Solution Approach

The solution uses dynamic programming with memoization to find the minimum cost. Here's how the implementation works:

Main Function Structure: We define a recursive function dfs(i) that calculates the minimum cost to split the array starting from index i. The answer to our problem is dfs(0).

Base Case: When i >= n (reached or passed the end of array), we return 0 since there's nothing left to split.

Recursive Case: For each starting position i, we try all possible ending positions j from i to n-1:

1. Initialize tracking variables:

   • cnt = Counter(): A hash table to track frequency of each number in current subarray
   • one = 0: Count of numbers appearing exactly once
   • ans = inf: Store the minimum cost found

2. Extend the subarray: For each j from i to n-1:

   • Add nums[j] to our frequency counter: cnt[nums[j]] += 1
   • Update the one counter:
     • If cnt[nums[j]] == 1: A new unique number, so one += 1
     • If cnt[nums[j]] == 2: A number is no longer unique, so one -= 1

3. Calculate cost for this split:

   • Length of current subarray: j - i + 1
   • Length of trimmed subarray: (j - i + 1) - one
   • Importance value: k + (j - i + 1) - one
   • Total cost: k + j - i + 1 - one + dfs(j + 1)

4. Track minimum: Update ans = min(ans, current_cost)

Memoization: The @cache decorator automatically memoizes the results of dfs(i). This prevents recalculating the same subproblem multiple times, reducing time complexity from exponential to O(n²).

Time Complexity: O(n²) - We have n states for dfs(i), and for each state, we iterate through at most n positions.

Space Complexity: O(n) - For the recursion stack and memoization cache.

Example Walkthrough

Let's walk through the solution with nums = [1,2,1,2,1] and k = 2.

We start with dfs(0), which needs to find the minimum cost to split the entire array starting from index 0.

From index 0, we try different ending positions:

Option 1: Subarray [0,0] = [1]

• cnt = {1: 1}, one = 1 (number 1 appears once)
• Trimmed length = 1 - 1 = 0
• Importance = k + 0 = 2 + 0 = 2
• Total cost = 2 + dfs(1)

Option 2: Subarray [0,1] = [1,2]

• Start with cnt = {1: 1}, one = 1
• Add nums[1]=2: cnt = {1: 1, 2: 1}, one = 2 (both appear once)
• Trimmed length = 2 - 2 = 0
• Importance = k + 0 = 2 + 0 = 2
• Total cost = 2 + dfs(2)

Option 3: Subarray [0,2] = [1,2,1]

• Start with cnt = {1: 1, 2: 1}, one = 2
• Add nums[2]=1: cnt = {1: 2, 2: 1}, one = 1 (only 2 appears once now)
• Trimmed length = 3 - 1 = 2
• Importance = k + 2 = 2 + 2 = 4
• Total cost = 4 + dfs(3)

Option 4: Subarray [0,3] = [1,2,1,2]

• Continue from previous: add nums[3]=2
• cnt = {1: 2, 2: 2}, one = 0 (no numbers appear once)
• Trimmed length = 4 - 0 = 4
• Importance = k + 4 = 2 + 4 = 6
• Total cost = 6 + dfs(4)

Option 5: Subarray [0,4] = [1,2,1,2,1]

• Continue: add nums[4]=1
• cnt = {1: 3, 2: 2}, one = 0 (still no numbers appear once)
• Trimmed length = 5 - 0 = 5
• Importance = k + 5 = 2 + 5 = 7
• Total cost = 7 + dfs(5) = 7 + 0 = 7

Now let's evaluate dfs(2) (needed for Option 2):

• Subarray [2,2] = [1]: importance = 2, total = 2 + dfs(3)
• Subarray [2,3] = [1,2]: importance = 2, total = 2 + dfs(4)
• Subarray [2,4] = [1,2,1]: importance = 4, total = 4 + 0 = 4

And dfs(3):

• Subarray [3,3] = [2]: importance = 2, total = 2 + dfs(4)
• Subarray [3,4] = [2,1]: importance = 2, total = 2 + 0 = 2

And dfs(4):

• Subarray [4,4] = [1]: importance = 2, total = 2 + 0 = 2

Working backwards with memoization:

• dfs(4) = 2
• dfs(3) = min(2 + 2, 2) = 2
• dfs(2) = min(2 + 2, 2 + 2, 4) = 4
• dfs(1) = min(2 + 4, 2 + 2, 4 + 0, 5) = 4
• dfs(0) = min(2 + 4, 2 + 4, 4 + 2, 6 + 2, 7) = 6

The minimum cost is 6, achieved by splitting as [1] [2,1,2,1] or [1,2] [1,2,1].

Time and Space Complexity

Time Complexity: O(n^2)

The algorithm uses dynamic programming with memoization. The dfs(i) function is called for each possible starting position from 0 to n-1, resulting in at most n unique states. For each state dfs(i), the function iterates through all positions from i to n-1, performing O(n-i) operations. In the worst case, this gives us:

• State 0: iterates through positions 0 to n-1 → n operations
• State 1: iterates through positions 1 to n-1 → n-1 operations
• ...
• State n-1: iterates through position n-1 → 1 operation

The total operations sum up to n + (n-1) + ... + 1 = n(n+1)/2 = O(n^2).

Within each iteration, operations like updating the counter and checking conditions take O(1) time, so they don't affect the overall complexity.

Space Complexity: O(n)

The space complexity consists of:

1. Recursion stack: In the worst case, the recursion depth can be O(n) when each subarray contains only one element.
2. Memoization cache: The @cache decorator stores results for at most n different states (one for each starting position i).
3. Counter object: Within each recursive call, a Counter is used which can store at most O(n) distinct elements in the worst case, but only one Counter exists at any point in the recursion path.

The dominant factor is the recursion stack depth plus the memoization storage, both of which are O(n), resulting in an overall space complexity of O(n).

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

Common Pitfalls

1. Incorrect Unique Element Count Update Logic

The Pitfall: A common mistake is incorrectly updating the unique_count when elements transition between different frequency states. Developers often forget to handle the case when an element's frequency increases beyond 2, leading to incorrect trimmed subarray calculations.

Incorrect Implementation:

# WRONG: Only considering transitions between 1 and 2 if frequency_map[current_element] == 1:  unique_count += 1 elif frequency_map[current_element] == 2:  unique_count -= 1 # Missing: What if frequency goes from 2 to 3, 3 to 4, etc.?

Why It's Wrong: When an element appears 3 or more times, the unique count shouldn't change, but some might mistakenly try to update it further or reset it.

Correct Solution: The provided code is actually correct - it only updates unique_count when:

• Frequency changes from 0→1 (element becomes unique): increment
• Frequency changes from 1→2 (element no longer unique): decrement
• Frequency changes from 2→3+ (no change needed): do nothing

2. Off-by-One Error in Importance Value Calculation

The Pitfall: Miscalculating the importance value by confusing the trimmed length formula.

Incorrect Calculation:

# WRONG: Subtracting unique count from k instead of subarray length importance_score = k - unique_count current_cost = importance_score + dp(end_idx + 1)

Correct Solution:

# CORRECT: Importance = k + (trimmed_length) # Where trimmed_length = total_length - unique_elements subarray_length = end_idx - start_idx + 1 trimmed_length = subarray_length - unique_count importance_score = k + trimmed_length current_cost = importance_score + dp(end_idx + 1)

3. Using Global Counter Instead of Local

The Pitfall: Using a single Counter object across all recursive calls, causing frequency counts to bleed between different subarray considerations.

Incorrect Implementation:

class Solution:  def minCost(self, nums: List[int], k: int) -> int:  frequency_map = Counter() # WRONG: Global counter   @cache  def dp(start_idx):  # Using the global frequency_map  for end_idx in range(start_idx, n):  frequency_map[nums[end_idx]] += 1 # Modifies global state  # ... rest of logic

Why It's Wrong: This would accumulate counts across different recursive paths, leading to incorrect frequency calculations.

Correct Solution: Create a new Counter for each call to dp():

def dp(start_idx):  frequency_map = Counter() # Local to this function call  unique_count = 0  # ... rest of logic

4. Forgetting to Reset or Clear Cache Between Test Cases

The Pitfall: When testing multiple cases or submitting to an online judge, the @cache decorator might retain values from previous test cases if the function is defined at class level.

Solution: Either:

1. Define the cached function inside the main method (as shown in the correct code)
2. Explicitly clear the cache if needed: dp.cache_clear()
3. Use manual memoization with a dictionary that's initialized within the method

The provided solution correctly handles this by defining the dp function inside the minCost method, ensuring a fresh cache for each problem instance.