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3251. Find the Count of Monotonic Pairs II

Description

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:

  1. ([0, 1, 1], [2, 2, 1])
  2. ([0, 1, 2], [2, 2, 0])
  3. ([0, 2, 2], [2, 1, 0])
  4. ([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

Solutions

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 $\sum_{j=0}^{\textit{nums}[n-1]} f[n-1][j]$.

When $i = 0$, we have $f[0][j] = 1$ for $0 \leq j \leq \textit{nums}[0]$.

When $i > 0$, we can calculate $f[i][j]$ based on $f[i-1][j']$. Since $\textit{arr1}$ is non-decreasing, $j' \leq j$. Additionally, since $\textit{arr2}$ is non-increasing, $\textit{nums}[i] - j \leq \textit{nums}[i - 1] - j'$. Thus, $j' \leq \min(j, j + \textit{nums}[i - 1] - \textit{nums}[i])$.

The answer is $\sum_{j=0}^{\textit{nums}[n-1]} f[n-1][j]$.

The time complexity is $O(n \times m)$, and the space complexity is $O(n \times m)$. Here, $n$ represents the length of the array $\textit{nums}$, and $m$ represents the maximum value in the array $\textit{nums}$.

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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

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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;
    }
}

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class Solution {
public:
    int countOfPairs(vector<int>& nums) {
        const int mod = 1e9 + 7;
        int n = nums.size();
        int m = *max_element(nums.begin(), nums.end());
        vector<vector<int>> f(n, vector<int>(m + 1));
        for (int j = 0; j <= nums[0]; ++j) {
            f[0][j] = 1;
        }
        vector<int> g(m + 1);
        for (int i