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| 1 | +''' |
| 2 | +Alice is attempting to type a specific string on her computer. However, she tends to be clumsy and may press a key for too long, resulting in a character being typed multiple times. |
| 3 | + |
| 4 | +You are given a string word, which represents the final output displayed on Alice's screen. You are also given a positive integer k. |
| 5 | + |
| 6 | +Return the total number of possible original strings that Alice might have intended to type, if she was trying to type a string of size at least k. |
| 7 | + |
| 8 | +Since the answer may be very large, return it modulo 109 + 7. |
| 9 | +''' |
| 10 | + |
| 11 | + |
| 12 | +class Solution: |
| 13 | + MOD = 10**9 + 7 |
| 14 | + |
| 15 | + def possibleStringCount(self, word: str, k: int) -> int: |
| 16 | + if not word: |
| 17 | + return 0 |
| 18 | + |
| 19 | + groups = [] |
| 20 | + count = 1 |
| 21 | + for i in range(1, len(word)): |
| 22 | + if word[i] == word[i - 1]: |
| 23 | + count += 1 |
| 24 | + else: |
| 25 | + groups.append(count) |
| 26 | + count = 1 |
| 27 | + groups.append(count) |
| 28 | + |
| 29 | + total = 1 |
| 30 | + for num in groups: |
| 31 | + total = (total * num) % self.MOD |
| 32 | + |
| 33 | + if k <= len(groups): |
| 34 | + return total |
| 35 | + |
| 36 | + dp = [0] * k |
| 37 | + dp[0] = 1 |
| 38 | + |
| 39 | + for num in groups: |
| 40 | + new_dp = [0] * k |
| 41 | + sum_val = 0 |
| 42 | + for s in range(k): |
| 43 | + if s > 0: |
| 44 | + sum_val = (sum_val + dp[s - 1]) % self.MOD |
| 45 | + if s > num: |
| 46 | + sum_val = (sum_val - dp[s - num - 1] + self.MOD) % self.MOD |
| 47 | + new_dp[s] = sum_val |
| 48 | + dp = new_dp |
| 49 | + |
| 50 | + invalid = sum(dp[len(groups):k]) % self.MOD |
| 51 | + return (total - invalid + self.MOD) % self.MOD |
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