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3361. Shift Distance Between Two Strings

  • Time: $O(26^2 + n) = O(n)$
  • Space: $O(26^2) = O(1)$
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class Solution {
 public:
  long long shiftDistance(string s, string t, vector<int>& nextCost,
                          vector<int>& previousCost) {
    long ans = 0;
    // prev[i][j] := the prev cost to shift from ('a' + i) to ('a' + j)
    vector<vector<long>> prev(26, vector<long>(26));
    // next[i][j] := the next cost to shift from ('a' + i) to ('a' + j)
    vector<vector<long>> next(26, vector<long>(26));

    for (int i = 0; i < 26; ++i) {
      long cost = 0;
      for (int j = 0; j < 26; ++j) {
        next[i][(i + j) % 26] = cost;
        cost += nextCost[(i + j) % 26];
      }
    }

    for (int i = 0; i < 26; ++i) {
      long cost = 0;
      for (int j = 0; j < 26; ++j) {
        prev[i][(i - j + 26) % 26] = cost;
        cost += previousCost[(i - j + 26) % 26];
      }
    }

    for (int i = 0; i < s.length(); ++i) {
      const int a = s[i] - 'a';
      const int b = t[i] - 'a';
      ans += min(next[a][b], prev[a][b]);
    }

    return ans;
  }
};

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class Solution {
  public long shiftDistance(String s, String t, int[] nextCost, int[] previousCost) {
    long ans = 0;
    // prev[i][j]: the prev cost to shift from ('a' + i) to ('a' + j)
    long[][] prev = new long[26][26];
    // next[i][j]: the next cost to shift from ('a' + i) to ('a' + j)
    long[][] next = new long[26][26];

    for (int i = 0; i < 26; ++i) {
      long cost = 0;
      for (int j = 0; j < 26; ++j) {
        next[i][(i + j) % 26] = cost;
        cost += nextCost[(i + j) % 26];
      }
    }

    for (int i = 0; i < 26; ++i) {
      long cost = 0;
      for (int j = 0; j < 26; ++j) {
        prev[i][(i - j + 26) % 26] = cost;
        cost += previousCost[(i - j + 26) % 26];
      }
    }

    for (int i = 0; i < s.length(); ++i) {
      final int a = s.charAt(i) - 'a';
      final int b = t.charAt(i) - 'a';
      ans += Math.min(next[a][b], prev[a][b]);
    }

    return ans;
  }
}

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class Solution:
  def shiftDistance(
      self,
      s: str,
      t: str,
      nextCost: list[int],
      previousCost: list[int]
  ) -> int:
    ans = 0
    # prev[i][j]: the prev cost to shift from ('a' + i) to ('a' + j)
    prev = [[0] * 26 for _ in range(26)]
    # next[i][j]: the next cost to shift from ('a' + i) to ('a' + j)
    next = [[0] * 26 for _ in range(26)]

    for i in range(26):
      cost = 0
      for j in range(26):
        next[i][(i + j) % 26] = cost
        cost += nextCost[(i + j) % 26]

    for i in range(26):
      cost = 0
      for j in range(26):
        prev[i][(i - j + 26) % 26] = cost
        cost += previousCost[(i - j + 26) % 26]

    for a, b in zip(s, t):
      i = ord(a) - ord('a')
      j = ord(b) - ord('a')
      ans += min(next[i][j], prev[i][j])

    return ans