Quickselect is a selection algorithm to find the k-th smallest/largest element in an unordered list. It uses the partition method in Quick Sort. The difference is, instead of recurring for both sides (after finding pivot), it recurs only for the part that contains the k-th smallest/largest element.
The time complexity is O(N) on average, and O(N^2) in the worst case.
Implementation
Quick select with elements sorted in ascending order.
// OJ: https://leetcode.com/problems/kth-largest-element-in-an-array/
// Author: github.com/lzl124631x
// Time: O(N) on averge, O(N^2) in the worst case
// Space: O(1)
class Solution {
int partition(vector<int> &A, int L, int R) {
int i = L, pivotIndex = L + rand() % (R - L + 1), pivot = A[pivotIndex];
swap(A[pivotIndex], A[R]);
for (int j = L; j < R; ++j) {
if (A[j] < pivot) swap(A[i++], A[j]);
}
swap(A[i], A[R]);
return i;
}
public:
int findKthLargest(vector<int>& A, int k) {
int L = 0, R = A.size() - 1;
k = A.size() - k + 1;
while (true) {
int M = partition(A, L, R);
if (M + 1 == k) return A[M];
if (M + 1 > k) R = M - 1;
else L = M + 1;
}
}
};
Quick select with elements sorted in descending order.
// OJ: https://leetcode.com/problems/kth-largest-element-in-an-array/
// Author: github.com/lzl124631x
// Time: O(N) on averge, O(N^2) in the worst case
// Space: O(1)
class Solution {
int partition(vector<int> &A, int L, int R) {
int i = L, pivotIndex = L + rand() % (R - L + 1), pivot = A[pivotIndex];
swap(A[pivotIndex], A[R]);
for (int j = L; j < R; ++j) {
if (A[j] > pivot) swap(A[i++], A[j]);
}
swap(A[i], A[R]);
return i;
}
public:
int findKthLargest(vector<int>& A, int k) {
int L = 0, R = A.size() - 1;
while (true) {
int M = partition(A, L, R);
if (M + 1 == k) return A[M];
if (M + 1 > k) R = M - 1;
else L = M + 1;
}
}
};
Or STL
// OJ: https://leetcode.com/problems/kth-largest-element-in-an-array/
// Author: github.com/lzl124631x
// Time: O(N) on average, O(N^2) in the worst case
// Space: O(1)
class Solution {
public:
int findKthLargest(vector<int>& A, int k) {
nth_element(begin(A), begin(A) + k - 1, end(A), greater<int>());
return A[k - 1];
}
};