Introduction

In this blog post, we will explore how to solve the LeetCode problem "373" 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 sorted in non-decreasing order and an integer k. Define a pair (u, v) which consists of one element from the first array and one element from the second array. Return the k pairs (u1, v1), (u2, v2), ..., (uk, vk) with the smallest sums. Example 1: Input: nums1 = [1,7,11], nums2 = [2,4,6], k = 3 Output: [[1,2],[1,4],[1,6]] Explanation: The first 3 pairs are returned from the sequence: [1,2],[1,4],[1,6],[7,2],[7,4],[11,2],[7,6],[11,4],[11,6] Example 2: Input: nums1 = [1,1,2], nums2 = [1,2,3], k = 2 Output: [[1,1],[1,1]] Explanation: The first 2 pairs are returned from the sequence: [1,1],[1,1],[1,2],[2,1],[1,2],[2,2],[1,3],[1,3],[2,3] Constraints: 1 <= nums1.length, nums2.length <= 105 -109 <= nums1[i], nums2[i] <= 109 nums1 and nums2 both are sorted in non-decreasing order. 1 <= k <= 104 k <= nums1.length * nums2.length

Explanation

• Priority Queue (Heap): Use a min-heap to store pairs and their sums. The heap will always maintain the pair with the smallest sum at the top.
  • Initialization: Start by adding pairs formed by the first element of nums1 with each element of nums2 to the heap.
  • Iteration: Extract the pair with the smallest sum from the heap. Add this pair to the result. Then, add the next possible pair involving the same element from nums2 to the heap if available.

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

Code

    import heapq

def kSmallestPairs(nums1, nums2, k):
    """
    Finds the k pairs with the smallest sums from two sorted integer arrays.

    Args:
        nums1: The first sorted integer array.
        nums2: The second sorted integer array.
        k: The number of pairs to return.

    Returns:
        A list of the k pairs with the smallest sums.
    """

    result = []
    heap = []  # Min-heap to store pairs and their sums
    visited = set()

    # Add initial pairs to the heap (first element of nums1 with each element of nums2)
    for i in range(min(len(nums2), k)):
        heapq.heappush(heap, (nums1[0] + nums2[i], 0, i)) # (sum, index in nums1, index in nums2)
        visited.add((0, i)) # Mark as visited

    while k > 0 and heap:
        sum_val, i, j = heapq.heappop(heap)
        result.append([nums1[i], nums2[j]])
        k -= 1

        # Add the next possible pair to the heap (increment nums1 index)
        if i + 1 < len(nums1) and (i + 1, j) not in visited:
            heapq.heappush(heap, (nums1[i + 1] + nums2[j], i + 1, j))
            visited.add((i + 1, j))

    return result