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3333. Find the Original Typed String II 👍

  • Time: $O(n + k^2)$
  • Space: $O(n)$
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class Solution {
 public:
  int possibleStringCount(string word, int k) {
    const vector<int> groups = getConsecutiveLetters(word);
    const int totalCombinations =
        accumulate(groups.begin(), groups.end(), 1L,
                   [](long acc, int group) { return acc * group % kMod; });
    if (k <= groups.size())
      return totalCombinations;

    // dp[j] := the number of ways to form strings of length j using
    // groups[0..i]
    vector<int> dp(k);
    dp[0] = 1;  // Base case: empty string

    for (int i = 0; i < groups.size(); ++i) {
      vector<int> newDp(k);
      int windowSum = 0;
      int group = groups[i];
      for (int j = i; j < k; ++j) {
        newDp[j] = (newDp[j] + windowSum) % kMod;
        windowSum = (windowSum + dp[j]) % kMod;
        if (j >= group)
          windowSum = (windowSum - dp[j - group] + kMod) % kMod;
      }
      dp = std::move(newDp);
    }

    const int invalidCombinations =
        accumulate(dp.begin(), dp.end(), 0,
                   [](int acc, int count) { return (acc + count) % kMod; });
    return (totalCombinations - invalidCombinations + kMod) % kMod;
  }

 private:
  static constexpr int kMod = 1'000'000'007;

  // Returns consecutive identical letters in the input string.
  // e.g. "aabbbc" -> [2, 3, 1].
  vector<int> getConsecutiveLetters(const string& word) {
    vector<int> groups;
    int group = 1;
    for (int i = 1; i < word.length(); ++i)
      if (word[i] == word[i - 1]) {
        ++group;
      } else {
        groups.push_back(group);
        group = 1;
      }
    groups.push_back(group);
    return groups;
  }
};

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class Solution {
  public int possibleStringCount(String word, int k) {
    List<Integer> groups = getConsecutiveLetters(word);
    final int totalCombinations =
        (int) groups.stream().mapToLong(Integer::longValue).reduce(1L, (a, b) -> a * b % MOD);
    if (k <= groups.size())
      return totalCombinations;

    // dp[j] := the number of ways to form strings of length j using
    // groups[0..i]
    int[] dp = new int[k];
    dp[0] = 1; // Base case: empty string

    for (int i = 0; i < groups.size(); ++i) {
      int[] newDp = new int[k];
      int windowSum = 0;
      int group = groups.get(i);
      for (int j = i; j < k; ++j) {
        newDp[j] = (newDp[j] + windowSum) % MOD;
        windowSum = (windowSum + dp[j]) % MOD;
        if (j >= group)
          windowSum = (windowSum - dp[j - group] + MOD) % MOD;
      }
      dp = newDp;
    }

    final int invalidCombinations = Arrays.stream(dp).reduce(0, (a, b) -> (a + b) % MOD);
    return (totalCombinations - invalidCombinations + MOD) % MOD;
  }

  private static final int MOD = 1_000_000_007;

  // Returns consecutive identical letters in the input string.
  // e.g. "aabbbc" -> [2, 3, 1].
  private List<Integer> getConsecutiveLetters(final String word) {
    List<Integer> groups = new ArrayList<>();
    int group = 1;
    for (int i = 1; i < word.length(); ++i)
      if (word.charAt(i) == word.charAt(i - 1)) {
        ++group;
      } else {
        groups.add(group);
        group = 1;
      }
    groups.add(group);
    return groups;
  }
}

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class Solution:
  def possibleStringCount(self, word: str, k: int) -> int:
    MOD = 1_000_000_007
    groups = self._getConsecutiveLetters(word)
    totalCombinations = functools.reduce(lambda subtotal, group:
                                         subtotal * group % MOD, groups)
    if k <= len(groups):
      return totalCombinations

    # dp[j] := the number of ways to form strings of length j using groups[0..i]
    dp = [0] * k
    dp[0] = 1  # Base case: empty string

    for i, group in enumerate(groups):
      newDp = [0] * k
      windowSum = 0
      for j in range(i, k):
        newDp[j] = (newDp[j] + windowSum) % MOD
        windowSum = (windowSum + dp[j]) % MOD
        if j >= group:
          windowSum = (windowSum - dp[j - group] + MOD) % MOD
      dp = newDp

    return (totalCombinations - sum(dp)) % MOD

  def _getConsecutiveLetters(self, word: str) -> list[int]:
    """
    Returns consecutive identical letters in the input string.
    e.g. "aabbbc" -> [2, 3, 1].
    """
    groups = []
    group = 1
    for i in range(1, len(word)):
      if word[i] == word[i - 1]:
        group += 1
      else:
        groups.append(group)
        group = 1
    groups.append(group)
    return groups