FindKPairsWithSmallestSums373.java

/**
 * 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]
 *
 */


public class FindKPairsWithSmallestSums373 {
    /**
     * https://discuss.leetcode.com/topic/50885/simple-java-o-klogk-solution-with-explanation
     */
    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;
    }

    /**
     * https://discuss.leetcode.com/topic/50529/java-9ms-heap-queue-solution-k-log-k
     */
    public List<int[]> kSmallestPairs2(int[] nums1, int[] nums2, int k) {
        List<int[]> ret = new ArrayList<>();
        if (nums1==null || nums2==null || nums1.length ==0 || nums2.length ==0) return ret;
        int len1 = nums1.length, len2=nums2.length;

        PriorityQueue<Pair> q = new PriorityQueue(k, new CompPair());
        for (int i=0; i<nums1.length && i<k ; i++) { // only need first k number in nums1 to start
            q.offer( new Pair(0, nums1[i],nums2[0]) );
        }
        for (int i=1; i<=k && !q.isEmpty(); i++) { // get the first k sums
            Pair p = q.poll();
            ret.add( p.pair );
            if (p.idx < len2 -1 ) { // get to next value in nums2
                int next = p.idx+1;
                q.offer( new Pair(next, p.pair[0], nums2[next]) );
            }
        }
        return ret;
    }

    class Pair{
        int[] pair;
        int idx; // current index to nums2
        long sum;
        Pair(int idx, int n1, int n2){
            this.idx = idx;
            pair = new int[]{n1, n2};
            sum = (long) n1 + (long) n2;
        }
    }

    class CompPair implements Comparator<Pair> {
        public int compare(Pair p1, Pair p2){
            return Long.compare(p1.sum, p2.sum);
        }
    }

    /**
     * https://discuss.leetcode.com/topic/52953/share-my-solution-which-beat-96-42
     */
    public List<int[]> kSmallestPairs3(int[] nums1, int[] nums2, int k) {
        PriorityQueue<Tuple> pq = new PriorityQueue<Tuple>();
        int m = nums1.length, n = nums2.length;
        List<int[]> res = new ArrayList<int[]>();
        if(nums1 == null || nums1.length == 0 || nums2 == null || nums2.length == 0 || k <= 0) return res;
        for(int j = 0; j <= n-1; j++) pq.offer(new Tuple(0, j, nums1[0]+nums2[j]));
        for(int i = 0; i < Math.min(k, m *n); i++) {
            Tuple t = pq.poll();
            res.add(new int[]{nums1[t.x], nums2[t.y]});
            if(t.x == m - 1) continue;
            pq.offer(new Tuple (t.x + 1, t.y, nums1[t.x + 1] + nums2[t.y]));
        }
        return res;
    }

    class Tuple implements Comparable<Tuple> {
        int x, y, val;
        public Tuple (int x, int y, int val) {
            this.x = x;
            this.y = y;
            this.val = val;
        }

        @Override
        public int compareTo (Tuple that) {
            return this.val - that.val;
        }
    }

}