LeetCode-in-Java

3250. Find the Count of Monotonic Pairs I

Hard

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] <= 50

Solution

public class Solution {
    public int countOfPairs(int[] nums) {
        int[] maxShift = new int[nums.length];
        maxShift[0] = nums[0];
        int currShift = 0;
        for (int i = 1; i < nums.length; i++) {
            currShift = Math.max(currShift, nums[i] - maxShift[i - 1]);
            maxShift[i] = Math.min(maxShift[i - 1], nums[i] - currShift);
            if (maxShift[i] < 0) {
                return 0;
            }
        }
        int[][] cases = getAllCases(nums, maxShift);
        return cases[nums.length - 1][maxShift[nums.length - 1]];
    }

    private int[][] getAllCases(int[] nums, int[] maxShift) {
        int[] currCases;
        int[][] cases = new int[nums.length][];
        cases[0] = new int[maxShift[0] + 1];
        for (int i = 0; i < cases[0].length; i++) {
            cases[0][i] = i + 1;
        }
        for (int i = 1; i < nums.length; i++) {
            currCases = new int[maxShift[i] + 1];
            currCases[0] = 1;
            for (int j = 1; j < currCases.length; j++) {
                int prevCases =
                        j < cases[i - 1].length
                                ? cases[i - 1][j]
                                : cases[i - 1][cases[i - 1].length - 1];
                currCases[j] = (currCases[j - 1] + prevCases) % (1_000_000_000 + 7);
            }
            cases[i] = currCases;
        }
        return cases;
    }
}

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