If the solution in the official editorial also manages the remaining number of colors in each row and column instead of scanning \(26\) elements each time, it takes \(O(H + W)\) time per round to determine if all cookies in each row and column have the same color, for a total of \(O((H + W)^2)\) time.

Sample Implementation (Python)

H, W = map(int, input().split())
c = [list(input()) for _ in range(H)]
A = 26

row = [[0] * A for _ in range(H)]
row_c = [0] * H
col = [[0] * A for _ in range(W)]
col_c = [0] * W
num_row, num_col = H, W

for i in range(H):
    for j in range(W):
        v = ord(c[i][j]) - ord('a')
        row[i][v] += 1
        if row[i][v] == 1:
            row_c[i] += 1
        col[j][v] += 1
        if col[j][v] == 1:
            col_c[j] += 1

while True:
    del_row, del_col = [], []
    for i in range(H):
        if row_c[i] == 1 and num_col >= 2:
            del_row.append(i)
    for j in range(W):
        if col_c[j] == 1 and num_row >= 2:
            del_col.append(j)
    if del_row == [] and del_col == []:
        break

    def remove(i, j):
        if c[i][j] != ' ':
            v = ord(c[i][j]) - ord('a')
            row[i][v] -= 1
            if row[i][v] == 0:
                row_c[i] -= 1
            col[j][v] -= 1
            if col[j][v] == 0:
                col_c[j] -= 1
            c[i][j] = ' '

    for i in del_row:
        for j in range(W):
            remove(i, j)
        num_row -= 1
    for j in del_col:
        for i in range(H):
            remove(i, j)
        num_col -= 1

print(num_row * num_col)

P.S. I proposed A, B, C, D, E, F.