LeetCode #3251 — HARD

Find the Count of Monotonic Pairs II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

You are given an array of positive integers nums of length n.

We call a pair of non-negative integer arrays (arr1, arr2) monotonic if:

The lengths of both arrays are n.
arr1 is monotonically non-decreasing, in other words, arr1[0] <= arr1[1] <= ... <= arr1[n - 1].
arr2 is monotonically non-increasing, in other words, arr2[0] >= arr2[1] >= ... >= arr2[n - 1].
arr1[i] + arr2[i] == nums[i] for all 0 <= i <= n - 1.

Return the count of monotonic pairs.

Since the answer may be very large, return it modulo 10⁹ + 7.

Example 1:

Input: nums = [2,3,2]

Output: 4

Explanation:

The good pairs are:

([0, 1, 1], [2, 2, 1])
([0, 1, 2], [2, 2, 0])
([0, 2, 2], [2, 1, 0])
([1, 2, 2], [1, 1, 0])

Example 2:

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

Output: 126

Constraints:

1 <= n == nums.length <= 2000
1 <= nums[i] <= 1000

Step 01

Brute Force Baseline

Problem summary: You are given an array of positive integers nums of length n. We call a pair of non-negative integer arrays (arr1, arr2) monotonic if: The lengths of both arrays are n. arr1 is monotonically non-decreasing, in other words, arr1[0] <= arr1[1] <= ... <= arr1[n - 1]. arr2 is monotonically non-increasing, in other words, arr2[0] >= arr2[1] >= ... >= arr2[n - 1]. arr1[i] + arr2[i] == nums[i] for all 0 <= i <= n - 1. Return the count of monotonic pairs. Since the answer may be very large, return it modulo 109 + 7.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math · Dynamic Programming

Example 1

[2,3,2]

Example 2

[5,5,5,5]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3251: Find the Count of Monotonic Pairs II
class Solution {
    public int countOfPairs(int[] nums) {
        final int mod = (int) 1e9 + 7;
        int n = nums.length;
        int m = Arrays.stream(nums).max().getAsInt();
        int[][] f = new int[n][m + 1];
        for (int j = 0; j <= nums[0]; ++j) {
            f[0][j] = 1;
        }
        int[] g = new int[m + 1];
        for (int i = 1; i < n; ++i) {
            g[0] = f[i - 1][0];
            for (int j = 1; j <= m; ++j) {
                g[j] = (g[j - 1] + f[i - 1][j]) % mod;
            }
            for (int j = 0; j <= nums[i]; ++j) {
                int k = Math.min(j, j + nums[i - 1] - nums[i]);
                if (k >= 0) {
                    f[i][j] = g[k];
                }
            }
        }
        int ans = 0;
        for (int j = 0; j <= nums[n - 1]; ++j) {
            ans = (ans + f[n - 1][j]) % mod;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3251: Find the Count of Monotonic Pairs II
func countOfPairs(nums []int) (ans int) {
	const mod int = 1e9 + 7
	n := len(nums)
	m := slices.Max(nums)
	f := make([][]int, n)
	for i := range f {
		f[i] = make([]int, m+1)
	}
	for j := 0; j <= nums[0]; j++ {
		f[0][j] = 1
	}
	g := make([]int, m+1)
	for i := 1; i < n; i++ {
		g[0] = f[i-1][0]
		for j := 1; j <= m; j++ {
			g[j] = (g[j-1] + f[i-1][j]) % mod
		}
		for j := 0; j <= nums[i]; j++ {
			k := min(j, j+nums[i-1]-nums[i])
			if k >= 0 {
				f[i][j] = g[k]
			}
		}
	}
	for j := 0; j <= nums[n-1]; j++ {
		ans = (ans + f[n-1][j]) % mod
	}
	return
}

# Accepted solution for LeetCode #3251: Find the Count of Monotonic Pairs II
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

// Accepted solution for LeetCode #3251: Find the Count of Monotonic Pairs II
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3251: Find the Count of Monotonic Pairs II
// class Solution {
//     public int countOfPairs(int[] nums) {
//         final int mod = (int) 1e9 + 7;
//         int n = nums.length;
//         int m = Arrays.stream(nums).max().getAsInt();
//         int[][] f = new int[n][m + 1];
//         for (int j = 0; j <= nums[0]; ++j) {
//             f[0][j] = 1;
//         }
//         int[] g = new int[m + 1];
//         for (int i = 1; i < n; ++i) {
//             g[0] = f[i - 1][0];
//             for (int j = 1; j <= m; ++j) {
//                 g[j] = (g[j - 1] + f[i - 1][j]) % mod;
//             }
//             for (int j = 0; j <= nums[i]; ++j) {
//                 int k = Math.min(j, j + nums[i - 1] - nums[i]);
//                 if (k >= 0) {
//                     f[i][j] = g[k];
//                 }
//             }
//         }
//         int ans = 0;
//         for (int j = 0; j <= nums[n - 1]; ++j) {
//             ans = (ans + f[n - 1][j]) % mod;
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3251: Find the Count of Monotonic Pairs II
function countOfPairs(nums: number[]): number {
    const mod = 1e9 + 7;
    const n = nums.length;
    const m = Math.max(...nums);
    const f: number[][] = Array.from({ length: n }, () => Array(m + 1).fill(0));
    for (let j = 0; j <= nums[0]; j++) {
        f[0][j] = 1;
    }
    const g: number[] = Array(m + 1).fill(0);
    for (let i = 1; i < n; i++) {
        g[0] = f[i - 1][0];
        for (let j = 1; j <= m; j++) {
            g[j] = (g[j - 1] + f[i - 1][j]) % mod;
        }
        for (let j = 0; j <= nums[i]; j++) {
            const k = Math.min(j, j + nums[i - 1] - nums[i]);
            if (k >= 0) {
                f[i][j] = g[k];
            }
        }
    }
    let ans = 0;
    for (let j = 0; j <= nums[n - 1]; j++) {
        ans = (ans + f[n - 1][j]) % mod;
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.