Given a string s and an array of integers cost where cost[i] is the cost of deleting the character i in s.
Return the minimum cost of deletions such that there are no two identical letters next to each other.
Notice that you will delete the chosen characters at the same time, in other words, after deleting a character, the costs of deleting other characters will not change.
Example 1:
Input: s = "abaac", cost = [1,2,3,4,5] Output: 3 Explanation: Delete the letter "a" with cost 3 to get "abac" (String without two identical letters next to each other).
Example 2:
Input: s = "abc", cost = [1,2,3] Output: 0 Explanation: You don't need to delete any character because there are no identical letters next to each other.
Example 3:
Input: s = "aabaa", cost = [1,2,3,4,1]
Output: 2
Explanation: Delete the first and the last character, getting the string ("aba").
Constraints:
s.length == cost.length1 <= s.length, cost.length <= 10^51 <= cost[i] <= 10^4scontains only lowercase English letters.
Solution: Group by group
For a group of same letters, delete all expect the one with the highest cost.
Time complexity: O(n)
Space complexity: O(1)
C++
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class Solution { public: int minCost(string s, vector<int>& cost) { int t = cost[0]; int m = cost[0]; int ans = 0; for (int i = 1; i < s.length(); ++i) { if (s[i] != s[i - 1]) { ans += t - m; t = m = 0; } t += cost[i]; m = max(m, cost[i]); } return ans + (t - m); } }; |
python3
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class Solution: def minCost(self, s: str, cost: List[int]) -> int: s = '*' + s + '*' cost = [0] + cost + [0] ans = t = m = 0 for i in range(1, len(s)): if s[i] != s[i - 1]: ans += t - m t = m = 0 t += cost[i] m = max(m, cost[i]) return ans |