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

Find the Count of Monotonic Pairs I (LeetCode #3250) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m) time complexity and O(n * m) space complexity. Key insight: Dynamic programming can significantly reduce the time complexity by avoiding redundant calculations.

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

The brute force approach for Find the Count of Monotonic Pairs I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible pairs of arrays that meet the monotonic conditions and checking if they satisfy the sum condition. This is straightforward but inefficient due to the large number of combinations. 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 I?

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

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

Always clarify the problem constraints and edge cases before diving into coding. Think about how you can optimize a brute-force solution and articulate your thought process. Practice explaining your code and logic as you write it; this simulates the interview environment.

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

Failing to consider edge cases where nums[i] is at its minimum or maximum. Overlooking the importance of prefix sums in dynamic programming solutions.

#3250

Find the Count of Monotonic Pairs I

Hard

ArrayMathDynamic ProgrammingCombinatoricsPrefix SumDynamic ProgrammingPrefix SumCombinatorial Counting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

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

The optimal solution uses dynamic programming to count the valid monotonic pairs efficiently, leveraging the constraints of the problem. We build upon previous results to avoid redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Initialize a 2D DP array where dp[i][s] represents the count of monotonic pairs of length i with arr1[i-1] = s.
2Step 2: For each index i from 1 to n, and for each possible value s from 0 to nums[i-1], update dp[i][s] based on the previous values in dp[i-1].
3Step 3: Use prefix sums to efficiently calculate the number of valid pairs for arr1 and arr2.

solution.py15 lines

1# Full working Python code
2def countMonotonicPairs(nums):
3    MOD = 10**9 + 7
4    n = len(nums)
5    dp = [[0] * 51 for _ in range(n + 1)]
6    dp[0][0] = 1
7
8    for i in range(1, n + 1):
9        prefix = [0] * 51
10        for s in range(51):
11            prefix[s] = (prefix[s - 1] + dp[i - 1][s]) % MOD if s > 0 else dp[i - 1][s]
12        for s in range(nums[i - 1] + 1):
13            dp[i][s] = prefix[s]
14
15    return sum(dp[n]) % MOD

ℹ

Complexity note: The complexity is O(n * m) where n is the length of nums and m is the maximum value in nums (which is 50). This is efficient due to the limited range of values.

1Dynamic programming can significantly reduce the time complexity by avoiding redundant calculations.
2Understanding monotonic properties helps in designing efficient algorithms.

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