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FindKPairsSmallestSums.java
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FindKPairsSmallestSums.java
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/*
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.
Find the k pairs (u1,v1),(u2,v2) ...(uk,vk) with the smallest sums.
Example 1:
Given nums1 = [1,7,11], nums2 = [2,4,6], k = 3
Return: [1,2],[1,4],[1,6]
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:
Given nums1 = [1,1,2], nums2 = [1,2,3], k = 2
Return: [1,1],[1,1]
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:
Given nums1 = [1,2], nums2 = [3], k = 3
Return: [1,3],[2,3]
All possible pairs are returned from the sequence:
[1,3],[2,3]
Analysis:
Basic idea: Use min_heap to keep track on next minimum pair sum, and we only need to maintain
K possible candidates in the data structure.
Some observations: For every numbers in nums1, its best partner(yields min sum) always strats
from nums2[0] since arrays are all sorted; And for a specific number in nums1, its next candidate
sould be [this specific number] + nums2[current_associated_index + 1], unless out of boundary;)
*/
class FindKPairsSmallestSums {
public List<int[]> kSmallestPairs(int[] nums1, int[] nums2, int k) {
PriorityQueue<int[]> queue = new PriorityQueue<>(new Comparator<int[]>() {
@Override
public int compare(int[] a, int[] b) {
return a[0] + a[1] - b[0] - b[1];
}
});
List<int[]> result= new ArrayList<>();
if(nums1.length==0 || nums2.length==0 || k==0) {
return result;
}
for(int i = 0; i < nums1.length && i < k; i++) {
//the third element in array is for tracking index of nums2
queue.offer(new int[]{nums1[i], nums2[0], 0});
}
while(k-- > 0 && !queue.isEmpty()) {
int[] curr = queue.poll();
result.add(new int[]{curr[0], curr[1]});
if(curr[2] == nums2.length - 1)
continue;
queue.offer(new int[]{curr[0], nums2[curr[2] + 1], curr[2] + 1});
}
return result;
}
}