Given two numbers, `hour` and `minutes`, return _the smaller angle (in degrees) formed between the_ `hour` _and the_ `minute` _hand_.

Answers within `10-5` of the actual value will be accepted as correct.

**Example 1:**

**Input:** hour = 12, minutes = 30
**Output:** 165

**Example 2:**

**Input:** hour = 3, minutes = 30
**Output:** 75

**Example 3:**

**Input:** hour = 3, minutes = 15
**Output:** 7.5

**Constraints:**

*   `1 <= hour <= 12`
*   `0 <= minutes <= 59`


def maxEqualFreq(nums):
    count, freq = {}, {}
    max_len, max_freq = 0, 0
    for i, num in enumerate(nums):
        count[num] = count.get(num, 0) + 1
        freq[count[num] - 1] = freq.get(count[num] - 1, 0) - 1
        freq[count[num]] = freq.get(count[num], 0) + 1

        max_freq = max(max_freq, count[num])
        if max_freq * (i + 1) == i or max_freq * (freq.get(max_freq - 1, 0) + 1) + (max_freq - 1) * freq.get(max_freq - 1, 0) == i:
            max_len = i + 1
    return max_len
