Problem Description

You are given three integers n, m, and k. An array arr is called k-even if there are exactly k indices such that, for each of these indices i (where (0 \leq i < n - 1)), the expression ((\text{arr}[i] \times \text{arr}[i + 1]) - \text{arr}[i] - \text{arr}[i + 1]) is even. The task is to return the number of possible k-even arrays of size n where all elements are within the range ([1, m]). Since the answer may be very large, return it modulo (10^9 + 7).

Intuition

To solve this problem, we need to determine the number of ways to fill an array of size n with numbers between 1 and m such that exactly k contiguous pairs within the array satisfy the given even condition.

Key Observations:

1. Parity of Expressions: The expression ((\text{arr}[i] \times \text{arr}[i + 1]) - \text{arr}[i] - \text{arr}[i + 1]) becomes even if the product (\text{arr}[i] \times \text{arr}[i + 1]) is odd, which happens only when both numbers are odd.

2. Counting Parities: There are (\text{cnt0} = \left\lfloor \frac{m}{2} \right\rfloor) even numbers and (\text{cnt1} = m - \text{cnt0}) odd numbers within the range [1, m].

3. Recursive Approach: We define a recursive function, dfs(i, j, k), representing the number of valid ways to fill up to the (i^{th}) position of the array with j needed pairs left to satisfy the even condition. The third parameter k indicates the parity of the last added element (0 for even and 1 for odd).

Recursive Steps:

• If j < 0, the configuration is invalid, so return 0.
• If i >= n, check if all needed pairs j are used up; if j == 0, it is a valid configuration, return 1; otherwise, return 0.
• Depending on whether the last number added was odd or even, calculate the possibilities of adding another odd or even number and continue.

This approach is efficiently implemented using memoization to store and reuse previously computed results, significantly optimizing the recursive calculations. The modulo operation ensures the result's constraints are satisfied.

Learn more about Dynamic Programming patterns.

Solution Approach

The problem is tackled using a recursive approach with memoization, which efficiently explores all possible configurations of the array while minimizing redundant calculations. Here's a detailed explanation of the solution implementation:

Recursive Function Design:

• Function Definition: We define dfs(i, j, k), where:

  • i is the current index in the array we are trying to fill.
  • j is the number of remaining pair indices that need to satisfy the even condition.
  • k indicates the parity of the last element added (0 for even and 1 for odd).

• Base Cases:

  • If j < 0, we've exceeded the required number of valid pairs, and thus return 0.
  • If i >= n, we've filled the array. If j equals 0, indicating all required pairs are valid, return 1; otherwise, return 0.

• Recursive Logic:

  • Consider the current number being added, checking if it's odd or even (k).
  • If the last element added is odd (k == 1), calculate the possible ways to continue with the next element being odd or even:
    • Add an odd: cnt1 * dfs(i + 1, j, 1) (where cnt1 is the count of odd numbers).
    • Add an even: cnt0 * dfs(i + 1, j - 1, 0) (where cnt0 is the count of even numbers and j-1 since a valid pair is formed).
  • Combine these possibilities and take the sum modulo (10^9 + 7).

Pre-computation:

• Calculate the number of odd and even numbers in the range ([1, m]) as cnt1 and cnt0 respectively.

• Modulo Operation: The operations are performed under modulo (10^9 + 7) to ensure numbers remain manageable and satisfy output constraints.

The algorithm efficiently combines recursive exploration with the use of a cached dfs utility to handle overlapping subproblems, thereby achieving optimal time complexity. The approach leverages depth-first search with memoization, enabling it to handle larger inputs within the given constraints.

Example Walkthrough

Let's walk through a simple example to better understand the solution approach.

Example:

Suppose we have n = 3, m = 3, and k = 2. We want to find how many 3-element arrays satisfy exactly 2 pairs with the given even condition.

Step-by-step Walkthrough:

1. Identify Odd and Even Numbers:

   • Given m = 3, the numbers [1, 2, 3] are possible.
   • Odd numbers: [1, 3] (count cnt1 = 2)
   • Even numbers: [2] (count cnt0 = 1)

2. Understand the Condition:

   • The expression becomes even if both numbers in the pair are odd.
   • We need exactly 2 such pairs formed from our array of size 3.

3. Define Recursive Steps:

   Use a function dfs(i, j, k):

   • i: Current index in the array.
   • j: Remaining pairs needed to satisfy the even condition.
   • k: Parity of the last number added (0 for even, 1 for odd).

4. Recursive Exploration Example:

   • Start from the first position: dfs(0, 2, -1) (begin with no parity set).

   • First Element:

     • Can be 1 (odd): dfs(1, 2, 1)
     • Can be 2 (even): dfs(1, 2, 0)
     • Can be 3 (odd): dfs(1, 2, 1)

   • Second Element (if first was odd):

     • If first was 1, and add 1: dfs(2, 2, 1) (No valid pair formed)
     • If first was 1, and add 2: dfs(2, 1, 0) (Valid pair: (1, 2) formed)
     • If first was 1, and add 3: dfs(2, 2, 1) (No valid pair formed)
     • Repeat for other possibilities similarly.

   • Check Base Cases:

     • If we reach i = n and j == 0, it means all k pairs are formed.

5. Final Calculation:

   • Summing up all valid configurations, apply modulo (10^9 + 7) to keep the result manageable.

This method explores all configurations recursively with state (index, remaining pairs, last parity), effectively utilizing memoization to store sub-problems results, thus optimizing the solution.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(n * k), since the depth-first search with memoization (dfs) is called for each of the n indices and up to k distinct values of j, making the number of dfs calls proportional to n * k. The memoization ensures that each unique state (i, j, k) is computed at most once.

The space complexity is O(n * k), which accounts for the space required to store the results of the dfs calls in the cache. This cache stores results for each state defined by the parameters i and j, leading to a maximum of n * k storage requirements.

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