Introduction

In this blog post, we will explore how to solve the LeetCode problem "3478" 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 two integer arrays, nums1 and nums2, both of length n, along with a positive integer k. For each index i from 0 to n - 1, perform the following: Find all indices j where nums1[j] is less than nums1[i]. Choose at most k values of nums2[j] at these indices to maximize the total sum. Return an array answer of size n, where answer[i] represents the result for the corresponding index i. Example 1: Input: nums1 = [4,2,1,5,3], nums2 = [10,20,30,40,50], k = 2 Output: [80,30,0,80,50] Explanation: For i = 0: Select the 2 largest values from nums2 at indices [1, 2, 4] where nums1[j] < nums1[0], resulting in 50 + 30 = 80. For i = 1: Select the 2 largest values from nums2 at index [2] where nums1[j] < nums1[1], resulting in 30. For i = 2: No indices satisfy nums1[j] < nums1[2], resulting in 0. For i = 3: Select the 2 largest values from nums2 at indices [0, 1, 2, 4] where nums1[j] < nums1[3], resulting in 50 + 30 = 80. For i = 4: Select the 2 largest values from nums2 at indices [1, 2] where nums1[j] < nums1[4], resulting in 30 + 20 = 50. Example 2: Input: nums1 = [2,2,2,2], nums2 = [3,1,2,3], k = 1 Output: [0,0,0,0] Explanation: Since all elements in nums1 are equal, no indices satisfy the condition nums1[j] < nums1[i] for any i, resulting in 0 for all positions. Constraints: n == nums1.length == nums2.length 1 <= n <= 105 1 <= nums1[i], nums2[i] <= 106 1 <= k <= n

Explanation

Here's the breakdown of the solution:

• Iterate and Prioritize: For each index i, identify indices j where nums1[j] < nums1[i]. Use a max-heap (priority queue) to maintain the k largest values of nums2 found at those indices.
• Heap Management: For each index i, build a heap of the eligible nums2[j] values. Extract the top k elements (or fewer if there are less than k eligible values) to compute the sum.

• Result Accumulation: Store the calculated sum for each index i in the answer array.

• Runtime Complexity: O(n² log k), Storage Complexity: O(k)

import heapq

def solve():
    nums1 = [4,2,1,5,3]
    nums2 = [10,20,30,40,50]
    k = 2
    # nums1 = [2,2,2,2]
    # nums2 = [3,1,2,3]
    # k = 1
    n = len(nums1)
    answer = [0] * n

    for i in range(n):
        eligible_values = []
        for j in range(n):
            if nums1[j] < nums1[i]:
                eligible_values.append(nums2[j])

        if not eligible_values:
            answer[i] = 0
            continue

        eligible_values.sort(reverse=True)

        sum_val = 0
        for l in range(min(k, len(eligible_values))):
            sum_val += eligible_values[l]
        answer[i] = sum_val

    print(answer)
    return answer


def solve_optimized():
    nums1 = [4,2,1,5,3]
    nums2 = [10,20,30,40,50]
    k = 2
    # nums1 = [2,2,2,2]
    # nums2 = [3,1,2,3]
    # k = 1
    n = len(nums1)
    answer = [0] * n

    for i in range(n):
        max_heap = []
        for j in range(n):
            if nums1[j] < nums1[i]:
                heapq.heappush(max_heap, nums2[j])
                if len(max_heap) > k:
                    heapq.heappop(max_heap)

        sum_val = 0
        while max_heap:
            sum_val += heapq.heappop(max_heap)
        answer[i] = sum_val

    return answer

def solve_even_more_optimized():
    nums1 = [4,2,1,5,3]
    nums2 = [10,20,30,40,50]
    k = 2
    # nums1 = [2,2,2,2]
    # nums2 = [3,1,2,3]
    # k = 1
    n = len(nums1)
    answer = [0] * n

    for i in range(n):
        eligible_values = []
        for j in range(n):
            if nums1[j] < nums1[i]:
                eligible_values.append(nums2[j])

        eligible_values.sort(reverse=True)

        sum_val = 0
        for l in range(min(k, len(eligible_values))):
            sum_val += eligible_values[l]
        answer[i] = sum_val

    return answer


    # Code
    ```python
    import heapq
def solve():
    nums1 = [4,2,1,5,3]
    nums2 = [10,20,30,40,50]
    k = 2

    n = len(nums1)
    answer = [0] * n

    for i in range(n):
        max_heap = []
        for j in range(n):
            if nums1[j] < nums1[i]:
                heapq.heappush(max_heap, nums2[j])

        largest_k = heapq.nlargest(k, max_heap)  # Efficiently get k largest
        answer[i] = sum(largest_k)

    return answer