Problem Description

You have an array nums of positive integers with length n. Your task is to find how many ways you can split this array into two arrays (arr1 and arr2) that satisfy specific monotonic properties.

A pair of non-negative integer arrays (arr1, arr2) is considered monotonic if:

1. Both arrays have the same length n as the original array
2. arr1 is non-decreasing: arr1[0] <= arr1[1] <= ... <= arr1[n-1]
3. arr2 is non-increasing: arr2[0] >= arr2[1] >= ... >= arr2[n-1]
4. At each position i, the sum equals the original value: arr1[i] + arr2[i] = nums[i]

For example, if nums = [2, 3, 2], one valid monotonic pair could be:

• arr1 = [0, 1, 2] (non-decreasing)
• arr2 = [2, 2, 0] (non-increasing)
• At each position: 0+2=2, 1+2=3, 2+0=2

The problem asks you to count all possible monotonic pairs that can be formed. Since the number of valid pairs can be very large, return the result modulo 10^9 + 7.

The solution uses dynamic programming where f[i][j] represents the number of valid monotonic pairs for the first i+1 elements where arr1[i] = j. The key insight is that for a valid split:

• Since arr1 must be non-decreasing, if arr1[i] = j, then arr1[i-1] <= j
• Since arr2 must be non-increasing and arr2[i] = nums[i] - j, we need nums[i] - j <= nums[i-1] - arr1[i-1]

These constraints help determine which previous states can lead to each current state, allowing us to build up the solution incrementally using prefix sums for optimization.

Intuition

The key observation is that once we fix the value of arr1[i], the value of arr2[i] is automatically determined since arr1[i] + arr2[i] = nums[i]. So instead of tracking both arrays, we only need to track the possible values of arr1.

Let's think about what makes a valid split. If we choose arr1[i] = j, then arr2[i] = nums[i] - j. For this to be valid:

• arr1 must be non-decreasing, so arr1[i-1] <= j
• arr2 must be non-increasing, so arr2[i-1] >= arr2[i], which means nums[i-1] - arr1[i-1] >= nums[i] - j

Rearranging the second constraint: arr1[i-1] <= j + nums[i-1] - nums[i]

Combining both constraints, we get: arr1[i-1] <= min(j, j + nums[i-1] - nums[i])

This tells us that if we want arr1[i] = j, then arr1[i-1] can be any value from 0 to min(j, j + nums[i-1] - nums[i]).

This naturally leads to a dynamic programming approach where we track the number of ways to form valid splits ending at each position with specific values of arr1[i]. We define f[i][j] as the number of valid monotonic pairs for the first i+1 elements where arr1[i] = j.

For the base case, when i = 0, we can choose any value from 0 to nums[0] for arr1[0] (ensuring arr2[0] = nums[0] - arr1[0] is non-negative).

For transitions, the number of ways to reach state f[i][j] is the sum of all valid previous states f[i-1][j'] where j' <= min(j, j + nums[i-1] - nums[i]).

Since we're summing over a range of previous states, we can optimize this using prefix sums to avoid recalculating the same sums repeatedly. This reduces the time complexity from O(n × m²) to O(n × m), where m is the maximum value in nums.

Learn more about Math, Dynamic Programming, Combinatorics and Prefix Sum patterns.

Solution Approach

The solution implements dynamic programming with prefix sum optimization as outlined in the reference approach.

We define f[i][j] to represent the number of monotonic array pairs for the subarray [0, ..., i] where arr1[i] = j. The final answer will be sum(f[n-1][j]) for all valid j.

Initialization:

• Create a 2D DP table f with dimensions n × (m + 1), where n = len(nums) and m = max(nums)
• For the base case when i = 0, set f[0][j] = 1 for all 0 ≤ j ≤ nums[0], since we can choose any value from 0 to nums[0] for arr1[0]

State Transition: For each position i from 1 to n-1:

1. First, compute the prefix sum array s from the previous row: s = list(accumulate(f[i-1]))

   • This allows us to quickly query the sum of f[i-1][0] to f[i-1][k] as s[k]

2. For each possible value j from 0 to nums[i]:

   • Calculate the maximum valid previous value: k = min(j, j + nums[i-1] - nums[i])
   • If k ≥ 0, then f[i][j] = s[k] % mod
   • This represents the sum of all f[i-1][j'] where j' ≤ k

The constraint k = min(j, j + nums[i-1] - nums[i]) ensures:

• arr1[i-1] ≤ j (non-decreasing property of arr1)
• arr2[i-1] ≥ arr2[i] (non-increasing property of arr2)

Computing the Answer: The final answer is the sum of all valid splits ending at position n-1:

sum(f[-1][:nums[-1] + 1]) % mod

This sums up f[n-1][j] for all j from 0 to nums[n-1], representing all possible values of arr1[n-1] that result in valid monotonic pairs.

Time Complexity: O(n × m) where n is the length of nums and m is the maximum value in nums Space Complexity: O(n × m) for the DP table

Example Walkthrough

Let's walk through the solution with nums = [2, 3, 2].

Setup:

• n = 3, m = max(nums) = 3
• We create a DP table f of size 3 × 4 (to accommodate values 0 to 3)
• mod = 10^9 + 7

Step 1: Initialize base case (i = 0)

For nums[0] = 2, we can choose arr1[0] to be any value from 0 to 2:

• If arr1[0] = 0, then arr2[0] = 2 - 0 = 2 ✓
• If arr1[0] = 1, then arr2[0] = 2 - 1 = 1 ✓
• If arr1[0] = 2, then arr2[0] = 2 - 2 = 0 ✓

So f[0] = [1, 1, 1, 0] (indices represent j = 0, 1, 2, 3)

Step 2: Process i = 1 (nums[1] = 3)

First, compute prefix sum: s = [1, 2, 3, 3] (cumulative sum of f[0])

For each possible j from 0 to 3:

• j = 0:

  • k = min(0, 0 + 2 - 3) = min(0, -1) = -1
  • Since k < 0, f[1][0] = 0

• j = 1:

  • k = min(1, 1 + 2 - 3) = min(1, 0) = 0
  • f[1][1] = s[0] = 1
  • This counts: arr1[0] = 0, arr1[1] = 1

• j = 2:

  • k = min(2, 2 + 2 - 3) = min(2, 1) = 1
  • f[1][2] = s[1] = 2
  • This counts: arr1[0] ∈ {0, 1}, arr1[1] = 2

• j = 3:

  • k = min(3, 3 + 2 - 3) = min(3, 2) = 2
  • f[1][3] = s[2] = 3
  • This counts: arr1[0] ∈ {0, 1, 2}, arr1[1] = 3

So f[1] = [0, 1, 2, 3]

Step 3: Process i = 2 (nums[2] = 2)

Prefix sum: s = [0, 1, 3, 6]

For each possible j from 0 to 2:

• j = 0:

  • k = min(0, 0 + 3 - 2) = min(0, 1) = 0
  • f[2][0] = s[0] = 0

• j = 1:

  • k = min(1, 1 + 3 - 2) = min(1, 2) = 1
  • f[2][1] = s[1] = 1
  • This represents: arr1 = [0, 1, 1], arr2 = [2, 2, 1]

• j = 2:

  • k = min(2, 2 + 3 - 2) = min(2, 3) = 2
  • f[2][2] = s[2] = 3
  • This represents three valid sequences:
    • arr1 = [0, 1, 2], arr2 = [2, 2, 0]
    • arr1 = [0, 2, 2], arr2 = [2, 1, 0]
    • arr1 = [1, 2, 2], arr2 = [1, 1, 0]

So f[2] = [0, 1, 3, 0]

Final Answer: Sum of f[2][j] for j from 0 to nums[2] = 2: f[2][0] + f[2][1] + f[2][2] = 0 + 1 + 3 = 4

The 4 valid monotonic pairs are:

1. arr1 = [0, 1, 1], arr2 = [2, 2, 1]
2. arr1 = [0, 1, 2], arr2 = [2, 2, 0]
3. arr1 = [0, 2, 2], arr2 = [2, 1, 0]
4. arr1 = [1, 2, 2], arr2 = [1, 1, 0]

Solution Implementation

Time and Space Complexity

The time complexity is O(n × m), where n is the length of the array nums and m is the maximum value in the array nums.

• The outer loop runs n times (from index 1 to n-1)
• For each iteration i, we compute the prefix sum using accumulate(f[i-1]), which takes O(m) time
• The inner loop runs up to nums[i] + 1 times, which is at most m + 1 times
• Inside the inner loop, operations are O(1)
• Therefore, the total time complexity is O(n × m)

The space complexity is O(n × m).

• The 2D array f has dimensions n × (m + 1), requiring O(n × m) space
• The prefix sum array s created in each iteration has size O(m), but this doesn't increase the overall space complexity
• Therefore, the total space complexity is O(n × m)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Integer Overflow When Computing Prefix Sums

The most common pitfall in this solution is forgetting to apply the modulo operation during intermediate calculations, particularly when computing prefix sums. The accumulate function in Python doesn't automatically apply modulo, which can lead to integer overflow issues even though Python handles big integers well - the values can become unnecessarily large and slow down computation.

Problem Code:

# This can cause values to grow very large
prefix_sum = list(accumulate(dp[i - 1]))

Solution: Apply modulo during the accumulation process:

def accumulate_with_mod(arr, mod):
    result = []
    total = 0
    for val in arr:
        total = (total + val) % mod
        result.append(total)
    return result

# Use this instead
prefix_sum = accumulate_with_mod(dp[i - 1], MOD)

2. Index Out of Bounds When Calculating max_prev_value

Another pitfall is not properly handling edge cases where max_prev_value could exceed the bounds of the prefix sum array. If nums[i-1] - nums[i] is large, j + nums[i-1] - nums[i] might exceed the maximum index of the previous dp row.

Problem Scenario: When nums[i-1] > nums[i] by a large margin, the calculated max_prev_value could be larger than max_value.

Solution: Ensure proper bounds checking:

max_prev_value = min(j, j + nums[i - 1] - nums[i], max_value)
if max_prev_value >= 0 and max_prev_value < len(prefix_sum):
    dp[i][j] = prefix_sum[max_prev_value] % MOD

3. Incorrect Base Case Initialization

A subtle pitfall is incorrectly initializing the base case. Some might initialize only dp[0][0] = 1 thinking we must start with arr1[0] = 0, but actually any value from 0 to nums[0] is valid for the first position.

Wrong Initialization:

dp[0][0] = 1  # Only one value initialized

Correct Initialization:

for value in range(nums[0] + 1):
    dp[0][value] = 1  # All valid values from 0 to nums[0]

4. Final Answer Calculation Error

When calculating the final answer, it's easy to mistakenly sum the entire last row of the dp table instead of only summing valid entries up to nums[-1].

Wrong:

return sum(dp[-1]) % MOD  # Sums all entries including invalid ones

Correct:

return sum(dp[-1][:nums[-1] + 1]) % MOD  # Only sum valid entries

These pitfalls can lead to incorrect results or runtime errors. Always ensure proper modulo operations, bounds checking, and careful attention to the problem constraints when implementing dynamic programming solutions.