Is Number of Ways to Paint Sheets easy or hard?

Number of Ways to Paint Sheets is classified as a Hard problem because it requires identifying the correct DP states and deriving the recurrence. Once the pattern is recognized, the final implementation becomes concise with O(n) time and constant space.

Number of Ways to Paint Sheets Python/Java solution

The solution can be implemented in Python, Java, C++, or Go using dynamic programming. Maintain counters representing valid states for the previous sheet and update them iteratively. Each update applies the recurrence and modulo operation to keep numbers within range.

How to solve Number of Ways to Paint Sheets in O(n)?

Define recurrence relations that represent how many valid colorings end in certain patterns. Instead of storing a full DP table, keep a few counters describing valid configurations up to the previous sheet. Update those counters for each new sheet using the allowed color transitions, applying modulo arithmetic when necessary.

What is the best approach for Number of Ways to Paint Sheets?

The best approach uses dynamic programming with state compression. Instead of tracking every color choice for every sheet, maintain aggregated states representing valid transitions between sheets. This reduces the runtime to O(n) with constant space while still counting all valid configurations using modular arithmetic.

Is Number of Ways to Paint Sheets asked at Google/Amazon/Meta?

Problems involving counting valid painting configurations commonly appear in interviews at companies like Google, Amazon, and Meta. They test dynamic programming skills, recurrence derivation, and the ability to optimize combinatorial counting problems.

What data structure is used in Number of Ways to Paint Sheets?

The main technique relies on dynamic programming arrays or a few running variables for state compression. No complex data structures are required; the focus is on recurrence relations and efficient counting using arithmetic operations.

What is the time complexity of Number of Ways to Paint Sheets?

The optimized dynamic programming solution runs in O(n) time where n is the number of sheets. Each step updates a small set of state variables derived from the previous sheet. Space complexity can be reduced to O(1) because only the previous state is required.

Description Hints 4

3802. Number of Ways to Paint Sheets

Hard66.7% acceptancePremium elsewhere, free here

You are given an integer n representing the number of sheets.

You are also given an integer array limit of size m, where limit[i] is the maximum number of sheets that can be painted using color i.

You must paint all n sheets under the following conditions:

Exactly two distinct colors are used.
Each color must cover a single contiguous segment of sheets.
The number of sheets painted with color i cannot exceed limit[i].

Return an integer denoting the number of distinct ways to paint all sheets. Since the answer may be large, return it modulo 10⁹ + 7.

Note: Two ways differ if at least one sheet is painted with a different color.

Example 1:

Input: n = 4, limit = [3,1,2]

Output: 6

Explanation:

For each ordered pair (i, j), where color i is used for the first segment and color j for the second segment (i != j), a split of x and 4 - x is valid if 1 <= x <= limit[i] and 1 <= 4 - x <= limit[j].

Valid pairs and counts are:

(0, 1): x = 3
(0, 2): x = 2, 3
(1, 0): x = 1
(2, 0): x = 1, 2

Therefore, there are 6 valid ways in total.

Example 2:

Input: n = 3, limit = [1,2]

Output: 2

Explanation:

For each ordered pair (i, j), where color i is used for the first segment and color j for the second segment (i != j), a split of x and 3 - x is valid if 1 <= x <= limit[i] and 1 <= 3 - x <= limit[j].

Valid pairs and counts are:

(0, 1): x = 1
(1, 0): x = 2

Hence, there are 2 valid ways in total.

Example 3:

Input: n = 3, limit = [2,2]

Output: 4

Explanation:

Valid pairs and counts are:

(0, 1): x = 1, 2
(1, 0): x = 1, 2

Therefore, there are 4 valid ways in total.

Constraints:

2 <= n <= 10⁹
2 <= m == limit.length <= 10⁵
1 <= limit[i] <= 10⁹

Hints open one at a time

Hint 1

Sort limit and use binary search (lower_bound) to compute num_ge(t), the number of colors with limit >= t.

Hint 2

For a split x, count ordered pairs (i, j) with i != j as num_ge(x) * num_ge(n - x) - num_ge(max(x, n - x)).

Hint 3

Note that num_ge(x) and num_ge(n - x) change only when x crosses 1, n - 1, L + 1, or n - L for some L in limit.

Hint 4

Collect all such critical x values, sort and deduplicate them, and treat the answer as constant between consecutive values.

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1class Solution:
  def solve(self):
      pass


Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

4
[3,1,2]

Input

3
[1,2]

Input

3
[2,2]

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