How do you solve Find the Count of Monotonic Pairs II on LeetCode?

Find the Count of Monotonic Pairs II (LeetCode #3251) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m) time complexity and O(m) space complexity. Key insight: Understanding monotonic sequences is crucial.

What is the brute force solution for Find the Count of Monotonic Pairs II?

The brute force approach for Find the Count of Monotonic Pairs II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible pairs of monotonic arrays and checking if they satisfy the conditions. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n * m).

What algorithm or data structure is used to solve Find the Count of Monotonic Pairs II?

Find the Count of Monotonic Pairs II uses the following concepts: Array, Math, Dynamic Programming, Combinatorics, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find the Count of Monotonic Pairs II?

Explain your thought process clearly. Consider edge cases and ask clarifying questions. Practice combinatorial problems to build intuition.

What are common mistakes when solving Find the Count of Monotonic Pairs II?

Not considering edge cases with minimum and maximum values. Overlooking the modulo operation in large counts.

#3251

Find the Count of Monotonic Pairs II

Hard

ArrayMathDynamic ProgrammingCombinatoricsPrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n * m)Space O(m)

The optimal approach uses combinatorial mathematics to count the valid pairs without generating them explicitly. We leverage the properties of monotonic sequences and prefix sums to efficiently calculate the count.

⚙️

Algorithm

3 steps

1Step 1: Create a prefix sum array to store the number of ways to split each nums[i] into two non-negative parts.
2Step 2: Use combinatorial counting to determine how many valid pairs can be formed for each index based on the prefix sums.
3Step 3: Sum the counts for all indices and return the result modulo 10^9 + 7.

solution.py12 lines

1# Full working Python code
2
3MOD = 10**9 + 7
4
5def countMonotonicPairs(nums):
6    n = len(nums)
7    dp = [0] * (max(nums) + 1)
8    dp[0] = 1
9    for num in nums:
10        for j in range(num, -1, -1):
11            dp[j] = (dp[j] + dp[j - 1]) % MOD
12    return dp[nums[-1]] % MOD

ℹ

Complexity note: Here, n is the length of the nums array and m is the maximum value in nums. This is much more efficient than the brute force method as we avoid generating all pairs explicitly.

1Understanding monotonic sequences is crucial.
2Combinatorial counting can simplify complex problems.

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