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KthLargeStream.cpp
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82 lines (65 loc) · 2.58 KB
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/*
Design a class to find the kth largest element in a stream. Note that it is the kth largest element in the sorted order, not the kth distinct element.
Implement KthLargest class:
KthLargest(int k, int[] nums) Initializes the object with the integer k and the stream of integers nums.
int add(int val) Appends the integer val to the stream and returns the element representing the kth largest element in the stream.
Example 1:
Input
["KthLargest", "add", "add", "add", "add", "add"]
[[3, [4, 5, 8, 2]], [3], [5], [10], [9], [4]]
Output
[null, 4, 5, 5, 8, 8]
Explanation
KthLargest kthLargest = new KthLargest(3, [4, 5, 8, 2]);
kthLargest.add(3); // return 4
kthLargest.add(5); // return 5
kthLargest.add(10); // return 5
kthLargest.add(9); // return 8
kthLargest.add(4); // return 8
*/
// This is a typicaly heap question.The biggere question is what heap do we use/
// While both heaps would yield the correct answer we want to produce an anwer with the brtter complexity.
// The answer would be a minheap
// Let num = [3,4,5,10,2] k=3
// | 3 |
// | 4 | <---------This would be the minheap after storing the first 3 element in it with the smallest elemeyt at the top.
// |__5__|
// Now the what we do is our main goal is to make the kth larget element at the top of the heap so extraction takes o(1)time excluding the heapify time.
// We would build the heap by storing the larget k numbers in the heao with the smallest of them at the top yielding the kth
// largest number.
// | 4 |
// | 5 | <---------as 10 was greater than the top of the heap (i.e 3) we popped it and stored 10 in it.
// |__10_|
// Remember Our Goal here is the store the k largest elements into the heap.
// | 4 |
// | 5 | <---------as 2 wasnt greater than heap.top() we just dont do anything.
// |__10_|
// Thus our final answer is heap.top() -> 4.
#include<bits/stdc++.h>
using namespace std;
class KthLargest {
public:
priority_queue<int,vector<int>,greater<int>> minheap;
int m_k;
KthLargest(int k, vector<int>& nums) {
m_k=k;
for(int x:nums)add(x);
}
int add(int val) {
if(minheap.size()<m_k)minheap.push(val);
else
{
if(val>minheap.top())
{
minheap.pop();
minheap.push(val);
}
}
return minheap.top();
}
};
/**
* Your KthLargest object will be instantiated and called as such:
* KthLargest* obj = new KthLargest(k, nums);
* int param_1 = obj->add(val);
*/