#3250

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Find the Count of Monotonic Pairs I

Compute the number of monotonic pairs in an integer array using state transition dynamic programming efficiently.

array math dynamic programming combinatorics prefix sum

Problem Statement

Given an array nums of n positive integers, count all pairs of non-negative integer arrays (arr1, arr2) that satisfy monotonic constraints. A pair is considered monotonic if each element of arr1 and arr2 maintains the required ordering, as defined by the problem.

Return the total number of monotonic pairs possible for the given nums array. The constraints are 1 <= n <= 2000 and 1 <= nums[i] <= 50, and results must account for all valid combinations efficiently.

Examples

Example 1

Input: nums = [2,3,2]

Output: 4

The good pairs are:

Example 2

Input: nums = [5,5,5,5]

Output: 126

Example details omitted.

Constraints

1 <= n == nums.length <= 2000
1 <= nums[i] <= 50

Solution Approach

Dynamic Programming State Definition

Define dp[i][s] as the number of monotonic pairs of length i where the last element of arr1 is s. This captures all valid transitions and allows building solutions incrementally while enforcing monotonicity.

Iterative Transition

For each element in nums, iterate over possible states s and update dp[i][s] using previous dp[i-1][prev] values where prev <= s. Use prefix sums to accelerate cumulative counts and avoid recomputation.

Result Aggregation

After processing all elements, sum dp[n][s] over all valid s to get the total number of monotonic pairs. Ensure boundary conditions and combinatorial counts are applied correctly to handle repeated numbers.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity depends on n and the maximum nums[i] value due to DP iterations and prefix sums, roughly O(n * maxValue). Space complexity is O(n * maxValue) for the DP table.

What Interviewers Usually Probe

Check if candidate correctly defines dp[i][s] and understands state transitions.
Observe whether prefix sums are used to optimize the cumulative counts in DP.
Watch for correct handling of repeated elements and monotonic constraints.

Common Pitfalls or Variants

Common pitfalls

Forgetting to enforce monotonicity when updating DP states.
Inefficiently iterating without prefix sums, leading to timeouts.
Incorrectly summing final DP values, missing some valid pairs.

Follow-up variants

Counting strictly increasing monotonic pairs instead of non-decreasing.
Finding monotonic triples or larger tuples with similar DP approach.
Applying the DP state transition pattern to other bounded integer arrays with combinatorial constraints.

FAQ

What is the main idea behind counting monotonic pairs in this problem?

The core idea is using a DP table dp[i][s] where each state represents the number of monotonic pairs ending with a specific value, ensuring transitions maintain monotonicity.

Can this problem handle arrays with maximum length 2000 efficiently?

Yes, with state transition DP and prefix sums, the algorithm scales to n = 2000 without timing out.

Why do we use prefix sums in the DP solution?

Prefix sums allow cumulative updates efficiently, avoiding O(n^2) inner loops when summing valid previous states.

Are repeated numbers in nums handled differently?

Repeated numbers are naturally handled in the DP state; the transitions include counts from previous identical values without violating monotonicity.

Is the state transition dynamic programming pattern common in combinatorial array problems?

Yes, this pattern efficiently counts sequences or pairs under constraints and is a key approach for monotonic pair counting like in this problem.

terminal

Solution

Solution 1: Dynamic Programming + Prefix Sum Optimization

We define $f[i][j]$ to represent the number of monotonic array pairs for the subarray $[0, \ldots, i]$ where $arr1[i] = j$. Initially, $f[i][j] = 0$, and the answer is $\sum_{j=0}^{\textit{nums}[n-1]} f[n-1][j]$.

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class Solution:
    def countOfPairs(self, nums: List[int]) -> int:
        mod = 10**9 + 7
        n, m = len(nums), max(nums)
        f = [[0] * (m + 1) for _ in range(n)]
        for j in range(nums[0] + 1):
            f[0][j] = 1
        for i in range(1, n):
            s = list(accumulate(f[i - 1]))
            for j in range(nums[i] + 1):
                k = min(j, j + nums[i - 1] - nums[i])
                if k >= 0:
                    f[i][j] = s[k] % mod
        return sum(f[-1][: nums[-1] + 1]) % mod

Continue Practicing

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