Description
You are given a sorted integer array arr
containing 1
and prime numbers, where all the integers of arr
are unique. You are also given an integer k
.
For every i
and j
where 0 <= i < j < arr.length
, we consider the fraction arr[i] / arr[j]
.
Return the kth
smallest fraction considered. Return your answer as an array of integers of size 2
, where answer[0] == arr[i]
and answer[1] == arr[j]
.
Example 1:
Input: arr = [1,2,3,5], k = 3 Output: [2,5] Explanation: The fractions to be considered in sorted order are: 1/5, 1/3, 2/5, 1/2, 3/5, and 2/3. The third fraction is 2/5.
Example 2:
Input: arr = [1,7], k = 1 Output: [1,7]
Constraints:
2 <= arr.length <= 1000
1 <= arr[i] <= 3 * 104
arr[0] == 1
arr[i]
is a prime number fori > 0
.- All the numbers of
arr
are unique and sorted in strictly increasing order. 1 <= k <= arr.length * (arr.length - 1) / 2
Code
Min Heap
Time Complexity: , Space Complexity:
class Solution {
struct cmp {
bool operator() (const pair<double, pair<int, int>>& p1, const pair<double, pair<int, int>>& p2) {
return p1.first > p2.first;
}
};
public:
vector<int> kthSmallestPrimeFraction(vector<int>& arr, int k) {
priority_queue<pair<double, pair<int, int>>, vector<pair<double, pair<int, int>>>, cmp> min_heap;
// total number of elements to be pushed into heap
int size = arr.size();
size = size * (size - 1);
size = size / 2;
// size - (k - 1) = (size - k + 1) is the number of elements
// we want to pop out from the heap in order
// to make the top of the heap the kth smallest prime
// after the for loop ends
for(int i = 0; i < arr.size(); i++) {
for(int j = i + 1; j < arr.size(); j++) {
min_heap.push({((double)arr[i] / (double)arr[j]), {i, j}});
if(min_heap.size() > (size - k + 1)) {
min_heap.pop();
}
}
}
auto res = min_heap.top();
vector<int> ans;
ans.push_back(arr[res.second.first]);
ans.push_back(arr[res.second.second]);
return ans;
}
};
Min Heap
Time Complexity: , Space Complexity:
使用 Merge k Sorted Lists 的觀念,以 [1, 7, 23, 29, 47]
為例,就是以下四條 list。
1/47 < 1/29 < 1/23 < 1/7
7/47 < 7/29 < 7/23
23/47 < 23/29
29/47
class Solution {
struct cmp {
bool operator() (const pair<double, pair<int, int>>& p1, const pair<double, pair<int, int>>& p2) {
return p1.first > p2.first;
}
};
public:
vector<int> kthSmallestPrimeFraction(vector<int>& arr, int k) {
priority_queue<pair<double, pair<int, int>>, vector<pair<double, pair<int, int>>>, cmp> min_heap;
int n = arr.size();
for(int i = 0; i < arr.size() - 1; i++) {
min_heap.push({((double)arr[i] / (double)arr[n-1]), {i, n-1}});
}
for(int i = 0; i < k - 1; i++) {
auto res = min_heap.top();
min_heap.pop();
int idx1 = res.second.first;
int idx2 = res.second.second;
if(idx2 - 1 > idx1) {
min_heap.push({((double)arr[idx1] / (double)arr[idx2 - 1]), {idx1, idx2 - 1}});
}
}
vector<int> ans;
auto res = min_heap.top();
ans.push_back(arr[res.second.first]);
ans.push_back(arr[res.second.second]);
return ans;
}
};
Binary Search
Time Complexity: , Space Complexity:
相同概念:Kth Smallest Element in a Sorted Matrix