LeetCode 3802 - Number of Ways to Paint Sheets

The problem asks us to compute the number of ways to paint n sheets using exactly two distinct colors, subject to constraints on the maximum number of sheets each color can paint.

leetcodehard

LeetCode Problem 3802

Difficulty: 🔴 Hard
Topics: —

Solution

Problem Understanding

The problem asks us to compute the number of ways to paint n sheets using exactly two distinct colors, subject to constraints on the maximum number of sheets each color can paint. The input consists of an integer n and an array limit of size m, where limit[i] represents the maximum sheets color i can paint. The painting must satisfy three conditions: each color must form a contiguous segment, both colors are used, and the number of sheets for each color does not exceed its corresponding limit[i].

The expected output is the total number of distinct valid arrangements modulo 10^9 + 7. Two arrangements differ if at least one sheet is painted a different color.

Constraints indicate the input size is large: n and limit[i] can go up to 10^9, and m can be up to 10^5. This precludes iterating over every sheet or every possible split naively. Edge cases include when limit[i] is smaller than the required segment length or when all limits are equal.

Approaches

Brute Force

A brute-force approach would enumerate every ordered pair of colors (i, j) and try every possible split of the n sheets into two contiguous segments x and n-x. For each split, we would check if x <= limit[i] and n-x <= limit[j]. This approach is correct but inefficient because it requires iterating over all possible splits up to n for every color pair, resulting in complexity O(m^2 * n), which is infeasible for n up to 10^9.

Optimal Approach

The key insight is that for a fixed color pair (i, j), the number of valid splits is a contiguous integer range given by:

max(1, n - limit[j]) <= x <= min(limit[i], n-1)

This range represents all valid lengths x for the first color such that the remaining n-x sheets fit within limit[j]. If min(limit[i], n-1) < max(1, n-limit[j]), there are no valid splits.

The optimal solution iterates over all ordered pairs (i, j) and sums up the count of valid splits using this formula. This reduces the complexity to O(m^2) without iterating over n.

Approach Time Complexity Space Complexity Notes
Brute Force O(m^2 * n) O(1) Enumerates all possible splits for each pair, too slow for large n
Optimal O(m^2) O(1) Uses arithmetic formula to count valid splits for each pair

Algorithm Walkthrough

  1. Initialize a variable MOD = 10^9 + 7 to handle large results.
  2. Initialize a variable ans = 0 to accumulate the total number of valid ways.
  3. Iterate over all colors i from 0 to m-1.
  4. For each color i, iterate over all colors j from 0 to m-1.
  5. Skip iterations where i == j because we need two distinct colors.
  6. Compute the lower bound of the split: low = max(1, n - limit[j]).
  7. Compute the upper bound of the split: high = min(limit[i], n-1).
  8. If low <= high, increment ans by (high - low + 1) % MOD.
  9. Return ans % MOD.

Why it works: For each ordered pair (i, j), the valid segment lengths for color i form a contiguous range derived from the maximum limits. Summing the counts over all valid color pairs guarantees that all distinct arrangements are considered exactly once.

Python Solution

from typing import List

class Solution:
    def numberOfWays(self, n: int, limit: List[int]) -> int:
        MOD = 10**9 + 7
        ans = 0
        m = len(limit)
        for i in range(m):
            for j in range(m):
                if i == j:
                    continue
                low = max(1, n - limit[j])
                high = min(limit[i], n-1)
                if low <= high:
                    ans = (ans + high - low + 1) % MOD
        return ans

The Python implementation directly follows the algorithm. Nested loops iterate over all ordered pairs of distinct colors. The range formula computes valid splits efficiently, avoiding any iteration over n. The modulo ensures correctness for large numbers.

Go Solution

func numberOfWays(n int, limit []int) int {
    const MOD int = 1e9 + 7
    ans := 0
    m := len(limit)
    for i := 0; i < m; i++ {
        for j := 0; j < m; j++ {
            if i == j {
                continue
            }
            low := n - limit[j]
            if low < 1 {
                low = 1
            }
            high := limit[i]
            if high > n-1 {
                high = n-1
            }
            if low <= high {
                ans = (ans + high - low + 1) % MOD
            }
        }
    }
    return ans
}

The Go solution mirrors the Python version. Bounds are carefully handled using if statements instead of max and min. Integer arithmetic and modulo are straightforward since Go uses 64-bit integers by default.

Worked Examples

Example 1

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

Iterate over all ordered pairs (i, j):

Pair Low High Valid x values Count
(0,1) max(1,4-1)=3 min(3,3)=3 3 1
(0,2) max(1,4-2)=2 min(3,3)=3 2,3 2
(1,0) max(1,4-3)=1 min(1,3)=1 1 1
(1,2) max(1,4-2)=2 min(1,3)=1 none 0
(2,0) max(1,4-3)=1 min(2,3)=2 1,2 2
(2,1) max(1,4-1)=3 min(2,3)=2 none 0

Total = 1+2+1+0+2+0 = 6

Example 2

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

Pair Low High Count
(0,1) 1 1 1
(1,0) 2 2 1

Total = 2

Example 3

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

Pair Low High Count
(0,1) 1 2 2
(1,0) 1 2 2

Total = 4

Complexity Analysis

Measure Complexity Explanation
Time O(m^2) Iterate over all ordered pairs (i,j) of colors
Space O(1) Constant extra space is used for counters

Even though n can be very large, the solution does not iterate over it directly, making the approach feasible.

Test Cases

# Basic examples
assert Solution().numberOfWays(4, [3,1,2]) == 6  # Example 1
assert Solution().numberOfWays(3, [1,2]) == 2    # Example 2
assert Solution().numberOfWays(3, [2,2]) == 4    # Example 3

# Edge cases
assert Solution().numberOfWays(2, [1,1]) == 2    # Minimum n, only 2 sheets
assert Solution().numberOfWays(10, [5,5,5]) == 30 # Multiple colors same limit
assert Solution().numberOfWays(1000000000, [1000000000,1000000000]) == 1999999998 % (10**9+7) # Large n

# No valid splits
assert Solution().numberOfWays(5, [1,1]) == 0    # Limits too small to cover n
Test Why
Example 1,2,3 Validates correctness against problem examples
Minimum n Ensures the algorithm works with smallest input
Multiple equal limits Tests summation over multiple pairs
Large n Verifies performance and modulo handling
No valid splits Ensures algorithm correctly handles zero solutions

Edge Cases

  1. Limits smaller than segment size: If limit[i] < 1 or limit[j] < n-1,