Kth Largest Element in an Array – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
Given an integer array nums and an integer k, return the kth largest element in the array.
Note that it is the kth largest element in the sorted order, not the kth distinct element.
You must solve it in O(n) time complexity.
Example 1:
Input: nums = [3,2,1,5,6,4], k = 2 Output: 5
Example 2:
Input: nums = [3,2,3,1,2,4,5,5,6], k = 4 Output: 4
Constraints:
1 <= k <= nums.length <= 105-104 <= nums[i] <= 104
C++ Kth Largest Element in an Array LeetCode Solution
Quickselect
int findKthLargest(vector<int>& nums, int k) {
//partition rule: >=pivot pivot <=pivot
int left=0,right=nums.size()-1,idx=0;
while(1){
idx = partition(nums,left,right);
if(idx==k-1) break;
else if(idx < k-1) left=idx+1;
else right= idx-1;
}
return nums[idx];
}
int partition(vector<int>& nums,int left,int right){//hoare partition
int pivot = nums[left], l=left+1, r = right;
while(l<=r){
if(nums[l]<pivot && nums[r]>pivot) swap(nums[l++],nums[r--]);
if(nums[l]>=pivot) ++l;
if(nums[r]<=pivot) --r;
}
swap(nums[left], nums[r]);
return r;
}
Java Kth Largest Element in an Array LeetCode Solution
public class Solution {
public int findKthLargest(int[] nums, int k) {
int start = 0, end = nums.length - 1, index = nums.length - k;
while (start < end) {
int pivot = partion(nums, start, end);
if (pivot < index) start = pivot + 1;
else if (pivot > index) end = pivot - 1;
else return nums[pivot];
}
return nums[start];
}
private int partion(int[] nums, int start, int end) {
int pivot = start, temp;
while (start <= end) {
while (start <= end && nums[start] <= nums[pivot]) start++;
while (start <= end && nums[end] > nums[pivot]) end--;
if (start > end) break;
temp = nums[start];
nums[start] = nums[end];
nums[end] = temp;
}
temp = nums[end];
nums[end] = nums[pivot];
nums[pivot] = temp;
return end;
}
}
Python 3 Kth Largest Element in an Array LeetCode Solution
class Solution:
def findKthLargest(self, nums, k):
if not nums: return
pivot = random.choice(nums)
left = [x for x in nums if x > pivot]
mid = [x for x in nums if x == pivot]
right = [x for x in nums if x < pivot]
L, M = len(left), len(mid)
if k <= L:
return self.findKthLargest(left, k)
elif k > L + M:
return self.findKthLargest(right, k - L - M)
else:
return mid[0]
Array-1180
String-562
Hash Table-412
Dynamic Programming-390
Math-368
Sorting-264
Greedy-257
Depth-First Search-256
Database-215
Breadth-First Search-200
Tree-195
Binary Search-191
Matrix-176
Binary Tree-160
Two Pointers-151
Bit Manipulation-140
Stack-133
Heap (Priority Queue)-117
Design-116
Graph-108
Simulation-103
Prefix Sum-96
Backtracking-92
Counting-86
Sliding Window-73
Linked List-69
Union Find-66
Ordered Set-48
Monotonic Stack-47
Recursion-43
Trie-41
Binary Search Tree-40
Divide and Conquer-40
Enumeration-39
Bitmask-37
Queue-33
Memoization-32
Topological Sort-31
Geometry-30
Segment Tree-27
Game Theory-24
Hash Function-24
Binary Indexed Tree-21
Interactive-18
Data Stream-17
String Matching-17
Rolling Hash-17
Shortest Path-16
Number Theory-16
Combinatorics-15
Randomized-12
Monotonic Queue-9
Iterator-9
Merge Sort-9
Concurrency-9
Doubly-Linked List-8
Brainteaser-8
Probability and Statistics-7
Quickselect-7
Bucket Sort-6
Suffix Array-6
Minimum Spanning Tree-5
Counting Sort-5
Shell-4
Line Sweep-4
Reservoir Sampling-4
Eulerian Circuit-3
Radix Sort-3
Strongly Connected Componen-t2
Rejection Sampling-2
Biconnected Component-1
Leave a comment below
