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373. Find K Pairs with Smallest Sums 👍

  • Time: $O(k\log k)$
  • Space: $O(k)$
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struct T {
  int i;
  int j;
  int sum;  // nums1[i] + nums2[j]
};

class Solution {
 public:
  vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2,
                                     int k) {
    vector<vector<int>> ans;
    auto compare = [&](const T& a, const T& b) { return a.sum > b.sum; };
    priority_queue<T, vector<T>, decltype(compare)> minHeap(compare);

    for (int i = 0; i < k && i < nums1.size(); ++i)
      minHeap.emplace(i, 0, nums1[i] + nums2[0]);

    while (!minHeap.empty() && ans.size() < k) {
      const auto [i, j, _] = minHeap.top();
      minHeap.pop();
      ans.push_back({nums1[i], nums2[j]});
      if (j + 1 < nums2.size())
        minHeap.emplace(i, j + 1, nums1[i] + nums2[j + 1]);
    }

    return ans;
  }
};

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class T {
  public int i;
  public int j;
  public int sum; // nums1[i] + nums2[j]
  public T(int i, int j, int sum) {
    this.i = i;
    this.j = j;
    this.sum = sum;
  }
}

class Solution {
  public List<List<Integer>> kSmallestPairs(int[] nums1, int[] nums2, int k) {
    List<List<Integer>> ans = new ArrayList<>();
    Queue<T> minHeap = new PriorityQueue<>((a, b) -> Integer.compare(a.sum, b.sum));

    for (int i = 0; i < Math.min(k, nums1.length); ++i)
      minHeap.offer(new T(i, 0, nums1[i] + nums2[0]));

    while (!minHeap.isEmpty() && ans.size() < k) {
      final int i = minHeap.peek().i;
      final int j = minHeap.poll().j;
      ans.add(Arrays.asList(nums1[i], nums2[j]));
      if (j + 1 < nums2.length)
        minHeap.offer(new T(i, j + 1, nums1[i] + nums2[j + 1]));
    }

    return ans;
  }
}

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class Solution:
  def kSmallestPairs(self, nums1: list[int],
                     nums2: list[int],
                     k: int) -> list[list[int]]:
    minHeap = []

    for i in range(min(k, len(nums1))):
      heapq.heappush(minHeap, (nums1[i] + nums2[0], i, 0))

    ans = []
    while minHeap and len(ans) < k:
      _, i, j = heapq.heappop(minHeap)
      ans.append([nums1[i], nums2[j]])
      if j + 1 < len(nums2):
        heapq.heappush(minHeap, (nums1[i] + nums2[j + 1], i, j + 1))

    return ans