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Find the Count of Monotonic Pairs II

Difficulty: Hard

Problem 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.
  • arr2 is monotonically non-increasing.
  • 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 10^9 + 7.

Key Insights

  • The problem requires constructing two arrays under specific monotonic constraints.
  • The sum constraint (arr1[i] + arr2[i] == nums[i]) suggests a need to explore combinations of values for arr1 and arr2 that respect the bounds set by nums.
  • The constraints on monotonicity imply that we will need to maintain counts of valid configurations as we iterate through the possible values for the elements in the arrays.

Space and Time Complexity

Time Complexity: O(n * max(nums))
Space Complexity: O(max(nums))

Solution

To solve the problem, we can use dynamic programming along with combinatorial counting. The main idea is to iterate over each num in nums and determine how many valid pairs (arr1[i], arr2[i]) can be formed such that:

  1. arr1[i] is non-decreasing.
  2. arr2[i] is non-increasing. We will maintain a count of valid combinations using a prefix sum approach to efficiently calculate the counts for each index based on previous indices.

Code Solutions

def countMonotonicPairs(nums):
    MOD = 10**9 + 7
    n = len(nums)
    max_num = max(nums)
    
    # Count of ways to form arr1 and arr2 up to each number
    dp = [[0] * (max_num + 1) for _ in range(n + 1)]
    
    # Base case: one way to form an empty array
    for j in range(max_num + 1):
        dp[0][j] = 1
    
    for i in range(1, n + 1):
        total = 0
        for j in range(nums[i - 1] + 1):
            total = (total + dp[i - 1][j]) % MOD
            dp[i][j] = total
    
    count = 0
    for j in range(max_num + 1):
        count = (count + dp[n][j]) % MOD
    
    return count

# Example usage
print(countMonotonicPairs([2, 3, 2]))  # Output: 4
print(countMonotonicPairs([5, 5, 5, 5]))  # Output: 126
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