Return the power of the string.

Example 1:

Input: s = "leetcode"
Output: 2
Explanation: The substring "ee" is of length 2 with the character 'e' only.

Example 2:

Input: s = "abbcccddddeeeeedcba"
Output: 5
Explanation: The substring "eeeee" is of length 5 with the character 'e' only.

Example 3:

Input: s = "triplepillooooow"
Output: 5

Example 4:

Input: s = "hooraaaaaaaaaaay"
Output: 11

Example 5:

Input: s = "tourist"
Output: 1

Constraints:

• 1 <= s.length <= 500
• s contains only lowercase English letters.

Solution: Run length encoding?

Time complexity: O(n)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  int maxPower(string s) {
    char p = '?';
    int ans = 0;
    int cur = 0;
    for (char c : s) {
      if (c != p) cur = 0;      
      p = c;
      ans = max(ans, ++cur);
    }
    return ans;
  }
};

The post 花花酱 LeetCode 1446. Consecutive Characters appeared first on Huahua's Tech Road.

Write an iterator that iterates through a run-length encoded sequence.

The iterator is initialized by RLEIterator(int[] A), where A is a run-length encoding of some sequence.  More specifically, for all even i, A[i] tells us the number of times that the non-negative integer value A[i+1] is repeated in the sequence.

The iterator supports one function: next(int n), which exhausts the next n elements (n >= 1) and returns the last element exhausted in this way.  If there is no element left to exhaust, next returns -1 instead.

For example, we start with A = [3,8,0,9,2,5], which is a run-length encoding of the sequence [8,8,8,5,5].  This is because the sequence can be read as “three eights, zero nines, two fives”.

Example 1:

Input: ["RLEIterator","next","next","next","next"], [[[3,8,0,9,2,5]],[2],[1],[1],[2]]
Output: [null,8,8,5,-1]
Explanation: 
RLEIterator is initialized with RLEIterator([3,8,0,9,2,5]).
This maps to the sequence [8,8,8,5,5].
RLEIterator.next is then called 4 times:

.next(2) exhausts 2 terms of the sequence, returning 8.  The remaining sequence is now [8, 5, 5].

.next(1) exhausts 1 term of the sequence, returning 8.  The remaining sequence is now [5, 5].

.next(1) exhausts 1 term of the sequence, returning 5.  The remaining sequence is now [5].

.next(2) exhausts 2 terms, returning -1.  This is because the first term exhausted was 5,
but the second term did not exist.  Since the last term exhausted does not exist, we return -1.

Note:

1. 0 <= A.length <= 1000
2. A.length is an even integer.
3. 0 <= A[i] <= 10^9
4. There are at most 1000 calls to RLEIterator.next(int n) per test case.
5. Each call to RLEIterator.next(int n) will have 1 <= n <= 10^9.

Solution: Simulation

Time complexity: O(|A|)

Space complexity: O(|A|)

// Author: Huahua, 4 ms
class RLEIterator {
public:
  RLEIterator(vector<int> A): A_(std::move(A)), i_(0) {}

  int next(int n) {
    while (n && i_ < A_.size()) {
      if (n >= A_[i_]) {
        n -= A_[i_];
        i_ += 2;
        if (n == 0) return A_[i_ - 1];
      } else {
        A_[i_] -= n;
        return A_[i_ + 1];
      }        
    }
    return -1;
  }
private:
  int i_;
  vector<int> A_;
};

// Author: Huahua, 73 ms
class RLEIterator {
  private int[] A;
  private int i;
  public RLEIterator(int[] A) {
    this.A = A;
    this.i = 0;
  }

  public int next(int n) {
    while (n >= 0 && i < A.length) {
      if (n >= A[i]) {
        n -= A[i];
        i += 2;
        if (n == 0) return A[i - 1];
      } else {
        A[i] -= n;
        return A[i + 1];
      }
    }
    return -1;
  }
}

# Author: Huahua, 44 ms
class RLEIterator:
  def __init__(self, A):
    self.A = A
    self.i = 0

  def next(self, n):
    while n != 0 and self.i < len(self.A):
      if n >= self.A[self.i]:
        n -= self.A[self.i]
        self.i += 2
        if n == 0: return self.A[self.i - 1]
      else:
        self.A[self.i] -= n
        return self.A[self.i + 1]
    return -1

The post 花花酱 LeetCode 900. RLE Iterator appeared first on Huahua's Tech Road.