The median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle values.
For examples, if arr = [2,3,4], the median is 3.
For examples, if arr = [1,2,3,4], the median is (2 + 3) / 2 = 2.5.
You are given an integer array nums and an integer k. There is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position.
Return the median array for each window in the original array. Answers within 10-5 of the actual value will be accepted.
Input: nums = [1,2,3,4,2,3,1,4,2], k = 3
Output: [2.00000,3.00000,3.00000,3.00000,2.00000,3.00000,2.00000]
Constraints:
1 <= k <= nums.length <= 105
-231 <= nums[i] <= 231 - 1
Intuition
Approach:
Maintain two heaps to divide the window elements into two almost equal
Parts,
Then take top elements as median
But then, one thing to maintain is balance of the heaps,
One problem may arise, elements to be deleted at bottom of the heap
Maintain them in map to be deleted later, if they arrive at top later
Links
Video Links
Approach 1:
Sliding Window + HEap
C++
class Solution {
public:
vector<double> medianSlidingWindow(vector<int>& nums, int k) {
vector<double> medians;
int n = nums.size();
unordered_map<int,int> mp; // for late deletion
priority_queue<int> maxh; // max heap for lower half
priority_queue<int, vector<int>, greater<int>> minh; // min heap for upper half
for(int i=0;i<k;i++) {
maxh.push(nums[i]);
}
for(int i=0;i<(k/2);i++) {
minh.push(maxh.top());
maxh.pop();
}
// always try to main the middle element on the top of maxh heap
// if we have even elements in both the heaps, median is avg of top of both heaps
for(int i=k;i<n;i++) {
if(k&1) { // if k is odd, we will have median on the top of maxheap
medians.push_back(maxh.top()*1.0);
}
else {
medians.push_back(((double)maxh.top()+(double)minh.top())/2);
}
int p=nums[i-k], q=nums[i]; //
// we need to remove p and add q;
// we will delete p when it will come on the top
// to keep track we will maintain map
int balance = 0; // keep heaps in balance, for correct ans
// we decrese balance when remove elements from maxh, so basically if balance<0 it means maxheap has lesser elemets the minheap
// removing p or adding p to map to delete it later
if(p<=maxh.top()) { // p is in maxheap
balance--;
if(p==maxh.top())
maxh.pop();
else
mp[p]++;
}
else { // p is min heap
balance++;
if(p == minh.top())
minh.pop();
else
mp[p]++;
}
// inserting q to the correct heap
if(!maxh.empty() and q<=maxh.top()) { // pushing q to maxheap
maxh.push(q);
balance++;
}
else { // pushing q to minheap
minh.push(q);
balance--;
}
// balancing both the heaps
if(balance<0) {
maxh.push(minh.top());
minh.pop();
}
else if(balance>0) {
minh.push(maxh.top());
maxh.pop();
}
// removing top elements if they exist in our map(late deletion)
while(!maxh.empty() and mp[maxh.top()]) {
mp[maxh.top()]--;
maxh.pop();
}
while(!minh.empty() and mp[minh.top()]) {
mp[minh.top()]--;
minh.pop();
}
}
if(k&1) {
medians.push_back(maxh.top()*1.0);
}
else {
medians.push_back(((double)maxh.top()+(double)minh.top())/2.0);
}
return medians;
}
};