public static String isValid(String s) {
    // Step 1: Count character frequencies
    int[] freq = new int[26];
    for (char c : s.toCharArray()) {
        freq[c - 'a']++;
    }

    // Step 2: Find the maximum frequency
    int maxFreq = 0;
    for (int f : freq) {
        if (f > maxFreq) {
            maxFreq = f;
        }
    }

    // Step 3: Count occurrences of each frequency
    int[] freqCounts = new int[maxFreq + 1];
    for (int f : freq) {
        if (f > 0) {
            freqCounts[f]++;
        }
    }

    // Step 4: Analyze frequency counts
    int uniqueFreqs = 0; // Count of unique(non zero) frequencies
    boolean canRemoveOne = false; // Whether we can remove one character to make it valid

    for (int i = 1; i < freqCounts.length; i++) {
        if (freqCounts[i] > 0) {
            uniqueFreqs++;
            // Check if we can remove one character to balance frequencies
            if (i < freqCounts.length - 1 && freqCounts[i + 1] == 1) {
                canRemoveOne = true;
            }
        }
    }
// case 1: uniqueFreqs == 1 means: All characters have the same frequency
    // example: "aabbccdd" all appear twice. The only distinct frequency is 2.
    // case 2: There are 2 distinct values and one of them is "1" and it occurs only once
    // example: aaaabbbbc  the occurance of 4 (a and b) is 2 and the occurance of 1 is(c) is one. Remove the only letter which appears once
    // case 3: Frequencies differ by 1, and the higher frequency appears only once example: aaaabbbbccccc
    
		// Step 5: Return the result based on the cases
    if (uniqueFreqs == 1 || (uniqueFreqs == 2 && (freqCounts[1] == 1 || canRemoveOne))) {
        return "YES";
    }

    return "NO";
}