How do you solve Find Good Days to Rob the Bank on LeetCode?

Find Good Days to Rob the Bank (LeetCode #2100) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Precomputation can significantly reduce repeated calculations.

What is the brute force solution for Find Good Days to Rob the Bank?

The brute force approach for Find Good Days to Rob the Bank is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each day individually to see if it meets the criteria for being a good day to rob the bank. This means checking the guards' counts for both the 'time' days before and after each day, which can be quite slow. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Good Days to Rob the Bank?

Find Good Days to Rob the Bank uses the following concepts: Array, Dynamic Programming, Prefix Sum. The recommended patterns to study are: Array, Dynamic Programming.

What are the interview tips for Find Good Days to Rob the Bank?

Explain your thought process clearly while coding. Consider edge cases before finalizing your approach. Discuss potential optimizations even if you start with a brute force solution.

What are common mistakes when solving Find Good Days to Rob the Bank?

Not handling edge cases like time = 0 correctly. Overcomplicating the checks instead of using arrays to store results.

#2100

Find Good Days to Rob the Bank

Medium

ArrayDynamic ProgrammingPrefix SumArrayDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution precomputes the number of consecutive non-increasing days before each day and non-decreasing days after each day. This allows us to determine if a day is good in constant time.

⚙️

Algorithm

4 steps

1Step 1: Create two arrays, 'left' and 'right', to store the counts of non-increasing and non-decreasing days.
2Step 2: Fill the 'left' array by iterating through the security array to count consecutive non-increasing days.
3Step 3: Fill the 'right' array similarly for non-decreasing days.
4Step 4: Iterate through each day and check if both 'left[i] >= time' and 'right[i] >= time' to determine good days.

solution.py11 lines

1def goodDaysToRobBank(security, time):
2    n = len(security)
3    left = [0] * n
4    right = [0] * n
5    for i in range(1, n):
6        if security[i] <= security[i - 1]:
7            left[i] = left[i - 1] + 1
8    for i in range(n - 2, -1, -1):
9        if security[i] <= security[i + 1]:
10            right[i] = right[i + 1] + 1
11    return [i for i in range(time, n - time) if left[i] >= time and right[i] >= time]

ℹ

Complexity note: The complexity is linear because we traverse the security array a constant number of times (twice) to build the left and right arrays.

1Precomputation can significantly reduce repeated calculations.
2Understanding the conditions for good days helps in optimizing the solution.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.