Fill values of a dict in increase order given the initial key

Question

I am looking for the best way to do some task. Hope you can help me to improve what I have so far.

I have two dicts with same keys, in the dict ('other_dict') I need to find the key which match some value and then that key will be the key with value 0 in t_position dict and then fill the rest of the t_position dict keys with values from 1 to 5 in increasing order. Lets say that the value is in key "V3", then its position will be 0, "V4" : 1, "V5" : 2, "V0" : 3 ... "V2" : 5

t_positions = {"V0": 6, "V1": 6, "V2": 6, "V3": 6, "V4": 6, "V5": 6}
other_dict = {"V0": 0, "V1": 0, "V2": 0, "V3": 1, "V4": 0, "V5": 0}
value_to_find = 1

def pos_detector():
    dict_keys = ["V0", "V1", "V2", "V3", "V4", "V5"]
    value_index = None

    for key, value in other_dict.items():
        if value == value_to_find:
            t_positions[key] = 0
            value_index = dict_keys.index(key)
            break
    up_counter = value_index + 1
    down_counter = 0
    position_counter = 1

    while up_counter <= 5:
        t_positions[dict_keys[up_counter]] = position_counter 
        up_counter += 1
        position_counter += 1

    while down_counter < value_index:
        t_positions[dict_keys[down_counter]] = position_counter
        down_counter += 1
        position_counter += 1

pos_detector()

print(t_positions)

Thanks

*EDIT: Now there is a minimal reproducible example. I can think in other ways to do it, but the key of the question is about time, I am looking to the fastest solution, because I will use this function a lot of times.