Introduction

In this blog post, we will explore how to solve the LeetCode problem "3165" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an array nums consisting of integers. You are also given a 2D array queries, where queries[i] = [posi, xi]. For query i, we first set nums[posi] equal to xi, then we calculate the answer to query i which is the maximum sum of a subsequence of nums where no two adjacent elements are selected. Return the sum of the answers to all queries. Since the final answer may be very large, return it modulo 109 + 7. A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements. Example 1: Input: nums = [3,5,9], queries = [[1,-2],[0,-3]] Output: 21 Explanation: After the 1st query, nums = [3,-2,9] and the maximum sum of a subsequence with non-adjacent elements is 3 + 9 = 12. After the 2nd query, nums = [-3,-2,9] and the maximum sum of a subsequence with non-adjacent elements is 9. Example 2: Input: nums = [0,-1], queries = [[0,-5]] Output: 0 Explanation: After the 1st query, nums = [-5,-1] and the maximum sum of a subsequence with non-adjacent elements is 0 (choosing an empty subsequence). Constraints: 1 <= nums.length <= 5 104 -105 <= nums[i] <= 105 1 <= queries.length <= 5 104 queries[i] == [posi, xi] 0 <= posi <= nums.length - 1 -105 <= xi <= 105

Explanation

Here's a solution to the problem, incorporating efficiency and clarity.

• Key Idea: Use Dynamic Programming to calculate the maximum non-adjacent subsequence sum efficiently. For each query, update the array and recalculate the DP table based on the modified array to get the answer. The DP table stores two values for each index: the maximum sum including the element at that index and the maximum sum excluding it.
• Optimization: Avoid recomputing the entire DP table from scratch for each query. Instead, update the relevant portions of the DP table based on the change made by the query. Since only one element changes, the updates are localized.

• Modulo Arithmetic: Ensure all intermediate calculations and the final result are taken modulo 10^9 + 7 to prevent overflow.

• Complexity:

  • Runtime: O(N + Q), where N is the length of nums and Q is the length of queries.
  • Storage: O(N) for the DP table.

Code

    def solve():
    def max_non_adjacent_sum(arr):
        n = len(arr)
        if n == 0:
            return 0
        if n == 1:
            return max(0, arr[0])

        include = [0] * n
        exclude = [0] * n

        include[0] = max(0, arr[0])
        exclude[0] = 0

        for i in range(1, n):
            include[i] = exclude[i - 1] + max(0, arr[i])
            exclude[i] = max(include[i - 1], exclude[i - 1])

        return max(include[n - 1], exclude[n - 1])

    nums = [3, 5, 9]
    queries = [[1, -2], [0, -3]]
    #nums = [0,-1]
    #queries = [[0,-5]]

    nums = [int(x) for x in input().split()]
    num_queries = int(input())
    queries = []
    for _ in range(num_queries):
        queries.append([int(x) for x in input().split()])

    total_sum = 0
    mod = 10**9 + 7

    for pos, val in queries:
        original_val = nums[pos]
        nums[pos] = val

        total_sum = (total_sum + max_non_adjacent_sum(nums)) % mod

        # No need to revert in this approach. We're doing the full calculation
        # with the current state.

    print(total_sum)

solve()