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1737. Change Minimum Characters to Satisfy One of Three Conditions 👎

  • Time: $O(|\texttt{a}| + |\texttt{b}|)$
  • Space: $O(128) = O(1)$
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class Solution {
 public:
  int minCharacters(string a, string b) {
    const int m = a.length();
    const int n = b.length();
    vector<int> countA(128);
    vector<int> countB(128);

    for (const char c : a)
      ++countA[c];

    for (const char c : b)
      ++countB[c];

    int ans = INT_MAX;
    int prevA = 0;  // # of chars in a <= c
    int prevB = 0;  // # of chars in b <= c

    for (char c = 'a'; c <= 'z'; ++c) {
      // Condition 3
      ans = min(ans, m + n - countA[c] - countB[c]);
      // Condition 1 and 2
      if (c > 'a')
        ans = min({ans, m - prevA + prevB, n - prevB + prevA});
      prevA += countA[c];
      prevB += countB[c];
    }

    return ans;
  }
};

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class Solution {
  public int minCharacters(String a, String b) {
    final int m = a.length();
    final int n = b.length();
    int[] countA = new int[128];
    int[] countB = new int[128];

    for (final char c : a.toCharArray())
      ++countA[c];

    for (final char c : b.toCharArray())
      ++countB[c];

    int ans = Integer.MAX_VALUE;
    int prevA = 0; // # of chars in a <= c
    int prevB = 0; // # of chars in b <= c

    for (char c = 'a'; c <= 'z'; ++c) {
      // Condition 3
      ans = Math.min(ans, m + n - countA[c] - countB[c]);
      // Condition 1 and 2
      if (c > 'a')
        ans = Math.min(ans, Math.min(m - prevA + prevB, n - prevB + prevA));
      prevA += countA[c];
      prevB += countB[c];
    }

    return ans;
  }
}