You are given a 0-indexed array of strings nums
, where each string is of equal length and consists of only digits.
You are also given a 0-indexed 2D integer array queries
where queries[i] = [ki, trimi]
. For each queries[i]
, you need to:
- Trim each number in
nums
to its rightmosttrimi
digits. - Determine the index of the
kith
smallest trimmed number innums
. If two trimmed numbers are equal, the number with the lower index is considered to be smaller. - Reset each number in
nums
to its original length.
Return an array answer
of the same length as queries
, where answer[i]
is the answer to the ith
query.
Note:
- To trim to the rightmost
x
digits means to keep removing the leftmost digit, until onlyx
digits remain. - Strings in
nums
may contain leading zeros.
Input: nums = ["102","473","251","814"], queries = [[1,1],[2,3],[4,2],[1,2]] Output: [2,2,1,0] Explanation: 1. After trimming to the last digit, nums = ["2","3","1","4"]. The smallest number is 1 at index 2. 2. Trimmed to the last 3 digits, nums is unchanged. The 2nd smallest number is 251 at index 2. 3. Trimmed to the last 2 digits, nums = ["02","73","51","14"]. The 4th smallest number is 73. 4. Trimmed to the last 2 digits, the smallest number is 2 at index 0. Note that the trimmed number "02" is evaluated as 2.
Input: nums = ["24","37","96","04"], queries = [[2,1],[2,2]] Output: [3,0] Explanation: 1. Trimmed to the last digit, nums = ["4","7","6","4"]. The 2nd smallest number is 4 at index 3. There are two occurrences of 4, but the one at index 0 is considered smaller than the one at index 3. 2. Trimmed to the last 2 digits, nums is unchanged. The 2nd smallest number is 24.
1 <= nums.length <= 100
1 <= nums[i].length <= 100
nums[i]
consists of only digits.- All
nums[i].length
are equal. 1 <= queries.length <= 100
queries[i].length == 2
1 <= ki <= nums.length
1 <= trimi <= nums[i].length
Follow up: Could you use the Radix Sort Algorithm to solve this problem? What will be the complexity of that solution?
class Solution:
def smallestTrimmedNumbers(self, nums: List[str], queries: List[List[int]]) -> List[int]:
nums = [(nums[i], i) for i in range(len(nums))]
trims = {}
answer = [0] * len(queries)
for i in range(len(queries)):
if queries[i][1] not in trims:
trims[queries[i][1]] = []
trims[queries[i][1]].append((i, queries[i][0] - 1))
for trim in range(1, len(nums[0][0]) + 1):
count = [0] * 10
tmp = [("", 0)] * len(nums)
for num, _ in nums:
count[ord(num[-trim]) - 48] += 1
for i in range(1, 10):
count[i] += count[i - 1]
for num, i in nums[::-1]:
count[ord(num[-trim]) - 48] -= 1
tmp[count[ord(num[-trim]) - 48]] = (num, i)
for i, k in trims.get(trim, []):
answer[i] = tmp[k][1]
nums = tmp
return answer