1. Abridged Problem Statement  
Given integers N, M (3 ≤ N, M ≤ 200), paint an N×M grid with red (‘0’) and blue (‘#’) so that every 3×2 or 2×3 sub-rectangle contains exactly two blue cells. Among all valid paintings, use as few blue cells as possible. If there is no solution, print “No solution”; otherwise output any valid grid.

2. Detailed Editorial  

Goal: For every 3×2 or 2×3 block of cells, there must be exactly two ‘#’, and globally we want to minimize the total number of ‘#’.

Key Observation:  
If we paint exactly those cells (i,j) for which (i + j + offset) mod 3 = 0 as blue and all others red, then in any 3×2 or 2×3 block exactly two of the six cells satisfy (i + j) mod 3 = constant. Hence this pattern meets the local constraint. Why?  
- Consider any 3 consecutive rows and 2 consecutive columns: the six sums i+j form each residue class mod 3 exactly twice.  
- A similar argument holds for 2×3 blocks.

Thus the valid colorings are exactly the three “diagonal modulo-3” patterns, parameterized by offset ∈ {0,1,2}. Each has roughly ⌈N·M/3⌉ blue cells, but due to boundary effects one offset may use fewer blues. We simply try all three offsets, count the blues, pick the one with minimal count, and output that pattern.

Algorithm:  
1. Read N, M.  
2. For offset in {0,1,2}:  
   a. Count how many positions (i,j) with (i+j+offset)%3==0.  
3. Choose the offset that gives the smallest count.  
4. Construct the grid: mark ‘#’ at (i,j) satisfying (i+j+best_offset)%3==0, else ‘0’.  
5. Print the grid.

Time complexity: O(3·N·M) = O(N·M) which is fine for N,M ≤ 200.

3. Provided C++ Solution with Detailed Comments  

```cpp
#include <bits/stdc++.h>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n, m;
    cin >> n >> m;                       // Read dimensions N and M

    int best_offset = 0;                 // will store offset giving fewest blues
    int min_count = INT_MAX;             // best (minimum) number of blues found

    // Try all three diagonal patterns (offset = 0,1,2)
    for(int offset = 0; offset < 3; offset++) {
        int count = 0;                   // count of blues for this offset

        // Loop over all cells, count those satisfying the diagonal condition
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                // If (i+j+offset) mod 3 == 0, we would paint it blue
                if ((i + j + offset) % 3 == 0) {
                    count++;
                }
            }
        }
        // Update best_offset if we found a smaller blue count
        if (count < min_count) {
            min_count = count;
            best_offset = offset;
        }
    }

    // Construct the final grid using the best offset
    vector<string> grid(n, string(m, '0'));  // initialize all cells red ('0')
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < m; j++) {
            // Paint blue '#' where the diagonal condition holds
            if ((i + j + best_offset) % 3 == 0) {
                grid[i][j] = '#';
            }
        }
    }

    // Output the grid, one row per line
    for(int i = 0; i < n; i++) {
        cout << grid[i] << '\n';
    }
    return 0;
}
```

4. Python Solution with Detailed Comments  

```python
def main():
    import sys
    input = sys.stdin.readline

    # Read N, M
    n, m = map(int, input().split())

    best_offset = 0
    min_count = n * m + 1  # initialize to larger than maximum possible

    # Find which offset yields fewest blue cells
    for offset in range(3):
        cnt = 0
        for i in range(n):
            # Instead of checking every j, we can note that for fixed i,
            # j runs from 0 to m-1. We still check each one here for clarity.
            for j in range(m):
                if (i + j + offset) % 3 == 0:
                    cnt += 1
        if cnt < min_count:
            min_count = cnt
            best_offset = offset

    # Build the grid using the optimal offset
    grid = []
    for i in range(n):
        row = []
        for j in range(m):
            # Mark '#' if it satisfies the chosen diagonal pattern
            if (i + j + best_offset) % 3 == 0:
                row.append('#')
            else:
                row.append('0')
        grid.append(''.join(row))

    # Print the resulting flag
    print('\n'.join(grid))


if __name__ == "__main__":
    main()
```

5. Compressed Editorial  
Assign each cell (i,j) a value (i+j+offset)%3 and color it blue exactly when this value is 0. This guarantees every 3×2 or 2×3 block has exactly two blues (each residue appears twice), and by testing offsets 0,1,2 we pick the one minimizing blue count.