Since the arrays are sorted, the smallest sum is always nums1[0] + nums2[0]. We use a Min-Heap to explore potential pairs. We start with pairs (nums1[i], nums2[0]) for all i (up to k). When we pop a pair (nums1[i], nums2[j]), the next candidate from that nums1 element is (nums1[i], nums2[j+1]).

1. Initialize a Min-Heap.
2. Push (nums1[i] + nums2[0], i, 0) for i from 0 to min(len(nums1), k) - 1.
3. While k > 0 and heap is not empty:
   • Pop (sum, i, j).
   • Add [nums1[i], nums2[j]] to result.
   • If j + 1 < len(nums2), push (nums1[i] + nums2[j+1], i, j+1) to heap.
   • Decrement k.
4. Return result.

• Time Complexity: O(K log K).
• Space Complexity: O(K).

import heapq

def k_smallest_pairs(nums1, nums2, k):
    if not nums1 or not nums2: return []
    
    h = []
    # Initial candidates: (sum, index_in_nums1, index_in_nums2)
    for i in range(min(len(nums1), k)):
        heapq.heappush(h, (nums1[i] + nums2[0], i, 0))
        
    res = []
    while h and k > 0:
        val, i, j = heapq.heappop(h)
        res.append([nums1[i], nums2[j]])
        
        if j + 1 < len(nums2):
            heapq.heappush(h, (nums1[i] + nums2[j+1], i, j + 1))
            
        k -= 1
        
    return res