Created
June 22, 2022 07:58
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class Solution { | |
int getPivotIndex(int& l, int& r) { | |
return l + (r - l) / 2; | |
//return l + rand() % (r - l + 1); | |
} | |
int partition(vector<int>& A, int& l, int& r) { | |
int i = l, pivotIndex = getPivotIndex(l, r), 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: | |
// Quickselect, O(n^2) worst case, O(n) average case, O(1) space if we can modify the input directly | |
int findKthLargest(vector<int>& A, int& k) { | |
int l = 0, r = A.size() - 1, mid; | |
while (l < r) { | |
mid = partition(A, l, r); | |
if (mid + 1 == k) return A[mid]; | |
if (mid + 1 > k) r = mid - 1; | |
else l = mid + 1; | |
} | |
return A[l]; | |
} | |
// Priority queue (min heap), O(n log k) time and O(k) space | |
int findKthLargestMinHeap(const vector<int>& nums, int& k) { | |
priority_queue<int, vector<int>, greater<int>> pq(begin(nums), begin(nums) + k); | |
int n = nums.size(); | |
for (int i = k; i < n; i++) | |
if (nums[i] > pq.top()) { | |
pq.pop(); | |
pq.push(nums[i]); | |
} | |
return pq.top(); | |
} | |
}; |
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