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Diffstat (limited to 'kth_smallest_amount_single_denom.py')
-rw-r--r-- | kth_smallest_amount_single_denom.py | 50 |
1 files changed, 50 insertions, 0 deletions
diff --git a/kth_smallest_amount_single_denom.py b/kth_smallest_amount_single_denom.py new file mode 100644 index 0000000..36f1e93 --- /dev/null +++ b/kth_smallest_amount_single_denom.py @@ -0,0 +1,50 @@ +# 3116. Kth Smallest Amount With Single Denomination Combination +from collections import defaultdict +from itertools import combinations +from math import lcm + +""" +you are given an integer array 'coins' representing coins of different +denominations and an integer 'k'. you have an infinite number of coins of +each denomination. however, you are not allowed to combine coins of different +denominations. return the k'th smallest amount that can be made using these +coins. +""" + + +class Solution(object): + def findKthSmallest(self, coins, k): + """ + :type coins: List[int] + :type k: int + :rtype: int + """ + n, dic = len(coins), defaultdict(list) + for i in range(1, n + 1): + for comb in combinations(coins, i): + dic[len(comb)].append(lcm(*comb)) + + def cnt(dic, target): + res = 0 + for i in range(1, n + 1): + for j in dic[i]: + res += target // j * pow(-1, i + 1) + return res + + start, end = min(coins), min(coins) * k + while start + 1 < end: + mid = (start + end) // 2 + if cnt(dic, mid) >= k: + end = mid + else: + start = mid + if cnt(dic, start) >= k: + return start + else: + return end + + +if __name__ == "__main__": + obj = Solution() + print(obj.findKthSmallest(coins=[3, 6, 9], k=3)) + print(obj.findKthSmallest(coins=[5, 2], k=7)) |