Problem Description

The task is to find the maximum value from a triplet in an array of integers. A triplet (i, j, k) is defined by three conditions:

1. The indices i, j, and k follow an increasing sequence (i < j < k).
2. The values at those indices in the array also follow an increasing sequence (nums[i] < nums[j] < nums[k]).
3. The value of the triplet is calculated using the formula nums[i] - nums[j] + nums[k].

One needs to return the highest possible value from any triplet in the array that satisfies the conditions above.

Intuition

To find the maximum value of a triplet (i, j, k), we need to consider two main aspects:

1. Maximizing nums[i] with the condition nums[i] < nums[j].
2. Maximizing nums[k] with the condition nums[k] > nums[j].

To get the largest possible value for the triplet, we aim to pick the largest nums[i] that is less than nums[j] and the largest nums[k] that is greater than nums[j]. Since nums[j] is the constant middle value in our triplet calculation (nums[i] - nums[j] + nums[k]), we can traverse the array and keep track of possible nums[i] and nums[k] for each j.

To approach the solution, we'll implement two main strategies:

1. We prepare a suffix array right such that right[i] holds the maximum value to the right of i, including i. This helps us quickly fetch the maximum k value for any given j.

2. We use an ordered set to keep track of all the values to the left of j encountered so far. This set will help us efficiently query for the largest value less than nums[j] which serves as a candidate for nums[i].

The solution is a blend of dynamic programming (with the use of the right array) and binary search (using the ordered set to quickly find the largest element smaller than nums[j]). We loop through the array only once and, at each step, take logarithmic time to update our ordered set and find the needed value, leading to an efficient algorithm for the problem.

Solution Approach

In the given solution, we make use of three primary components:

1. A suffix array named right
2. An ordered set from the sortedcontainers library
3. A single pass loop to consider each nums[j]

Suffix Array right

The right array is a dynamic programming technique. We start from the end of the array and move towards the start, filling right[i] with the maximum value encountered from index i to the end of the array. This ensures that for any index j, right[j+1] will provide the maximum value available for nums[k] where k > j.

Here is the implementation:

right = [nums[-1]] * n
for i in range(n - 2, -1, -1):
    right[i] = max(nums[i], right[i + 1])

We initialize the right array to be of the same length as nums, filled with the last element of nums. We then traverse the array from the second-to-last element to the first, updating each right[i] to be the maximum of nums[i] and right[i+1].

Ordered Set

The ordered set from the sortedcontainers library acts similar to a balanced binary search tree. It maintains the sorted order of elements and provides efficient operations to add elements and to search for an element by its order.

Here's how it's used:

sl = SortedList([nums[0]])
for j in range(1, n - 1):
    if right[j + 1] > nums[j]:
        i = sl.bisect_left(nums[j]) - 1
        if i >= 0:
            ans = max(ans, sl[i] - nums[j] + right[j + 1])
    sl.add(nums[j])

We initialize our ordered set sl with the first element of nums and start our one pass through the array, considering each j from 1 to n - 2. For each j, we check if there's a valid k by confirming that right[j + 1] (the largest nums[k] after j) is greater than nums[j]. If so, we use sl.bisect_left(nums[j]) which gives us an index one greater than the index of the largest number less than nums[j] in sl. We then decrease it by 1 to get the correct index for nums[i]. If such an index exists (i >= 0), we calculate the value of the current triplet and update our answer ans if it's greater than the previous maximum.

The ordered set guarantees logarithmic time complexity for both the addition and the bisect_left operation.

With the combination of the suffix array right and the ordered set sl, we're able to find the maximum triplet value efficiently as we only make one pass through the array and maintain a set of potential i values for each j, from which we can quickly determine the maximum possible value of nums[i].

Example Walkthrough

Let's assume our given array of integers is nums = [3, 5, 1, 6, 4]. We want to find the maximum value of a triplet (i, j, k) such that i < j < k and nums[i] < nums[j] < nums[k] using the value calculated by nums[i] - nums[j] + nums[k].

We initialize our three primary components:

1. The suffix array right is created to store the maximum values to the right of each index, including the index itself.
2. An ordered set sl from the sortedcontainers library.
3. An answer variable ans initialized to minus infinity to store the maximum value of the triplets found.

Let's walk through the implementation step by step with our array nums = [3, 5, 1, 6, 4]:

Step 1: Create the Suffix Array

We start by populating our right array with the maximum value to the right of each index in nums.

right = [4, 4, 6, 6, 4]  # Initialized with the max to the right of each index

The right array would be filled as follows:

• Start from nums[4] = 4, so right[4] is 4.
• Check max(nums[3], right[4]), which is max(6, 4) = 6, so right[3] is 6.
• Following the pattern, right[2] is 6, right[1] is 6, and right[0] is 6.

We fill right from nums[-1] to nums[0] by taking the maximum of the current element and the next element in right.

Step 2: Initialize the Ordered Set and Iterate

Next, we initialize our ordered set sl with the first element of nums.

sl = SortedList([3])
ans = float('-inf')  # This represents negative infinity to start with no valid max.

Step 3: Iterate through nums

Now, we iterate from 1 to len(nums) - 2 because these are the potential j positions for a valid triplet.

For j = 1, nums[j] = 5.

• We check right[j + 1] which is 6, and since 6 > 5, we proceed.
• Find the largest value less than 5 in sl using sl.bisect_left(5) - 1 which gives 0.
• The element at 0 in sl is 3, so we calculate the triplet value 3 - 5 + 6 = 4.
• Since ans is -inf, we update ans with 4.
• Add nums[j] to sl.

Continue this process for j = 2, 3, and by the end of this single pass, we have considered all valid triplets.

Assuming we update ans appropriately through each iteration, at the end of this process, ans would hold the maximum triplet value. With the nums array given, the maximum triplet value we compute would be 1 - 5 + 6 = 2.

With this efficient approach utilizing the right array for the maximum k value and the ordered set for the maximum i value corresponding to each j, we can determine the highest possible value from any triplet following the conditions.

This walk-through demonstrates the solution approach on a small example and how it translates into finding the desired maximum value of the triplet in an array of integers.

Solution Implementation

Time and Space Complexity

The time complexity of the given code can be analyzed by looking at each segment of the code:

1. Building the right array: The array is created by iterating over the nums array from right to left once, which has a complexity of O(n).

2. The main for-loop runs from j = 1 to n - 1: Within the loop, the following operations occur:

   • Checking if right[j + 1] > nums[j]: This is an O(1) operation.
   • Executing sl.bisect_left(nums[j]): The SortedList uses a binary search technique to find the leftmost insertion point which takes O(log n) time.
   • Conditionally updating ans based on the i index: This is an O(1) operation.
   • Finally, sl.add(nums[j]): Adding an element to the SortedList which maintains a sorted order takes O(log n) time.

Considering the loop runs (n - 2) times and the most expensive operations inside are O(log n), the complexity of the loop becomes O(n * log n).

Combining the complexity of building the right array and the main loop, the overall time complexity remains O(n * log n) as the loop's complexity dominates.

The space complexity analysis includes:

1. The right array: As it stores n elements, it has a complexity of O(n).

2. The SortedList sl: In the worst case, this will hold n elements, resulting in a space complexity of O(n).

Hence, the overall space complexity is also O(n) since these two data structures are the dominant factors in space usage, and their complexities are additive.

In conclusion, the code exhibits a time complexity of O(n * log n) and a space complexity of O(n).

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