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3458. Select K Disjoint Special Substrings 👍

  • Time: $O(n)$
  • Space: $O(26) = O(1)$
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class Solution {
 public:
  bool maxSubstringLength(string s, int k) {
    const int n = s.length();
    vector<int> first(26, n);
    vector<int> last(26, -1);
    vector<char> seenOrder;
    // dp[i] := the maximum disjoint special substrings for the first i letters
    vector<int> dp(n + 1);

    for (int i = 0; i < n; ++i) {
      const char c = s[i];
      const int a = c - 'a';
      if (first[a] == n) {
        first[a] = i;
        seenOrder.push_back(c);
      }
      last[a] = i;
    }

    for (const char c : seenOrder) {
      const int a = c - 'a';
      for (int j = first[a]; j < last[a]; ++j) {
        const int b = s[j] - 'a';
        first[a] = min(first[a], first[b]);
        last[a] = max(last[a], last[b]);
      }
    }

    for (int i = 0; i < n; ++i) {
      const char c = s[i];
      const int a = c - 'a';
      if (last[a] != i || (first[a] == 0 && i == n - 1))
        dp[i + 1] = dp[i];
      else  // Start a new special substring.
        dp[i + 1] = max(dp[i], 1 + dp[first[a]]);
    }

    return dp[n] >= k;
  }
};

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class Solution {
  public boolean maxSubstringLength(String s, int k) {
    final int n = s.length();
    int[] first = new int[26];
    int[] last = new int[26];
    // dp[i] := the maximum disjoint special substrings for the first i letters
    int[] dp = new int[n + 1];
    List<Character> seenOrder = new ArrayList<>();

    Arrays.fill(first, n);
    Arrays.fill(last, -1);

    for (int i = 0; i < n; ++i) {
      final char c = s.charAt(i);
      final int a = c - 'a';
      if (first[a] == n) {
        first[a] = i;
        seenOrder.add(c);
      }
      last[a] = i;
    }

    for (final char c : seenOrder) {
      final int a = c - 'a';
      for (int j = first[a]; j < last[a]; ++j) {
        final int b = s.charAt(j) - 'a';
        first[a] = Math.min(first[a], first[b]);
        last[a] = Math.max(last[a], last[b]);
      }
    }

    for (int i = 0; i < n; i++) {
      final char c = s.charAt(i);
      final int a = c - 'a';
      if (last[a] != i || (first[a] == 0 && i == n - 1))
        dp[i + 1] = dp[i];
      else // Start a new special substring.
        dp[i + 1] = Math.max(dp[i], 1 + dp[first[a]]);
    }

    return dp[n] >= k;
  }
}

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class Solution:
  def maxSubstringLength(self, s: str, k: int) -> bool:
    n = len(s)
    first = [n] * 26
    last = [-1] * 26
    # dp[i] := the maximum disjoint special substrings for the first i letters
    dp = [0] * (n + 1)
    seenOrder = []

    for i, c in enumerate(s):
      a = ord(c) - ord('a')
      if first[a] == n:
        first[a] = i
        seenOrder.append(c)
      last[a] = i

    for c in seenOrder:
      a = ord(c) - ord('a')
      for j in range(first[a], last[a]):
        b = ord(s[j]) - ord('a')
        first[a] = min(first[a], first[b])
        last[a] = max(last[a], last[b])

    for i, c in enumerate(s):
      a = ord(c) - ord('a')
      if last[a] != i or (first[a] == 0 and i == n - 1):
        dp[i + 1] = dp[i]
      else:  # Start a new special substring.
        dp[i + 1] = max(dp[i], 1 + dp[first[a]])

    return dp[n] >= k