Question

2565. Subsequence With the Minimum Score

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Description

You are given two strings s and t.

You are allowed to remove any number of characters from the string t.

The score of the string is 0 if no characters are removed from the string t, otherwise:

• Let left be the minimum index among all removed characters.
• Let right be the maximum index among all removed characters.

Then the score of the string is right - left + 1.

Return _the minimum possible score to make _t_ a subsequence of _s_._

A subsequence of a string is a new string that is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (i.e., "ace" is a subsequence of "abcde" while "aec" is not).

 

Example 1:

Input: s = "abacaba", t = "bzaa"
Output: 1
Explanation: In this example, we remove the character "z" at index 1 (0-indexed).
The string t becomes "baa" which is a subsequence of the string "abacaba" and the score is 1 - 1 + 1 = 1.
It can be proven that 1 is the minimum score that we can achieve.

Example 2:

Input: s = "cde", t = "xyz"
Output: 3
Explanation: In this example, we remove characters "x", "y" and "z" at indices 0, 1, and 2 (0-indexed).
The string t becomes "" which is a subsequence of the string "cde" and the score is 2 - 0 + 1 = 3.
It can be proven that 3 is the minimum score that we can achieve.

 

Constraints:

• 1 <= s.length, t.length <= 105
• s and t consist of only lowercase English letters.

Solutions

Solution 1: Prefix and Suffix Preprocessing + Binary Search

According to the problem, we know that the range of the index to delete characters is [left, right]. The optimal approach is to delete all characters within the range [left, right]. In other words, we need to delete a substring from string $t$, so that the remaining prefix of string $t$ can match the prefix of string $s$, and the remaining suffix of string $t$ can match the suffix of string $s$, and the prefix and suffix of string $s$ do not overlap. Note that the match here refers to subsequence matching.

Therefore, we can preprocess to get arrays $f$ and $g$, where $f[i]$ represents the minimum number of characters in the prefix $t[0,..i]$ of string $t$ that match the first $[0,..f[i]]$ characters of string $s$; similarly, $g[i]$ represents the maximum number of characters in the suffix $t[i,..n-1]$ of string $t$ that match the last $[g[i],..n-1]$ characters of string $s$.

The length of the deleted characters has monotonicity. If the condition is satisfied after deleting a string of length $x$, then the condition is definitely satisfied after deleting a string of length $x+1$. Therefore, we can use the method of binary search to find the smallest length that satisfies the condition.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Where $n$ is the length of string $t$.

class Solution:
  def minimumScore(self, s: str, t: str) -> int:
    def check(x):
      for k in range(n):
        i, j = k - 1, k + x
        l = f[i] if i >= 0 else -1
        r = g[j] if j < n else m + 1
        if l < r:
          return True
      return False

    m, n = len(s), len(t)
    f = [inf] * n
    g = [-1] * n
    i, j = 0, 0
    while i < m and j < n:
      if s[i] == t[j]:
        f[j] = i
        j += 1
      i += 1
    i, j = m - 1, n - 1
    while i >= 0 and j >= 0:
      if s[i] == t[j]:
        g[j] = i
        j -= 1
      i -= 1

    return bisect_left(range(n + 1), True, key=check)

class Solution {
  private int m;
  private int n;
  private int[] f;
  private int[] g;

  public int minimumScore(String s, String t) {
    m = s.length();
    n = t.length();
    f = new int[n];
    g = new int[n];
    for (int i = 0; i < n; ++i) {
      f[i] = 1 << 30;
      g[i] = -1;
    }
    for (int i = 0, j = 0; i < m && j < n; ++i) {
      if (s.charAt(i) == t.charAt(j)) {
        f[j] = i;
        ++j;
      }
    }
    for (int i = m - 1, j = n - 1; i >= 0 && j >= 0; --i) {
      if (s.charAt(i) == t.charAt(j)) {
        g[j] = i;
        --j;
      }
    }
    int l = 0, r = n;
    while (l < r) {
      int mid = (l + r) >> 1;
      if (check(mid)) {
        r = mid;
      } else {
        l = mid + 1;
      }
    }
    return l;
  }

  private boolean check(int len) {
    for (int k = 0; k < n; ++k) {
      int i = k - 1, j = k + len;
      int l = i >= 0 ? f[i] : -1;
      int r = j < n ? g[j] : m + 1;
      if (l < r) {
        return true;
      }
    }
    return false;
  }
}

class Solution {
public:
  int minimumScore(string s, string t) {
    int m = s.size(), n = t.size();
    vector<int> f(n, 1e6);
    vector<int> g(n, -1);
    for (int i = 0, j = 0; i < m && j < n; ++i) {
      if (s[i] == t[j]) {
        f[j] = i;
        ++j;
      }
    }
    for (int i = m - 1, j = n - 1; i >= 0 && j >= 0; --i) {
      if (s[i] == t[j]) {
        g[j] = i;
        --j;
      }
    }

    auto check = [&](int len) {
      for (int k = 0; k < n; ++k) {
        int i = k - 1, j = k + len;
        int l = i >= 0 ? f[i] : -1;
        int r = j < n ? g[j] : m + 1;
        if (l < r) {
          return true;
        }
      }
      return false;
    };

    int l = 0, r = n;
    while (l < r) {
      int mid = (l + r) >> 1;
      if (check(mid)) {
        r = mid;
      } else {
        l = mid + 1;
      }
    }
    return l;
  }
};

func minimumScore(s string, t string) int {
  m, n := len(s), len(t)
  f := make([]int, n)
  g := make([]int, n)
  for i := range f {
    f[i] = 1 << 30
    g[i] = -1
  }
  for i, j := 0, 0; i < m && j < n; i++ {
    if s[i] == t[j] {
      f[j] = i
      j++
    }
  }
  for i, j := m-1, n-1; i >= 0 && j >= 0; i-- {
    if s[i] == t[j] {
      g[j] = i
      j--
    }
  }
  return sort.Search(n+1, func(x int) bool {
    for k := 0; k < n; k++ {
      i, j := k-1, k+x
      l, r := -1, m+1
      if i >= 0 {
        l = f[i]
      }
      if j < n {
        r = g[j]
      }
      if l < r {
        return true
      }
    }
    return false
  })
}