1 def smallest_int_optimized(digits):
 2     # Step 1: Count frequencies - O(N)
 3     counts = [0] * 10
 4     for d in digits:
 5         counts[d] += 1
 6     
 7     # Step 2: Build string from 1 to 9 - O(1) (fixed number of digits)
 8     res = []
 9     for d in range(1, 10):
10         res.append(str(d) * counts[d])
11         
12     return "".join(res)

 1 from collections import Counter
 2 
 3 def find_min_greater_equal_counter(nums, lower_bound):
 4     s_bound = str(lower_bound)
 5     n = len(s_bound)
 6     # 统计每个数字出现的频率
 7     counts = Counter(nums)
 8     # 获取去重后的有序数字列表
 9     unique_digits = sorted(counts.keys())
10     
11     def solve(index, is_greater):
12         if index == n:
13             return ""
14 
15         target = int(s_bound[index])
16         
17         for d in unique_digits:
18             if counts[d] > 0:
19                 # 剪枝：如果还没超越 bound 且当前数字太小，跳过
20                 if not is_greater and d < target:
21                     continue
22                 
23                 # 尝试放置数字 d
24                 counts[d] -= 1
25                 res = solve(index + 1, is_greater or (d > target))
26                 
27                 if res is not None:
28                     return str(d) + res
29                 
30                 # 回溯
31                 counts[d] += 1
32         
33         return None
34 
35     return solve(0, False)