What is the space complexity of the following dynamic programming implementation of the boolean parenthesization problem? int count_bool_parenthesization(char *sym, char *op) int str_len = strlen(sym); int True[str_len][str_len],False[str_len][str_len]; int row,col,length,l; for(row = 0, col = 0; row < str_len; row++,col++) if(sym[row] == 'T') True[row][col] = 1; False[row][col] = 0; else True[row][col] = 0; False[row][col] = 1; for(length = 1; length < str_len; length++) for(row = 0, col = length; col < str_len; col++, row++) True[row][col] = 0; False[row][col] = 0; for(l = 0; l < length; l++) int pos = row + l; int t_row_pos = True[row][pos] + False[row][pos]; int t_pos_col = True[pos+1][col] + False[pos+1][col]; if(op[pos] == '|') False[row][col] += False[row][pos] * False[pos+1][col]; True[row][col] += t_row_pos * t_pos_col - False[row][pos] * False[pos+1][col];; if(op[pos] == '&') True[row][col] += True[row][pos] * True[pos+1][col]; False[row][col] += t_row_pos * t_pos_col - True[row][pos] * True[pos+1][col]; if(op[pos] == '^') True[row][col] += True[row][pos] * False[pos+1][col] + False[row][pos] * True[pos + 1][col]; False[row][col] += True[row][pos] * True[pos+1][col] + False[row][pos] * False[pos+1][col]; return True[0][str_len-1];

Answer: Option C