What is the time complexity of Maximum Sum of Subsequence With Non-adjacent Elements?

The optimal solution for Maximum Sum of Subsequence With Non-adjacent Elements runs in Unknown time and uses Unknown space.

What algorithm or data structure is used to solve Maximum Sum of Subsequence With Non-adjacent Elements?

Maximum Sum of Subsequence With Non-adjacent Elements uses the following concepts: Array, Divide and Conquer, Dynamic Programming, Segment Tree. The recommended patterns to study are: .

Maximum Sum of Subsequence With Non-adjacent Elements — LeetCode #3165 (Hard)

Tags: Array, Divide and Conquer, Dynamic Programming, Segment Tree

Related patterns:

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

The brute force approach involves generating all possible subsequences of the array and calculating their sums while ensuring no two adjacent elements are selected. This is simple but inefficient as it checks every combination.

The time complexity is O(n²) because we generate all subsequences, which takes exponential time, and checking each subsequence for adjacency is linear.

Step 1: For each query, update the specified position in the nums array.
Step 2: Generate all possible subsequences of the updated nums array.
Step 3: For each subsequence, check if it contains adjacent elements and calculate its sum.
Step 4: Keep track of the maximum sum found that adheres to the non-adjacency condition.
Step 5: Return the maximum sum for the query.

1. Start with nums = [3, 5, 9]. 2. First query: Update nums[1] to -2, nums = [3, -2, 9]. 3. Check subsequences: max sum = 12 (3 + 9). 4. Second query: Update nums[0] to -3, nums = [-3, -2, 9]. 5. Check subsequences: max sum = 9. 6. Total sum = 12 + 9 = 21.

Optimal Solution approach

Time complexity: Unknown. Space complexity: Unknown.

The optimal approach uses dynamic programming to calculate the maximum sum of non-adjacent elements efficiently. We maintain a running total of the maximum sums while updating the array based on queries.

Step 1: Initialize two variables, prev1 and prev2, to keep track of the maximum sums including and excluding the current element.
Step 2: For each query, update the specified position in the nums array.
Step 3: Iterate through the updated nums array, updating prev1 and prev2 based on the current element's value.
Step 4: After processing the entire array, the maximum sum is stored in prev1.
Step 5: Return the sum of maximum sums for all queries.

#3165

Maximum Sum of Subsequence With Non-adjacent Elements

Hard

Array↗Divide and Conquer↗Dynamic Programming↗Segment Tree↗

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	Unknown
Space	O(1)	Unknown

💡

Intuition

Time UnknownSpace Unknown

⚙️

Algorithm

5 steps

1Step 1: Initialize two variables, prev1 and prev2, to keep track of the maximum sums including and excluding the current element.
2Step 2: For each query, update the specified position in the nums array.
3Step 3: Iterate through the updated nums array, updating prev1 and prev2 based on the current element's value.
4Step 4: After processing the entire array, the maximum sum is stored in prev1.
5Step 5: Return the sum of maximum sums for all queries.

solution.py14 lines

1# Full working Python code
2MOD = 10**9 + 7
3
4def max_sum_subsequence(nums, queries):
5    total_sum = 0
6    for pos, x in queries:
7        nums[pos] = x
8        prev1, prev2 = 0, 0
9        for num in nums:
10            new_prev1 = max(prev1, prev2 + num)
11            prev2 = prev1
12            prev1 = new_prev1
13        total_sum = (total_sum + prev1) % MOD
14    return total_sum

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