // If even row if (i % 2 == 0) { // Traverse from left to // right for(int j = 0; j < c; j++) { // If color has been exhausted //, move to the next color if (arr.get(idx - 1) == 0) idx++; // Color the grid at // this position dp[i][j] = idx; // Reduce the color count arr.set(idx - 1, arr.get(idx - 1) - 1); } } else { // Traverse from right to // left for odd rows for(int j = c - 1; j >= 0; j--) { if (arr.get(idx - 1) == 0) idx++; dp[i][j] = idx; arr.set(idx - 1, arr.get(idx - 1) - 1); } } } // Print the grid for(int i = 0; i < r; ++i) { for(int j = 0; j < c; ++j) { System.out.print(dp[i][j] + " "); } System.out.println(); } } // Driver Code public static void main (String[] args){ int r = 3, c = 5; int n = 5; List<Integer> arr = Arrays.asList(1, 2, 3, 4, 5); solve(arr, r, c); } }// This code is contributed by offbeat |
Python3
# Python3 program to color a grid# such that all same color cells# are connected either# horizontally or verticallydef solve(arr, r, c): # Current color idx = 1 # Final grid dp = [[0 for i in range(c)] for i in range(r)] for i in range(r): # If even row if (i % 2 == 0): # Traverse from left to # right for j in range(c): # If color has been exhausted, # move to the next color if (arr[idx - 1] == 0): idx += 1 # Color the grid at # this position # print(i,j) dp[i][j] = idx # Reduce the color count arr[idx - 1] -= 1 else: # Traverse from right to # left for odd rows for j in range(c - 1, -1, -1): if (arr[idx - 1] == 0): idx += 1 dp[i][j] = idx arr[idx - 1] -= 1 # Print the grid for i in range(r): for j in range(c): print(dp[i][j], end = " ") print()# Driver codeif __name__ == '__main__': r = 3 c = 5 n = 5 arr = [ 1, 2, 3, 4, 5 ] solve(arr, r, c)# This code is contributed by mohit kumar 29 |
C#
// C# program to color a grid // such that all same color cells // are connected either // horizontally or vertically using System;using System.Collections.Generic;class GFG{ static void solve(List<int> arr, int r, int c) { // Current color int idx = 1; // Final grid int[,] dp = new int[r, c]; for(int i = 0; i < r; i++) { // If even row if (i % 2 == 0) { // Traverse from left to // right for(int j = 0; j < c; j++) { // If color has been exhausted, // move to the next color if (arr[idx - 1] == 0) idx++; // Color the grid at // this position dp[i, j] = idx; // Reduce the color count arr[idx - 1] = arr[idx - 1] - 1; } } else { // Traverse from right to // left for odd rows for(int j = c - 1; j >= 0; j--) { if (arr[idx - 1] == 0) idx++; dp[i, j] = idx; arr[idx - 1] = arr[idx - 1] - 1; } } } // Print the grid for(int i = 0; i < r; ++i) { for(int j = 0; j < c; ++j) { Console.Write(dp[i, j] + " "); } Console.Write('\n'); } } // Driver Code public static void Main (string[] args){ int r = 3, c = 5; //int n = 5; List<int> arr = new List<int>(); arr.Add(1); arr.Add(2); arr.Add(3); arr.Add(4); arr.Add(5); solve(arr, r, c); } }// This code is contributed by rutvik_56 |
Javascript
<script>// Javascript Program to Color a grid// such that all same color cells// are connected either// horizontally or verticallyfunction solve(arr,r,c){ // Current color var idx = 1; // final grid var dp = new Array(r); var i,j; for(i=0;i<r;i++) dp[i] = new Array(c); for(i = 0; i < r; i++) { // if even row if (i % 2 == 0) { // traverse from left to // right for(j = 0; j < c; j++) { // if color has been exhausted //, move to the next color if (arr[idx - 1] == 0) idx++; // color the grid at // this position dp[i][j] = idx; // reduce the color count arr[idx - 1]--; } } else { // traverse from right to // left for odd rows for (j = c - 1; j >= 0; j--) { if (arr[idx - 1] == 0) idx++; dp[i][j] = idx; arr[idx - 1]--; } } } // print the grid for(i = 0; i < r; ++i) { for(j = 0; j < c; ++j) { document.write(dp[i][j] + " "); } document.write("<br>"); }}// Driver code var r = 3, c = 5; var n = 5; var arr = [1, 2, 3, 4, 5]; solve(arr, r, c);</script> |
Output:
1 2 2 3 3 4 4 4 4 3 5 5 5 5 5
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