Link: https://leetcode-cn.com/problems/kth-smallest-element-in-a-sorted-matrix/
subject
Give you a n x n matrix, where the elements of each row and column are sorted in ascending order to find the k-smallest element in the matrix.
Note that it is the k-th smallest element after sorting, not the k-th different element.
Use case
Example 1:
Input: matrix = [[1,5,9],[10,11,13],[12,13,15]], k = 8
Output: 13
Explanation: the elements in the matrix are [1,5,9,10,11,12,13,13,15], and the 8th small element is 13
Example 2:
Input: matrix = [[-5]], k = 1
Output: - 5
Tips:
n == matrix.length
n == matrix[i].length
1 <= n <= 300
-109 <= matrix[i][j] <= 109
The topic data ensures that all rows and columns in the matrix are arranged in non decreasing order
1 <= k <= n2
thinking
Method 1: priority queue multi-channel merging
Each row or column is regarded as a chain, with a total of n chains. After that, the minimum value of each chain is input into the priority queue for multiplexing and merging
Every time the priority queue outputs a value, it will cycle through the next value in the linked list into the queue until the k-th value is found
class Solution {
public:
int kthSmallest(vector<vector<int>>& matrix, int k) {
struct point {
int val, x, y;
point(int val, int x, int y) : val(val), x(x), y(y) {}
bool operator> (const point& a) const { return this->val > a.val; }
};
priority_queue<point, vector<point>, greater<point>> que;
int n = matrix.size();
for (int i = 0; i < n; i++) {
que.emplace(matrix[i][0], i, 0);
}
for (int i = 0; i < k - 1; i++) {
point now = que.top();
que.pop();
if (now.y != n - 1) {
que.emplace(matrix[now.x][now.y + 1], now.x, now.y + 1);
}
}
return que.top().val;
}
};
Method binary search
For all kinds of ordered data structures, they can be transformed by binary search
The upper left corner of the matrix is the minimum value and the lower right corner of the matrix is the maximum value. First, find the middle value between the minimum and maximum values through bisection
Then, the number p not greater than the intermediate value in the count matrix is compared with the target value K. if p > = k, it indicates that ans is less than or equal to the intermediate value mid, so right=mid; If p < K, ans is greater than mid, left=mid+1;
The counting process can refer to Search two-dimensional matrix Count from the top right corner
class Solution {
public:
int kthSmallest(vector<vector<int>>& matrix, int k) {
int n=matrix.size();
auto count=[&](int x){
int i=0,j=n-1,cnt=0;
while(j>=0&&i<n){
if(matrix[i][j]>x){
j--;
}else{
cnt+=j+1;
i++;
}
}
return cnt;
};
int left=matrix[0][0],right=matrix[n-1][n-1];
while(left<right){
int mid =(right-left)/2+left;
int p=count(mid);
if(p>=k){
right=mid;
}else{
left=mid+1;
}
}
return left;
}
};
Similar questions:
subject
Given two integer arrays nums1 and nums2 arranged in ascending order, and an integer k.
Define a pair of values (u,v), where the first element is from nums1 and the second element is from nums2.
Please find the minimum number of k} pairs (u1,v1), (u2,v2) (uk,vk) .
Use case
Example 1:
Input: nums1 = [1,7,11], nums2 = [2,4,6], k = 3
Output: [1,2], [1,4], [1,6]
Explanation: returns the first 3 logarithms in 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: returns the first two logarithms in 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: it is also possible that all pairs in the sequence are returned: [1,3], [2,3]
Tips:
1 <= nums1.length, nums2.length <= 105
-109 <= nums1[i], nums2[i] <= 109
nums1 and nums2 are arranged in ascending order
1 <= k <= 1000
thinking
Method 1 multi channel merging
Put the sequence of the sum of each number in nums1[0] and nums2 into the priority queue, arrange it according to the size of sum, and then multiplex and merge
I wrote a single team directly, so there is still a problem of de duplication
class Solution {
public:
vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
int n1=nums1.size(),n2=nums2.size();
vector<vector<int>>ans;
auto cmp=[&](pair<int,int>&a,pair<int,int>&b){if(nums1[a.first]+nums2[a.second]==nums1[b.first]+nums2[b.second]){if(a.first==b.first)return a.second>b.second;else return a.first>b.first;}else return nums1[a.first]+nums2[a.second]>nums1[b.first]+nums2[b.second];};
priority_queue<pair<int,int>,vector<pair<int,int>>,decltype(cmp)>pq(cmp);
pq.push({0,0});
int cnt=0;
while(!pq.empty()&&cnt<k){
auto tmp=pq.top();
while(!pq.empty()&&tmp==pq.top()){
pq.pop();
}
int x=tmp.first,y=tmp.second;
ans.push_back({nums1[x],nums2[y]});
cnt++;
if(x+1<n1)pq.push({x+1,y});
if(y+1<n2)pq.push({x,y+1});
}
return ans;
}
};
After optimization and elimination of duplication
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-- > 0 && !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;
}
};
Method binary search
Ibid
class Solution {
public:
vector<vector<int>> kSmallestPairs(vector<int>& nums1, vector<int>& nums2, int k) {
int m = nums1.size();
int n = nums2.size();
auto count = [&](int target){
long long cnt = 0;
int start = 0;
int end = n - 1;
while (start < m && end >= 0) {
if (nums1[start] + nums2[end] > target) {
end--;
} else {
cnt += end + 1;
start++;
}
}
return cnt;
};
/*Bisection finds the size of the sum of the k-th smallest number pair*/
int left = nums1[0] + nums2[0];
int right = nums1.back() + nums2.back();
int pairSum = right;
while (left <= right) {
int mid = left + ((right - left) >> 1);
if (count(mid) < k) {
left = mid + 1;
} else {
pairSum = mid;
right = mid - 1;
}
}
vector<vector<int>> ans;
int pos = n - 1;
/*Found pairs less than the target pairSum*/
for (int i = 0; i < m; i++) {
while (pos >= 0 && nums1[i] + nums2[pos] >= pairSum) {
pos--;
}
for (int j = 0; j <= pos && k > 0; j++, k--) {
ans.push_back({nums1[i], nums2[j]});
}
}
/*Find pairs equal to the target value pairSum*/
pos = n - 1;
for (int i = 0; i < m && k > 0; i++) {
while (pos >= 0 && nums1[i] + nums2[pos] > pairSum) {
pos--;
}
for (int j = i; k > 0 && j >= 0 && nums1[j] + nums2[pos] == pairSum; j--, k--) {
ans.push_back({nums1[i], nums2[pos]});
}
}
return ans;
}
};