Color a grid such that all same color cells are connected either horizontally or vertically

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 optimised greedy algorithm.

  1. Create a new 2D array which will be our final grid. Let us call it dp[][].
  2. Traverse the color array A[]
  3. 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  
  4. If the quantity of a color is used up, move on to the next color greedily
  5. Arrow directions for filling the dp array:
    ------->
    <-------
    -------->
    <--------

    Example: R = 3, C = 5, N = 5, A = [1, 2, 3, 4, 5]

    1 2 2 3 3 
    [row 0 -> fill from left to right]
    4 4 4 4 3 
    [row 1 -> fill from right to left]
    5 5 5 5 5 [row 2 -> fill from left to right]

    Below is the implementation of the above approach:

    C++

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    // 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 code
    int main()
    {
        int r = 3, c = 5;
        int n = 5;
        vector<int> arr
            = { 1, 2, 3, 4, 5 };
        solve(arr, r, c);
        return 0;
    }

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    Python3

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    # Python3 program to color a grid
    # such that all same color cells
    # are connected either
    # horizontally or vertically
    def 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 code
    if __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

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    Output:

    1 2 2 3 3 
    4 4 4 4 3 
    5 5 5 5 5
    

    Time Complexity: O(R * C), where R = rows, C = columns
    Auxillary Space: O(R * C), where R = rows, C = columns

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