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3196. Maximize Total Cost of Alternating Subarrays

MediumArrayDynamic Programming
Leetcode Link

Problem Description

You are given an integer array nums with length n.

The cost of a subarray nums[l..r], where 0 <= l <= r < n, is defined as:

cost(l, r) = nums[l] - nums[l + 1] + ... + nums[r] * (-1)^(r - l)

Your task is to split nums into subarrays such that the total cost of the subarrays is maximized, ensuring each element belongs to exactly one subarray.

Formally, if nums is split into k subarrays, where k > 1, at indices i1, i2, ..., i(k - 1), where 0 <= i1 < i2 < ... < i(k - 1) < n - 1, then the total cost will be:

cost(0, i1) + cost(i1 + 1, i2) + ... + cost(i(k - 1) + 1, n - 1)

Return an integer denoting the maximum total cost of the subarrays after splitting the array optimally.

Note: If nums is not split into subarrays, i.e. k = 1, the total cost is simply cost(0, n - 1).

Intuition

The idea is to split the array into subarrays to maximize total cost, where each element in the array can potentially start a new subarray. A subarray's cost is defined by alternating additions and subtractions based on the positional difference, i.e., whether the last element of the subarray should be subtracted if its position makes the number of elements in the subarray odd.

The aim is to decide optimally where these splits should happen. At each element in the array, you have the choice to end a subarray there or continue adding to the current subarray. The function dfs(i, j) helps achieve this. It calculates the maximum possible cost, given whether the current number at position i can be flipped or not, controlled by the boolean j.

  • If j = 0, the current number cannot be flipped, meaning it should be added directly.
  • If j = 1, it can either be used as is or be flipped. This choice is encapsulated in picking between nums[i] + dfs(i + 1, 1) or -nums[i] + dfs(i + 1, 0) and taking the maximum.

To efficiently compute the maximum cost without recalculating subproblems multiple times, memoization is utilized to store previously computed results.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution employs a recursive approach with memoization to maximize the total cost of subarrays formed by splitting the given array nums. The method used is dynamic programming, specifically leveraging the memoization technique to optimize recursive function calls.

The core idea is encapsulated in a helper function dfs(i, j), where:

  • i is the current index in the nums array from which we are deciding whether to start a new subarray or continue the current subarray.
  • j is a binary indicator (0 or 1) that represents whether the current number can be flipped. Specifically, j = 0 means the current number in the subarray cannot be flipped (it's been an end of some subarray's range), while j = 1 allows flipping.

The recursion strategy works as follows:

  1. Base Case: If i is greater than or equal to the length of nums, return 0. This indicates that we have processed all numbers in the array.

  2. Recursive Step:

    • If the i-th number is not flipped (j = 1), two paths are considered:
      • Add the number as it is and continue the recursion: nums[i] + dfs(i + 1, 1)
      • Flip the number and continue the recursion: -nums[i] + dfs(i + 1, 0)
    • If the i-th number cannot be flipped (j = 0), it can only be added as is: nums[i] + dfs(i + 1, 1)

  3. Maximization: Choose the path (either flipped or non-flipped) that yields the maximum cost for the subarray and store that result.

To enhance efficiency, memoization is leveraged through python's @cache decorator from the functools module, thus avoiding recalculation of overlapping subproblems.

The algorithm is initiated with a call to dfs(0, 0), meaning the first number at index 0 starts processing with no ability to be flipped initially. The result is the maximum cost achieved by optimally splitting and flipping the elements of nums.

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Example Walkthrough

Consider a small example where nums = [2, 3, 1, 5].

Step-by-step Explanation:

  1. Initial State:

    • Start with the entire array [2, 3, 1, 5].
    • Call dfs(0, 0) to begin processing with the first element.

  2. Processing dfs(0, 0):

    • Since j = 0, we cannot flip nums[0] = 2.
    • Calculate the cost for including 2 without flipping: 2 + dfs(1, 1).

  3. Processing dfs(1, 1):

    • Here, j = 1, indicating we have a choice to flip nums[1] = 3.
    • Consider two paths:
      • Don't flip: 3 + dfs(2, 1)
      • Flip: -3 + dfs(2, 0)

  4. Processing dfs(2, 1) and dfs(2, 0):

    • For dfs(2, 1):
      • Choose to either not flip: 1 + dfs(3, 1)
      • Or flip: -1 + dfs(3, 0)
    • For dfs(2, 0):
      • Must add as is: 1 + dfs(3, 1)

  5. Processing dfs(3, 1) and dfs(3, 0):

    • For dfs(3, 1):
      • Either don't flip 5, leading to end with 5 + dfs(4, 1)
      • Or flip 5: -5 + dfs(4, 0)
    • For dfs(3, 0):
      • Only option is to add 5: 5 + dfs(4, 1)

  6. Reach Base Case:

    • dfs(4, *): Since index 4 is out of bounds, return 0.

  7. Backtracking and Maximizing:

    • Work backwards using stored values to determine the maximum path cost.
    • Compute and compare paths based on the accumulated costs at each step.

Result: The maximum cost path found through this strategy will be returned.

This example illustrates the dfs and memoization approach, showing how decisions at each step and their impacts on the potential paths are navigated to maximize the total cost of subarrays.

Solution Implementation

1from functools import cache
2from typing import List
3
4class Solution:
5    def maximumTotalCost(self, nums: List[int]) -> int:
6        # Recursive function with memoization to calculate the maximum total cost
7        @cache
8        def dfs(i: int, j: int) -> int:
9            # Base case: if we've reached beyond the last index, return 0
10            if i >= len(nums):
11                return 0
12
13            # Option 1: Add current number and continue with next index, setting j to 1
14            ans = nums[i] + dfs(i + 1, 1)
15
16            # Option 2: If j is 1, we can also subtract the current number and continue
17            if j == 1:
18                ans = max(ans, -nums[i] + dfs(i + 1, 0))
19
20            # Return the maximum of the options
21            return ans
22
23        # Start the dfs function with the initial index and j value as 0
24        return dfs(0, 0)
25
1class Solution {
2    private Long[][] memo; // Memoization table for storing results of subproblems
3    private int[] nums; // Array of numbers
4    private int length; // Length of the nums array
5
6    // Method to calculate the maximum total cost
7    public long maximumTotalCost(int[] nums) {
8        this.length = nums.length; // Set the length of the input array
9        this.nums = nums; // Assign input array to instance variable
10        this.memo = new Long[length][2]; // Initialize memoization table
11        return dfs(0, 0); // Call dfs starting from the first element and j = 0
12    }
13
14    // Depth-first search method with memoization
15    private long dfs(int i, int j) {
16        if (i >= length) { // Base case: if index i is beyond the array length, return 0
17            return 0;
18        }
19        if (memo[i][j] != null) { // If result is already computed, return it
20            return memo[i][j];
21        }
22        // Choose to take nums[i] and move to the next element with flag 1
23        memo[i][j] = nums[i] + dfs(i + 1, 1);
24
25        // If j is 1, consider taking -nums[i] instead and move to the next element with flag 0
26        if (j == 1) {
27            // Calculate the maximum cost between taking nums[i] and taking -nums[i]
28            memo[i][j] = Math.max(memo[i][j], -nums[i] + dfs(i + 1, 0));
29        }
30        return memo[i][j]; // Return the calculated value for memoization
31    }
32}
33
1class Solution {
2public:
3    // Function to calculate the maximum total cost
4    long long maximumTotalCost(std::vector<int>& nums) {
5        int n = nums.size();
6      
7        // 2D array to store intermediate results for dynamic programming
8        long long dp[n][2];
9      
10        // Initialize the array with a very low value (negative infinity equivalent)
11        std::fill(&dp[0][0], &dp[0][0] + sizeof(dp) / sizeof(long long), LLONG_MIN);
12
13        // Helper function for the depth-first search
14        // This is a lambda function that recursively calculates the maximum total cost
15        auto dfs = [&](auto&& dfsRef, int i, int j) -> long long {
16            // Base case: if we have tried all elements, return 0
17            if (i >= n) {
18                return 0;
19            }
20            // Return the cached result if it's already computed
21            if (dp[i][j] != LLONG_MIN) {
22                return dp[i][j];
23            }
24            // Option 1: add the current number and move to the next element with the 'positive' flag
25            dp[i][j] = nums[i] + dfsRef(dfsRef, i + 1, 1);
26            // Option 2: subtract the current number (only if flag j allows) and move with 'negative' flag
27            if (j) {
28                dp[i][j] = std::max(dp[i][j], -nums[i] + dfsRef(dfsRef, i + 1, 0));
29            }
30            // Return the computed maximum cost for the current state
31            return dp[i][j];
32        };
33
34        // Start the recursive DFS from the first element with the 'negative' flag
35        return dfs(dfs, 0, 0);
36    }
37};
38
1// Function to find the maximum total cost given an array of numbers
2function maximumTotalCost(nums: number[]): number {
3    const n = nums.length;
4
5    // Create a memoization table initialized with negative infinity
6    const f: number[][] = Array.from({ length: n }, () => Array(2).fill(-Infinity));
7
8    // Helper function for depth-first search
9    const dfs = (i: number, canSubtract: number): number => {
10        if (i >= n) {
11            // Base case: if index is out of bounds, return 0
12            return 0;
13        }
14
15        // Return cached result if it already exists
16        if (f[i][canSubtract] !== -Infinity) {
17            return f[i][canSubtract];
18        }
19
20        // Calculate maximum total cost by adding the current number
21        f[i][canSubtract] = nums[i] + dfs(i + 1, 1);
22
23        // Calculate maximum total cost by subtracting the current number if allowed
24        if (canSubtract) {
25            f[i][canSubtract] = Math.max(f[i][canSubtract], -nums[i] + dfs(i + 1, 0));
26        }
27
28        // Store the result in the memoization table
29        return f[i][canSubtract];
30    };
31
32    // Start the depth-first search from the first index and without allowing subtraction
33    return dfs(0, 0);
34}
35

Time and Space Complexity

The time complexity of the code is O(n). The function dfs uses memoization via the @cache decorator, ensuring that each recursive call with a particular set of arguments (i, j) is computed only once. The function iterates over the elements in nums, and each state (i, j) is computed in constant time, leading to a total complexity proportional to the number of elements in nums, which is n.

The space complexity of the code is O(n). This arises from the recursion stack depth of O(n) and the additional space used by the cache to store the results of subproblems, which is also O(n).

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

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Question 1 out of 10

Given a sorted array of integers and an integer called target, find the element that equals to the target and return its index. Select the correct code that fills the ___ in the given code snippet.

1def binary_search(arr, target):
2    left, right = 0, len(arr) - 1
3    while left ___ right:
4        mid = (left + right) // 2
5        if arr[mid] == target:
6            return mid
7        if arr[mid] < target:
8            ___ = mid + 1
9        else:
10            ___ = mid - 1
11    return -1
12
1public static int binarySearch(int[] arr, int target) {
2    int left = 0;
3    int right = arr.length - 1;
4
5    while (left ___ right) {
6        int mid = left + (right - left) / 2;
7        if (arr[mid] == target) return mid;
8        if (arr[mid] < target) {
9            ___ = mid + 1;
10        } else {
11            ___ = mid - 1;
12        }
13    }
14    return -1;
15}
16
1function binarySearch(arr, target) {
2    let left = 0;
3    let right = arr.length - 1;
4
5    while (left ___ right) {
6        let mid = left + Math.trunc((right - left) / 2);
7        if (arr[mid] == target) return mid;
8        if (arr[mid] < target) {
9            ___ = mid + 1;
10        } else {
11            ___ = mid - 1;
12        }
13    }
14    return -1;
15}
16

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