class Solution:
def minimumCost(self, cost: List[int]) -> int:
cost.sort(reverse=True)
res, i, N = 0, 0, len(cost)
while i < N:
res += sum(cost[i : i + 2])
i += 3
return res
1562B - Scenes From a Memory | 1521A - Nastia and Nearly Good Numbers |
208. Implement Trie | 1605B - Reverse Sort |
1607C - Minimum Extraction | 1604B - XOR Specia-LIS-t |
1606B - Update Files | 1598B - Groups |
1602B - Divine Array | 1594B - Special Numbers |
1614A - Divan and a Store | 2085. Count Common Words With One Occurrence |
2089. Find Target Indices After Sorting Array | 2090. K Radius Subarray Averages |
2091. Removing Minimum and Maximum From Array | 6. Zigzag Conversion |
1612B - Special Permutation | 1481. Least Number of Unique Integers after K Removals |
1035. Uncrossed Lines | 328. Odd Even Linked List |
1219. Path with Maximum Gold | 1268. Search Suggestions System |
841. Keys and Rooms | 152. Maximum Product Subarray |
337. House Robber III | 869. Reordered Power of 2 |
1593C - Save More Mice | 1217. Minimum Cost to Move Chips to The Same Position |
347. Top K Frequent Elements | 1503. Last Moment Before All Ants Fall Out of a Plank |