# 373. Find K Pairs with Smallest Sums

## Description

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]


Example 3:

Input: nums1 = [1,2], nums2 = [3], k = 3
Output: [[1,3],[2,3]]
Explanation: All possible pairs are returned from the sequence: [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

## Solutions

The value of num1[0]+num2[0] must be the smallest, so we first throw this value into the heap.

Then start traversing the next element.

The next value must be the smallest one from num1[1]+num2[0] and num1[0]+num2[1].

Loop all the combinations in turn until the extracted element is K.

## Variation Question

input: nums=[3,1,9], k=3

goal: find the k-th smallest diff of pairs from nums

pairs from nums: (1,3) (1,9) (3,9)

• so diffs are (2,8,6)
• sort diffs (2,6,8)
• so 3-rd diff is 8

optimized solution:

1. sort nums => get [1,3,9]
2. start from neighbours, with neighbour-distance=1 get => (1,3), (3,9)
3. if len(pairs) still less than k pairs, then keep trying with neighbour-distance=2 get => (1,9)
4. and then neighbour-distance=3, =4, so on
5. when len(pairs)==k , stop and ignore more further larger pair-diffs
• class Solution {
public List<List<Integer>> kSmallestPairs(int[] nums1, int[] nums2, int k) {
PriorityQueue<int[]> q = new PriorityQueue<>(Comparator.comparingInt(a -> a[0]));
for (int i = 0; i < Math.min(nums1.length, k); ++i) {
q.offer(new int[] {nums1[i] + nums2[0], i, 0});
}
List<List<Integer>> ans = new ArrayList<>();
while (!q.isEmpty() && k > 0) {
int[] e = q.poll();
--k;
if (e[2] + 1 < nums2.length) {
q.offer(new int[] {nums1[e[1]] + nums2[e[2] + 1], e[1], e[2] + 1});
}
}
return ans;
}
}

• class Solution {
public:
vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
auto cmp = [&nums1, &nums2](const pair<int, int>& a, const pair<int, int>& b) {
return nums1[a.first] + nums2[a.second] > nums1[b.first] + nums2[b.second];
};

int m = nums1.size();
int n = nums2.size();
vector<vector<int>> ans;
priority_queue<pair<int, int>, vector<pair<int, int>>, decltype(cmp)> pq(cmp);
for (int i = 0; i < min(k, m); i++)
pq.emplace(i, 0);
while (k-- && !pq.empty()) {
auto [x, y] = pq.top();
pq.pop();
ans.emplace_back(initializer_list<int>{nums1[x], nums2[y]});
if (y + 1 < n)
pq.emplace(x, y + 1);
}

return ans;
}
};

• from heapq import heapify

class Solution:
def kSmallestPairs(
self, nums1: List[int], nums2: List[int], k: int
) -> List[List[int]]:
'''
k could be a super large number

>>> [1,2,3][:99]
[1, 2, 3]
'''
q = [[u + nums2[0], i, 0] for i, u in enumerate(nums1[:k])] # still need '[u + nums2[0]', for q ordering
heapify(q)
ans = []
# both q and k should be checked
# because k can be super larger than nums1+nums2
while q and k > 0:
_, i, j = heappop(q)
ans.append([nums1[i], nums2[j]])
k -= 1
if j + 1 < len(nums2):
heappush(q, [nums1[i] + nums2[j + 1], i, j + 1])
return ans

############

import heapq

class Solution(object):
def kSmallestPairs(self, nums1, nums2, k):
"""
:type nums1: List[int]
:type nums2: List[int]
:type k: int
:rtype: List[List[int]]
"""
if not nums1 or not nums2:
return []
heap = [(nums1[0] + nums2[0], 0, 0)]
ans = []
visited = {(0, 0)}

while heap:
val, i, j = heapq.heappop(heap)
ans.append((nums1[i], nums2[j]))
k -= 1
if k == 0:
return ans
if i + 1 < len(nums1) and (i + 1, j) not in visited:
heapq.heappush(heap, (nums1[i + 1] + nums2[j], i + 1, j))
visited |= {(i + 1, j)}
if j + 1 < len(nums2) and (i, j + 1) not in visited:
heapq.heappush(heap, (nums1[i] + nums2[j + 1], i, j + 1))
visited |= {(i, j + 1)}
return ans

############

# variation question
# time complexity of O(n^2)
def kth_smallest_diff(nums, k):
# Step 1: Generate all possible pairs and calculate their differences
diffs = []
nums.sort()  # Sort nums first to make sure differences are calculated correctly
for i in range(len(nums)):
for j in range(i + 1, len(nums)):
diff = abs(nums[j] - nums[i])
diffs.append(diff)

# Step 2 & 3: Sort the list of differences
diffs.sort()

# Step 4: Return the k-th smallest difference
return diffs[k - 1]

# Example usage
nums = [3, 1, 9]
k = 3
print(kth_smallest_diff(nums, k))


• func kSmallestPairs(nums1, nums2 []int, k int) (ans [][]int) {
m, n := len(nums1), len(nums2)
h := hp{nil, nums1, nums2}
for i := 0; i < k && i < m; i++ {
h.data = append(h.data, pair{i, 0})
}
for h.Len() > 0 && len(ans) < k {
p := heap.Pop(&h).(pair)
i, j := p.i, p.j
ans = append(ans, []int{nums1[i], nums2[j]})
if j+1 < n {
heap.Push(&h, pair{i, j + 1})
}
}
return
}

type pair struct{ i, j int }
type hp struct {
data         []pair
nums1, nums2 []int
}

func (h hp) Len() int { return len(h.data) }
func (h hp) Less(i, j int) bool {
a, b := h.data[i], h.data[j]
return h.nums1[a.i]+h.nums2[a.j] < h.nums1[b.i]+h.nums2[b.j]
}
func (h hp) Swap(i, j int) { h.data[i], h.data[j] = h.data[j], h.data[i] }
func (h *hp) Push(v any)   { h.data = append(h.data, v.(pair)) }
func (h *hp) Pop() any     { a := h.data; v := a[len(a)-1]; h.data = a[:len(a)-1]; return v }