Introduction

In this blog post, we will explore how to solve the LeetCode problem "3080" 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 a 0-indexed array nums of size n consisting of positive integers. You are also given a 2D array queries of size m where queries[i] = [indexi, ki]. Initially all elements of the array are unmarked. You need to apply m queries on the array in order, where on the ith query you do the following: Mark the element at index indexi if it is not already marked. Then mark ki unmarked elements in the array with the smallest values. If multiple such elements exist, mark the ones with the smallest indices. And if less than ki unmarked elements exist, then mark all of them. Return an array answer of size m where answer[i] is the sum of unmarked elements in the array after the ith query. Example 1: Input: nums = [1,2,2,1,2,3,1], queries = [[1,2],[3,3],[4,2]] Output: [8,3,0] Explanation: We do the following queries on the array: Mark the element at index 1, and 2 of the smallest unmarked elements with the smallest indices if they exist, the marked elements now are nums = [1,2,2,1,2,3,1]. The sum of unmarked elements is 2 + 2 + 3 + 1 = 8. Mark the element at index 3, since it is already marked we skip it. Then we mark 3 of the smallest unmarked elements with the smallest indices, the marked elements now are nums = [1,2,2,1,2,3,1]. The sum of unmarked elements is 3. Mark the element at index 4, since it is already marked we skip it. Then we mark 2 of the smallest unmarked elements with the smallest indices if they exist, the marked elements now are nums = [1,2,2,1,2,3,1]. The sum of unmarked elements is 0. Example 2: Input: nums = [1,4,2,3], queries = [[0,1]] Output: [7] Explanation: We do one query which is mark the element at index 0 and mark the smallest element among unmarked elements. The marked elements will be nums = [1,4,2,3], and the sum of unmarked elements is 4 + 3 = 7. Constraints: n == nums.length m == queries.length 1 <= m <= n <= 105 1 <= nums[i] <= 105 queries[i].length == 2 0 <= indexi, ki <= n - 1

Explanation

Here's a breakdown of the approach, complexity, and the Python code:

• High-Level Approach:

  • Use a boolean array marked to keep track of which elements are marked.
  • For each query, first mark the element at indexi if it's not already marked.
  • Then, find the ki smallest unmarked elements and mark them. Calculate and store the sum of unmarked elements after each query.

• Complexity:

  • Runtime Complexity: O(n log n + m * n)
  • Storage Complexity: O(n)

• Code:

def solve():
    nums = [1,2,2,1,2,3,1]
    queries = [[1,2],[3,3],[4,2]]
    # nums = [1,4,2,3]
    # queries = [[0,1]]

    n = len(nums)
    m = len(queries)
    marked = [False] * n
    answer = []

    for indexi, ki in queries:
        if not marked[indexi]:
            marked[indexi] = True

        unmarked_indices = [i for i in range(n) if not marked[i]]

        # Find ki smallest unmarked elements and mark them
        smallest_unmarked = []
        for i in unmarked_indices:
            smallest_unmarked.append((nums[i], i))

        smallest_unmarked.sort()

        count = 0
        for val, idx in smallest_unmarked:
            if count < ki:
                if not marked[idx]:
                    marked[idx] = True
                    count += 1
            else:
                break


        # Calculate the sum of unmarked elements
        unmarked_sum = 0
        for i in range(n):
            if not marked[i]:
                unmarked_sum += nums[i]

        answer.append(unmarked_sum)

    return answer

def solve_efficient():
    nums = [1,2,2,1,2,3,1]
    queries = [[1,2],[3,3],[4,2]]
    # nums = [1,4,2,3]
    # queries = [[0,1]]

    n = len(nums)
    m = len(queries)
    marked = [False] * n
    answer = []

    for indexi, ki in queries:
        if not marked[indexi]:
            marked[indexi] = True

        unmarked_values_with_indices = []
        for i in range(n):
            if not marked[i]:
                unmarked_values_with_indices.append((nums[i], i))

        unmarked_values_with_indices.sort()

        marked_count = 0
        for value, index in unmarked_values_with_indices:
            if marked_count < ki:
                marked[index] = True
                marked_count += 1
            else:
                break

        unmarked_sum = 0
        for i in range(n):
            if not marked[i]:
                unmarked_sum += nums[i]
        answer.append(unmarked_sum)
    return answer


    # Code
    ```python
    def solve_optimal():    nums = [1,2,2,1,2,3,1]
    queries = [[1,2],[3,3],[4,2]]
    # nums = [1,4,2,3]
    # queries = [[0,1]]

    n = len(nums)
    m = len(queries)
    marked = [False] * n
    answer = []

    for indexi, ki in queries:
        if not marked[indexi]:
            marked[indexi] = True

        unmarked = []
        for i in range(n):
            if not marked[i]:
                unmarked.append((nums[i], i))

        unmarked.sort()

        count = 0
        for val, idx in unmarked:
            if count < ki:
                marked[idx] = True
                count += 1
            else:
                break

        unmarked_sum = 0
        for i in range(n):
            if not marked[i]:
                unmarked_sum += nums[i]
        answer.append(unmarked_sum)

    return answer