def isValid(s):
    # Write your code here
    char_freq: dict[str, int] = {}
    freq_count: dict[int, int] = {}
    for char in s:
        char_freq[char] = char_freq.get(char, 0) + 1
    for freq in char_freq.values():
        freq_count[freq] = freq_count.get(freq, 0) + 1
    # reject if more than 2 different frequencies (-1 in any cannot become uniform)
    if len(freq_count.keys()) > 2:
        return "NO"
    # accept if all frequencies are the same / uniform
    if len(freq_count.keys()) == 1:
        return "YES"
    freqs: list[int] = list(sorted(freq_count.keys()))
    big_freq = freqs[1]
    small_freq = freqs[0]
    # edge case for smaller freq is single char
    if small_freq == 1 and freq_count[small_freq] == 1:
        return "YES"
    # when there is only 1 character type with freq = big_freq, 
    # and removing 1 of that char will make the freq_count uniform
    if freq_count[big_freq] == 1 and big_freq == small_freq + 1:
        return "YES"
    return "NO"

public static String isValid(String s) {
// Write your code here
    Map<Character, Long> charFreq = s.codePoints().mapToObj(x -> (char) x).collect(Collectors.groupingBy(x -> x, Collectors.counting()));
    Map<Long, Long> freqCount = charFreq.values().stream().collect(Collectors.groupingBy(x -> x, Collectors.counting()));
    int keySetSize = freqCount.size();
    // accept if all frequencies are the same / uniform
    if (keySetSize < 2) {
        return "YES";
    }
    // reject if more than 2 different frequencies (-1 in any cannot become uniform)
    if (keySetSize > 2) {
        return "NO";
    }
    Long[] freqs = freqCount.keySet().toArray(new Long[0]);
    int bigFreqIndex = freqs[1] > freqs[0] ? 1 : 0;
    long smallFreq = freqs[1 - bigFreqIndex];
    long bigFreq = freqs[bigFreqIndex];
    // edge case for smaller freq is single char
    if (smallFreq == 1 && freqCount.get(smallFreq) == 1) {
        return "YES";
    }
    // when there is only 1 character type with freq = big_freq, 
    // and removing 1 of that char will make the freq_count uniform
    if (bigFreq == smallFreq + 1 && freqCount.get(bigFreq) == 1l) {
        return "YES";
    }
    return "NO";
}