How to put an incremental counter for each group in SQL (Snowflake)

I have the following table - essentially a product and its price between a set of start and end dates. The start and end dates are sorted in ascending order.

product start       end         price
A       08/04/2022  12/04/2022  29
A       13/04/2022  12/05/2022  45      
A       13/05/2022  25/12/2022  45
A       26/12/2022  16/05/2023  29
A       17/05/2023  23/05/2023  29
A       24/05/2023  31/12/9999  49

What I want is to have a counter that essentially checks if the price has changed from its previous row (for a particular product). If yes, increment the counter by 1, else copy the counter value from above.
Output should look like below:

product start       end         price  counter
A       08/04/2022  12/04/2022  29     1
A       13/04/2022  12/05/2022  45     2      
A       13/05/2022  25/12/2022  45     2
A       26/12/2022  16/05/2023  29     3
A       17/05/2023  23/05/2023  29     3
A       24/05/2023  31/12/9999  49     4

Any help will be highly appreciated. Thanks.

Snowflake supports CONDITIONAL_CHANGE_EVENT windowed function that is exactly designed for it:

> Returns a window event number for each row within a window partition when the value of the argument expr1 in the current row is different from the value of expr1 in the previous row. The window event number starts from 0 and is incremented by 1 to indicate the number of changes so far within that window.

SELECT *, CONDITIONAL_CHANGE_EVENT(price) 
          OVER(PARTITION BY product ORDER BY start_) + 1 AS counter
FROM tab
ORDER BY product, start_;

For input table:

ALTER SESSION SET DATE_INPUT_FORMAT = 'DD/MM/YYYY';

CREATE OR REPLACE TABLE tab(product TEXT, start_ DATE, end_ DATE, price INT)
AS
SELECT 'A','08/04/2022','12/04/2022', 29 UNION ALL
SELECT 'A','13/04/2022','12/05/2022', 45 UNION ALL      
SELECT 'A','13/05/2022','25/12/2022', 45 UNION ALL
SELECT 'A','26/12/2022','16/05/2023', 29 UNION ALL
SELECT 'A','17/05/2023','23/05/2023', 29 UNION ALL
SELECT 'A','24/05/2023','31/12/9999', 49;

Output:

You could use a running sum on a flag that checks if the previous value of the price is not equal to the current price value of a product:

with t as
(
  select *,
   case
     when lag(price, 1, price) over (partition by product order by start_dt) <> price
     then 1 else 0
   end as flag
 from tbl
)

select product, start_dt, end_dt, price,
(sum(flag) over (partition by product order by start_dt) + 1) counter
from t

demo (on SQL server)

Use rank() over.

select *, rank() over(order by price) counter
from mytable

rank() over - this window clause will start from 1, and if the next value is same, then it will continue and not add 1 to the rank value.

So, counter values will be 1,2,2,3,4 because 2nd and 3rd values are same.