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3443. Maximum Manhattan Distance After K Changes 👍

  • Time: $O(4n) = O(n)$
  • Space: $O(n)$
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class Solution {
 public:
  int maxDistance(string s, int k) {
    return max({flip(s, k, "NE"), flip(s, k, "NW"),  //
                flip(s, k, "SE"), flip(s, k, "SW")});
  }

 private:
  int flip(const string& s, int k, const string& direction) {
    int res = 0;
    int pos = 0;
    int opposite = 0;

    for (const char c : s) {
      if (direction.find(c) != string::npos) {
        ++pos;
      } else {
        --pos;
        ++opposite;
      }
      res = max(res, pos + 2 * min(k, opposite));
    }

    return res;
  }
};

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class Solution {
  public int maxDistance(String s, int k) {
    return Math.max(Math.max(flip(s, k, "NE"), flip(s, k, "NW")),
                    Math.max(flip(s, k, "SE"), flip(s, k, "SW")));
  }

  private int flip(String s, int k, String direction) {
    int res = 0;
    int pos = 0;
    int opposite = 0;

    for (final char c : s.toCharArray()) {
      if (direction.indexOf(c) >= 0) {
        ++pos;
      } else {
        --pos;
        ++opposite;
      }
      res = Math.max(res, pos + 2 * Math.min(k, opposite));
    }

    return res;
  }
}

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class Solution:
  def maxDistance(self, s: str, k: int) -> int:
    return max(self._flip(s, k, 'NE'), self._flip(s, k, 'NW'),
               self._flip(s, k, 'SE'), self._flip(s, k, 'SW'))

  def _flip(self, s: str, k: int, direction: str) -> int:
    res = 0
    pos = 0
    opposite = 0

    for c in s:
      if c in direction:
        pos += 1
      else:
        pos -= 1
        opposite += 1
      res = max(res, pos + 2 * min(k, opposite))

    return res