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2272. Substring With Largest Variance 👍

  • Time: $O(26^2n) = O(n)$
  • Space: $O(1)$
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class Solution {
 public:
  int largestVariance(string s) {
    int ans = 0;

    for (char a = 'a'; a <= 'z'; ++a)
      for (char b = 'a'; b <= 'z'; ++b)
        if (a != b)
          ans = max(ans, kadane(s, a, b));

    return ans;
  }

 private:
  // a := the letter with the higher frequency
  // b := the letter with the lower frequency
  int kadane(const string& s, char a, char b) {
    int ans = 0;
    int countA = 0;
    int countB = 0;
    bool canExtendPrevB = false;

    for (const char c : s) {
      if (c != a && c != b)
        continue;
      if (c == a)
        ++countA;
      else
        ++countB;
      if (countB > 0) {
        // An interval should contain at least one b.
        ans = max(ans, countA - countB);
      } else if (countB == 0 && canExtendPrevB) {
        // edge case: consider the previous b.
        ans = max(ans, countA - 1);
      }
      // Reset if the number of b > the number of a.
      if (countB > countA) {
        countA = 0;
        countB = 0;
        canExtendPrevB = true;
      }
    }

    return ans;
  }
};

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class Solution {
  public int largestVariance(String s) {
    int ans = 0;

    for (char a = 'a'; a <= 'z'; ++a)
      for (char b = 'a'; b <= 'z'; ++b)
        if (a != b)
          ans = Math.max(ans, kadane(s, a, b));

    return ans;
  }

  // a := the letter with the higher frequency
  // b := the letter with the lower frequency
  private int kadane(final String s, char a, char b) {
    int ans = 0;
    int countA = 0;
    int countB = 0;
    boolean canExtendPrevB = false;

    for (final char c : s.toCharArray()) {
      if (c != a && c != b)
        continue;
      if (c == a)
        ++countA;
      else
        ++countB;
      if (countB > 0) {
        // An interval should contain at least one b.
        ans = Math.max(ans, countA - countB);
      } else if (countB == 0 && canExtendPrevB) {
        // edge case: consider the previous b
        ans = Math.max(ans, countA - 1);
      }
      // Reset if the number of b > the number of a.
      if (countB > countA) {
        countA = 0;
        countB = 0;
        canExtendPrevB = true;
      }
    }

    return ans;
  }
}

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class Solution:
  def largestVariance(self, s: str) -> int:
    # a := the letter with the higher frequency
    # b := the letter with the lower frequency
    def kadane(a: str, b: str) -> int:
      ans = 0
      countA = 0
      countB = 0
      canExtendPrevB = False

      for c in s:
        if c != a and c != b:
          continue
        if c == a:
          countA += 1
        else:
          countB += 1
        if countB > 0:
          # An interval should contain at least one b.
          ans = max(ans, countA - countB)
        elif countB == 0 and canExtendPrevB:
          # edge case: consider the previous b.
          ans = max(ans, countA - 1)
        # Reset if the number of b > the number of a.
        if countB > countA:
          countA = 0
          countB = 0
          canExtendPrevB = True

      return ans

    return max(kadane(a, b)
               for a in string.ascii_lowercase
               for b in string.ascii_lowercase
               if a != b)