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| 1 | +''' |
| 2 | +You are given a string word and an integer k. |
| 3 | + |
| 4 | +We consider word to be k-special if |freq(word[i]) - freq(word[j])| <= k for all indices i and j in the string. |
| 5 | + |
| 6 | +Here, freq(x) denotes the frequency of the character x in word, and |y| denotes the absolute value of y. |
| 7 | + |
| 8 | +Return the minimum number of characters you need to delete to make word k-special. |
| 9 | +''' |
| 10 | +class Solution: |
| 11 | + def minimumDeletions(self, word: str, k: int) -> int: |
| 12 | + freqMap = defaultdict(int) |
| 13 | + for c in word: |
| 14 | + freqMap[c] += 1 |
| 15 | + |
| 16 | + frequencies = sorted(freqMap.values()) |
| 17 | + minDeletions = float('inf') |
| 18 | + n = len(frequencies) |
| 19 | + |
| 20 | + for i in range(n): |
| 21 | + base = frequencies[i] |
| 22 | + totalDeletions = 0 |
| 23 | + |
| 24 | + for j in range(i): |
| 25 | + totalDeletions += frequencies[j] |
| 26 | + |
| 27 | + for j in range(i, n): |
| 28 | + if frequencies[j] > base + k: |
| 29 | + totalDeletions += frequencies[j] - (base + k) |
| 30 | + |
| 31 | + if totalDeletions < minDeletions: |
| 32 | + minDeletions = totalDeletions |
| 33 | + |
| 34 | + return minDeletions |
| 35 | + |
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