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Find K Pairs with Smallest Sums – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
You are given two integer arrays nums1 and nums2 sorted in ascending 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] <= 109nums1andnums2both are sorted in ascending order.1 <= k <= 104
C++ Find K Pairs with Smallest Sums LeetCode Solution
class Solution {
public:
vector<pair<int, int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
vector<pair<int,int>> result;
if (nums1.empty() || nums2.empty() || k <= 0)
return result;
auto comp = [&nums1, &nums2](pair<int, int> a, pair<int, int> b) {
return nums1[a.first] + nums2[a.second] > nums1[b.first] + nums2[b.second];};
priority_queue<pair<int, int>, vector<pair<int, int>>, decltype(comp)> min_heap(comp);
min_heap.emplace(0, 0);
while(k-- > 0 && min_heap.size())
{
auto idx_pair = min_heap.top(); min_heap.pop();
result.emplace_back(nums1[idx_pair.first], nums2[idx_pair.second]);
if (idx_pair.first + 1 < nums1.size())
min_heap.emplace(idx_pair.first + 1, idx_pair.second);
if (idx_pair.first == 0 && idx_pair.second + 1 < nums2.size())
min_heap.emplace(idx_pair.first, idx_pair.second + 1);
}
return result;
}
};
Java Find K Pairs with Smallest Sums LeetCode Solution
public class Solution {
public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
PriorityQueue<int[]> que = new PriorityQueue<>((a,b)->a[0]+a[1]-b[0]-b[1]);
List<int[]> res = new ArrayList<>();
if(nums1.length==0 || nums2.length==0 || k==0) return res;
for(int i=0; i<nums1.length && i<k; i++) que.offer(new int[]{nums1[i], nums2[0], 0});
while(k-- > 0 && !que.isEmpty()){
int[] cur = que.poll();
res.add(new int[]{cur[0], cur[1]});
if(cur[2] == nums2.length-1) continue;
que.offer(new int[]{cur[0],nums2[cur[2]+1], cur[2]+1});
}
return res;
}
}
Python 3 Find K Pairs with Smallest Sums LeetCode Solution
from heapq import *
class Solution:
def kSmallestPairs(self, nums1, nums2, k):
if not nums1 or not nums2:
return []
visited = []
heap = []
output = []
heappush(heap, (nums1[0] + nums2[0], 0, 0))
visited.append((0, 0))
while len(output) < k and heap:
val = heappop(heap)
output.append((nums1[val[1]], nums2[val[2]]))
if val[1] + 1 < len(nums1) and (val[1] + 1, val[2]) not in visited:
heappush(heap, (nums1[val[1] + 1] + nums2[val[2]], val[1] + 1, val[2]))
visited.append((val[1] + 1, val[2]))
if val[2] + 1 < len(nums2) and (val[1], val[2] + 1) not in visited:
heappush(heap, (nums1[val[1]] + nums2[val[2] + 1], val[1], val[2] + 1))
visited.append((val[1], val[2] + 1))
return output
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