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3039. Apply Operations to Make String Empty 👍

  • Time: $O(n)$
  • Space: $O(26) = O(1)$
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class Solution {
 public:
  string lastNonEmptyString(string s) {
    string ans;
    vector<int> count(26);

    for (const char c : s)
      ++count[c - 'a'];

    const int maxFreq = ranges::max(count);

    for (int i = s.length() - 1; i >= 0; --i)
      if (count[s[i] - 'a']-- == maxFreq)
        ans += s[i];

    ranges::reverse(ans);
    return ans;
  }
};

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class Solution {
  public String lastNonEmptyString(String s) {
    StringBuilder sb = new StringBuilder();
    int[] count = new int[26];

    for (final char c : s.toCharArray())
      ++count[c - 'a'];

    final int maxFreq = Arrays.stream(count).max().getAsInt();

    for (int i = s.length() - 1; i >= 0; --i)
      if (count[s.charAt(i) - 'a']-- == maxFreq)
        sb.append(s.charAt(i));

    return sb.reverse().toString();
  }
}

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class Solution:
  def lastNonEmptyString(self, s: str) -> str:
    ans = []
    count = collections.Counter(s)
    maxFreq = max(count.values())

    for c in reversed(s):
      if count[c] == maxFreq:
        ans.append(c)
        count[c] -= 1

    return ''.join(reversed(ans))