Leetcode #2100: Find Good Days to Rob the Bank
In this guide, we solve Leetcode #2100 Find Good Days to Rob the Bank in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You and a gang of thieves are planning on robbing a bank. You are given a 0-indexed integer array security, where security[i] is the number of guards on duty on the ith day.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Dynamic Programming, Prefix Sum
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: security = [5,3,3,3,5,6,2], time = 2
Output: [2,3]
Explanation:
On day 2, we have security[0] >= security[1] >= security[2] <= security[3] <= security[4].
On day 3, we have security[1] >= security[2] >= security[3] <= security[4] <= security[5].
No other days satisfy this condition, so days 2 and 3 are the only good days to rob the bank.
Python Solution
class Solution:
def goodDaysToRobBank(self, security: List[int], time: int) -> List[int]:
n = len(security)
if n <= time * 2:
return []
left, right = [0] * n, [0] * n
for i in range(1, n):
if security[i] <= security[i - 1]:
left[i] = left[i - 1] + 1
for i in range(n - 2, -1, -1):
if security[i] <= security[i + 1]:
right[i] = right[i + 1] + 1
return [i for i in range(n) if time <= min(left[i], right[i])]
Complexity
The time complexity is O(n·m) (typical). The space complexity is O(n·m) or optimized.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.