What is the time complexity of the optimal greedy algorithm that finds the largest monotone increasing digits number less than or equal to n, where d is the number of digits in n?
medium🪤 Complexity Trap Q13 of Q15
Greedy Algorithms - Monotone Increasing Digits
What is the time complexity of the optimal greedy algorithm that finds the largest monotone increasing digits number less than or equal to n, where d is the number of digits in n?
AO(d), since it scans digits once from right to left and then sets trailing digits
BO(log n), since the number of digits d is proportional to log n
CO(d^2), because each decrement may cause re-checking previous digits multiple times
DO(n * d), where n is the input number and d is the number of digits
Step-by-Step Solution
Solution:
Step 1: Understand the relationship between digits and input size
The number of digits d in n is proportional to log10 n.
Step 2: Express complexity in terms of n
The algorithm runs in O(d) time, which is O(log n) when expressed in terms of n.
Final Answer:
Option B -> Option B
Quick Check:
Expressing complexity in terms of n, O(log n) is correct and more standard [OK]
Quick Trick:Algorithm runs in linear time relative to digit count, which is logarithmic in n [OK]
Common Mistakes:
MISTAKES
Confusing input size n with digit count d
Assuming repeated decrements cause quadratic time
Ignoring that digit count is log n
Trap Explanation:
PITFALL
Option A is correct in terms of digits d, but complexity is usually expressed in terms of input size n; thus, O(log n) is the best choice.
Interviewer Note:
CONTEXT
Checks if candidate understands complexity in terms of input size n, not just digit length.
Master "Monotone Increasing Digits" in Greedy Algorithms
3 interactive learning modes - each teaches the same concept differently
