Problem Description

You are given an integer array nums and a non-negative integer k. A sequence of integers seq is called good if there are at most k indices i in the range [0, seq.length - 2] such that seq[i] != seq[i + 1].

The task is to return the maximum possible length of a good subsequence of nums.

Intuition

To solve this problem, we use a dynamic programming approach. We want to find the longest good subsequence from the given array, subject to the constraint that there are at most k indices where consecutive elements differ.

The key is to use a 2D array f where f[i][h] represents the length of the longest good subsequence ending at index i with no more than h allowed differences between consecutive elements. Initially, we set f[i][h] = 1, as the shortest subsequence can be a single element.

For each element nums[i], we check all previous elements nums[j] where j < i. If nums[i] equals nums[j], we can extend any subsequence ending at j without costing any additional allowed differences. Hence, we update f[i][h] = max(f[i][h], f[j][h] + 1).

If nums[i] differs from nums[j], we can only extend the subsequence if we have some allowed differences left, i.e., if h > 0. In this case, we update f[i][h] = max(f[i][h], f[j][h - 1] + 1).

Finally, the answer is the maximum length found across all possible subsequences that use up to k allowed differences, which is max(f[i][k]).

Learn more about Dynamic Programming patterns.

Solution Approach

The solution employs a dynamic programming approach to determine the maximum length of a good subsequence given the constraints. Here's how the solution is implemented:

1. Initialization:

   • We start by determining the length of the input array nums and defining a 2D list f where f[i][h] tracks the longest good subsequence ending at index i, with h indicating the number of changes allowed. Each entry in f is initialized to 1 since the minimum subsequence length can be just one element.

2. Iterating Through nums:

   • We iterate over each element x in the array nums using its index i.
   • For every element x, we consider subsequences ending at each prior element y indexed by j.

3. Updating the DP Table:

   • Same Element Extension: If x (current element) is equal to y (previous element we are considering), then we can extend the subsequence ending at j without using any of the allowed k changes. Hence, we update f[i][h] = max(f[i][h], f[j][h] + 1) for the same number of changes.

   • Different Element Extension: If x is different from y, we can still extend the subsequence if we have remaining allowed changes (h > 0). In this case, update f[i][h] = max(f[i][h], f[j][h - 1] + 1). This uses one of the allowed changes to accommodate the difference between x and y.

4. Finding the Result:

   • After processing all elements and possible previous elements for each, the answer will be the maximum value in f[i][k] for any 0 ≤ i < n, which represents the longest sequence possible with up to k allowed changes.

This approach efficiently calculates the maximum length of a good subsequence by leveraging dynamic programming to balance potential subsequences and changes, providing the correct solution by exploring all possibilities within constraints.

Example Walkthrough

Let's illustrate the solution approach with a small example. Consider the input array nums = [1, 2, 2, 3, 1] and k = 1.

Step 1: Initialization

• Create a 2D array f. Suppose nums has length n = 5. The array f has dimensions 5 x 2 (since h ranges from 0 to k, inclusive), initialized to 1:

  f = [
    [1, 1],
    [1, 1],
    [1, 1],
    [1, 1],
    [1, 1]
  ]

Step 2: Iterating Through nums

• For i = 1 (nums[1] = 2):

  • Consider j = 0 (nums[0] = 1):
    • Since 2 ≠ 1, we can extend the subsequence if h > 0:
      • When h = 1: Update f[1][1] = max(f[1][1], f[0][0] + 1) = max(1, 2) = 2.

• For i = 2 (nums[2] = 2):

  • Consider j = 0 and j = 1:
    • nums[2] == nums[1] (2 == 2), so extend without additional change:
      • f[2][0] = max(f[2][0], f[1][0] + 1) = max(1, 2) = 2.
      • f[2][1] = max(f[2][1], f[1][1] + 1) = max(1, 3) = 3.
    • Again, 2 ≠ 1, we can extend using a change:
      • f[2][1] = max(f[2][1], f[0][0] + 1) = max(3, 2) = 3.

• For i = 3 (nums[3] = 3):

  • Consider j = 0, j = 1, and j = 2:
    • Different value: 3 ≠ 2, extend with change h = 1:
      • Update f[3][1] = max(f[3][1], f[2][0] + 1) = max(1, 3) = 3.
      • Update f[3][1] = max(f[3][1], f[1][0] + 1) = max(3, 2) = 3.

• For i = 4 (nums[4] = 1):

  • Consider j = 0, j = 1, j = 2, j = 3:
    • For j = 0 (1 == 1): f[4][0] = max(f[4][0], f[0][0] + 1) = max(1, 2) = 2.
    • Possible with change h = 1 for j = 3: f[4][1] = max(f[4][1], f[3][0] + 1) = max(1, 2) = 2.

Step 3: Finding the Result

• The answer is the maximum value in any f[i][k]. Calculating, max(f[i][1]) for each i gives 3.

Thus, the maximum possible length of a good subsequence is 3.

Solution Implementation

Time and Space Complexity

• Time Complexity: The given code has a time complexity of O(n^2 * k). This is because there are three nested loops:

  • The outer loop iterates n times through the nums list.
  • The second loop iterates k + 1 times, where k is the parameter given to the function.
  • The innermost loop iterates over a sublist of nums that can have up to n elements, leading to a potential n iterations.

  Combining these, the overall time complexity becomes O(n^2 * k).

• Space Complexity: The space complexity of the algorithm is O(n * k). This results from the f matrix, which contains n rows and k + 1 columns, resulting in a storage need of O(n * (k + 1)), which simplifies to O(n * k).

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