To find the median of an unsorted array, we can make a min-heap in O(nlogn) time for n elements, and then we can extract one by one n/2 elements to get the median. But this approach would take O(nlogn) time.
Can we do the same by some method in O(n) time? If we can, then please tell or suggest some method.
You can use the Median of Medians algorithm to find median of an unsorted array in linear time.
I have already upvoted the #dasblinkenlight answer since the Median of Medians algorithm in fact solves this problem in O(n) time. I only want to add that this problem could be solved in O(n) time by using heaps also. Building a heap could be done in O(n) time by using the bottom-up. Take a look to the following article for a detailed explanation Heap sort
Supposing that your array has N elements, you have to build two heaps: A MaxHeap that contains the first N/2 elements (or (N/2)+1 if N is odd) and a MinHeap that contains the remaining elements. If N is odd then your median is the maximum element of MaxHeap (O(1) by getting the max). If N is even, then your median is (MaxHeap.max()+MinHeap.min())/2 this takes O(1) also. Thus, the real cost of the whole operation is the heaps building operation which is O(n).
BTW this MaxHeap/MinHeap algorithm works also when you don't know the number of the array elements beforehand (if you have to resolve the same problem for a stream of integers for e.g). You can see more details about how to resolve this problem in the following article Median Of integer streams
Quickselect works in O(n), this is also used in the partition step of Quicksort.
The quick select algorithm can find the k-th smallest element of an array in linear (O(n)) running time. Here is an implementation in python:
import random
def partition(L, v):
smaller = []
bigger = []
for val in L:
if val < v: smaller += [val]
if val > v: bigger += [val]
return (smaller, [v], bigger)
def top_k(L, k):
v = L[random.randrange(len(L))]
(left, middle, right) = partition(L, v)
# middle used below (in place of [v]) for clarity
if len(left) == k: return left
if len(left)+1 == k: return left + middle
if len(left) > k: return top_k(left, k)
return left + middle + top_k(right, k - len(left) - len(middle))
def median(L):
n = len(L)
l = top_k(L, n / 2 + 1)
return max(l)
No, there is no O(n) algorithm for finding the median of an arbitrary, unsorted dataset.
At least none that I am aware of in 2022. All answers offered here are variations/combinations using heaps, Median of Medians, Quickselect, all of which are strictly O(nlogn).
See https://en.wikipedia.org/wiki/Median_of_medians and http://cs.indstate.edu/~spitla/abstract2.pdf.
The problem appears to be confusion about how algorithms are classified, which is according their limiting (worst case) behaviour. "On average" or "typically" O(n) with "worst case" O(f(n)) means (in textbook terms) "strictly O(f(n))". Quicksort for example, is often discussed as being O(nlogn) (which is how it typically performs), although it is in fact an O(n^2) algorithm because there is always some pathological ordering of inputs for which it can do no better than n^2 comparisons.
It can be done using Quickselect Algorithm in O(n), do refer to Kth order statistics (randomized algorithms).
As wikipedia says, Median-of-Medians is theoretically o(N), but it is not used in practice because the overhead of finding "good" pivots makes it too slow.
http://en.wikipedia.org/wiki/Selection_algorithm
Here is Java source for a Quickselect algorithm to find the k'th element in an array:
/**
* Returns position of k'th largest element of sub-list.
*
* #param list list to search, whose sub-list may be shuffled before
* returning
* #param lo first element of sub-list in list
* #param hi just after last element of sub-list in list
* #param k
* #return position of k'th largest element of (possibly shuffled) sub-list.
*/
static int select(double[] list, int lo, int hi, int k) {
int n = hi - lo;
if (n < 2)
return lo;
double pivot = list[lo + (k * 7919) % n]; // Pick a random pivot
// Triage list to [<pivot][=pivot][>pivot]
int nLess = 0, nSame = 0, nMore = 0;
int lo3 = lo;
int hi3 = hi;
while (lo3 < hi3) {
double e = list[lo3];
int cmp = compare(e, pivot);
if (cmp < 0) {
nLess++;
lo3++;
} else if (cmp > 0) {
swap(list, lo3, --hi3);
if (nSame > 0)
swap(list, hi3, hi3 + nSame);
nMore++;
} else {
nSame++;
swap(list, lo3, --hi3);
}
}
assert (nSame > 0);
assert (nLess + nSame + nMore == n);
assert (list[lo + nLess] == pivot);
assert (list[hi - nMore - 1] == pivot);
if (k >= n - nMore)
return select(list, hi - nMore, hi, k - nLess - nSame);
else if (k < nLess)
return select(list, lo, lo + nLess, k);
return lo + k;
}
I have not included the source of the compare and swap methods, so it's easy to change the code to work with Object[] instead of double[].
In practice, you can expect the above code to be o(N).
Let the problem be: finding the Kth largest element in an unsorted array.
Divide the array into n/5 groups where each group consisting of 5 elements.
Now a1,a2,a3....a(n/5) represent the medians of each group.
x = Median of the elements a1,a2,.....a(n/5).
Now if k<n/2 then we can remove the largets, 2nd largest and 3rd largest element of the groups whose median is greater than the x. We can now call the function again with 7n/10 elements and finding the kth largest value.
else if k>n/2 then we can remove the smallest ,2nd smallest and 3rd smallest element of the group whose median is smaller than the x. We can now call the function of again with 7n/10 elements and finding the (k-3n/10)th largest value.
Time Complexity Analysis:
T(n) time complexity to find the kth largest in an array of size n.
T(n) = T(n/5) + T(7n/10) + O(n)
if you solve this you will find out that T(n) is actually O(n)
n/5 + 7n/10 = 9n/10 < n
Notice that building a heap takes O(n) actually not O(nlogn), you can check this using amortized analysis or simply check in Youtube.
Extract-Min takes O(logn), therefore, extracting n/2 will take (nlogn/2) = O(nlogn) amortized time.
About your question, you can simply check at Median of Medians.
Related
It's quite clear that we see an O(n^2) algorithm to choose the second largest number, and an algorithm using tree style with O(n * Log(n)), but, with extra space cost， like below:
But, eh..., is there a in-place algorithm with time complexity O(n * Log(n)) to select the second largest number in an array/vector?
Yes, in fact you can do this with a single pass over the range without modifying it. Here's an example algorithm:
Let m and M be the second largest, and largest elements. Initialize them to the smallest possible values the input range could contain.
For each number n in the range, the new second largest number depends on the relative order between n, m and M. The 3 possible orderings are n < m < M, m < n < M, or m < M < n. The new second largest element must be m, n, and M respectively. Essentially, n must be clamped between m and M.
The new largest number can't be m, so it must be the larger of n and M.
Here's a demonstration in c++:
int m = 0, M = 0; // assuming a range with non-negative values
for (int n : v)
{
m = std::clamp(n, m, M);
M = std::max(n, M);
}
If you are looking for something very simple O(n):
int getSecondLargest(vector<int>& vec){
int firstLargest = INT_MIN, secondLargest = INT_MIN;
for(auto i: vec){
if(i >= firstLargest){
if(firstLargest != INT_MIN){
secondLargest = firstLargest;
}
firstLargest = i;
}
else if(i > secondLargest){
secondLargest = i;
}
}
return secondLargest;
}
nth_element:
Pros:
If tomorrow you want not the second largest but say fifth largest, you won't need much code changes. The above algorithm I presented won't help.
Cons:
If you are just looking for second largest, nth_element is an overkill. The swaps and/or writes are more as compared to the above algorithm I showed.
Why are you guys giving me O(n) when I am asking for O(nlogn)?
You can find various in-place O(nlogn) sorting algorithms. One of them is Block Sort.
No. I want it to solve it with a tree style and I want O(nlogn) and I want in place. Do you have something like that?
No. That is not possible. When you say in-place, you can't use extra space depending on n. Constant extra space is fine. But tree style would require O(logn) extra space.
I've been trying some coding algorithm exercises and one in particular topic has stood out to me. I've been trying to find out a good answer to this but I've been stuck in analysis paralysis. Let's say I have an array of unsorted integers and I want to determine the xth smallest element in this array.
I know of two options to go about this:
Option 1: Run a sort algorithm, sorting elements least to greatest and look up the xth element. To my understanding, the time complexity to this is O(n*log(n)) and O(1) space.
Option 2: Heapify the array, turning it into a min heap. Then pop() the top of the heap x times. To my understanding, this is O(n) + O(x*log(n)).
I can't tell which is optimal answer and maybe I fundamental misunderstanding of priority queues and when to use them. I've tried to measure runtime and I feel like I'm getting conflicting results. Maybe since with option 2, it depends on how big x is. And maybe there is a better way to go algo. If someone could help, I'd appreciate it!
Worst case time complexity of approach 2 should be O(n + n*log(n)), as maximum value of x = n.
For average case, time complexity = O(n + (1+2+3+....n)/n * log(n)) = O(n + (n+1)*log(n)).
Therefore approach 1 is more efficient than approach 2, but still not optimal.
PS: I would like you to have a look at quick select algorithm which works in O(n) on average case.
This algorithms complexity can revolve around two data points:
Value of x.
Value of n.
Space complexity
In both algos space complexity remains the O(1)
Time complexity
Approach 1
Best Case : O(nlog(n)) for sorting & O(1) for case x == 1;
Average Case : O(nlog(n)) if we consider all elements are unique &
O(x+nlog(n)) if there are duplicates.
Worst Case. : O(n+nlog(n)) for case x==n;
Approach 2:
Best Case : O(n) as just heapify would be require case x==1
Average Case : O(n + xlog(n))
Worst Case. : O(n+nlog(n)) case x==n;
Now Coming to the point to analyze this algo's in runtime.
In general below guidelines are to be followed.
1. Always test for larger values of n.
2. Have a good spread for values being tested(here x).
3. Do multiple iterations of the analysis with clean environment
(array created everytime before the experiment etc) & get the average of all
results.
4. Check for the any predefined functions code complexity for exact implementation.
In this case the sort(can be 2nlogn etc) & various heap operations code.
So if considered above all having idle values.
Method 2 should perform better than Method 1.
Although approach 1 will have less time complexity, but both of these algorithms will use auxiliary space,space complexity of std::sort is O(n). Another way of doing this ,in constant is to do is via binary search. You can do binary search for the xth element . Let l be the smallest element of the array and r be the largest, then time complexity will be O((nlog(r-l)).
int ans=l-1;
while(l<=r){
int mid=(l+r)/2;
int cnt=0;
for(int i=0;i<n;i++){
if(a[i]<=mid)
cnt++;
}
if(cnt<x){
ans=mid;
l=mid+1;
}
else
r=mid-1;
}
Now you can look for the smallest element larger than ans present in the array.
Time complexity-O(nlog(r-l))+O(n)(for the last step)
space complexity-O(1)
You can find xth element in O(n); there are also two simple heap algorithms that improve on your option 2 complexity. I'll start with the latter.
Simple heap algorithm №1: O(x + (n-x) log x) worst-case complexity
Create a max heap out of the first x elements; for each of the remaining elements, pop the max and push them instead:
import heapq
def findKthSmallest(nums: List[int], k: int) -> int:
heap = [-n for n in nums[:k]]
heapq.heapify(heap)
for num in nums[k:]:
if -num > heap[0]:
heapq.heapreplace(heap, -num)
return -heap[0]
Simple heap algorithm №2: O(n + x log x)
Turn the whole array into a min heap, and insert its root into an auxiliary min heap.
k-1 times pop an element from the second heap, and push back its children from the first heap.
Return the root of the second heap.
import heapq
def findKthSmallest(nums: List[int], k: int) -> int:
x = nums.copy()
heapq.heapify(x)
s = [(x[0], 0)] #auxiliary heap
for _ in range(k-1):
ind = heapq.heappop(s)[1]
if 2*ind+1 < len(x):
heapq.heappush(s, (x[2*ind+1], 2*ind+1))
if 2*ind+2 < len(x):
heapq.heappush(s, (x[2*ind+2], 2*ind+2))
return s[0][0]
Which of these is faster? It depends on values of x and n.
A more complicated Frederickson algorithm would allow you to find xth smallest element in a heap in O(x), but that would be overkill, since xth smallest element in unsorted array can be found in O(n) worst-case time.
Median-of-medians algorithm: O(n) worst-case time
Described in [1].
Quickselect algorithm: O(n) average time, O(n^2) worst-case time
def partition(A, lo, hi):
"""rearrange A[lo:hi+1] and return j such that
A[lo:j] <= pivot
A[j] == pivot
A[j+1:hi+1] >= pivot
"""
pivot = A[lo]
if A[hi] > pivot:
A[lo], A[hi] = A[hi], A[lo]
#now A[hi] <= A[lo], and A[hi] and A[lo] need to be exchanged
i = lo
j = hi
while i < j:
A[i], A[j] = A[j], A[i]
i += 1
while A[i] < pivot:
i += 1
j -= 1
while A[j] > pivot:
j -= 1
#now put pivot in the j-th place
if A[lo] == pivot:
A[lo], A[j] = A[j], A[lo]
else:
#then A[right] == pivot
j += 1
A[j], A[hi] = A[hi], A[j]
return j
def quickselect(A, left, right, k):
pivotIndex = partition(A, left, right)
if k == pivotIndex:
return A[k]
elif k < pivotIndex:
return quickselect(A, left, pivotIndex - 1, k)
else:
return quickselect(A, pivotIndex + 1, right, k)
Introselect: O(n) worst-case time
Basically, do quickselect, but if recursion gets too deep, switch to median-of-medians.
import numpy as np
def findKthSmallest(nums: List[int], k: int) -> int:
return np.partition(nums, k, kind='introselect')[k]
Rivest-Floyd algorithm: O(n) average time, O(n^2) worst-case time
Another way to speed up quickselect:
import math
C1 = 600
C2 = 0.5
C3 = 0.5
def rivest_floyd(A, left, right, k):
assert k < len(A)
while right > left:
if right - left > C1:
#select a random sample from A
N = right - left + 1
I = k - left + 1
Z = math.log(N)
S = C2 * math.exp(2/3 * Z) #sample size
SD = C3 * math.sqrt(Z * S * (N - S) / N) * math.copysign(1, I - N/2)
#select subsample such that kth element lies between newleft and newright most of the time
newleft = max(left, k - int(I * S / N + SD))
newright = min(right, k + int((N - I) * S / N + SD))
rivest_floyd(A, newleft, newright, k)
A[left], A[k] = A[k], A[left]
j = partition2(A, left, right)
if j <= k:
left = j+1
if k <= j:
right = j-1
return A[k]
[1]Cormen, Thomas H.; Leiserson, Charles E.; Rivest, Ronald L.; Stein, Clifford (2009) [1990]. Introduction to Algorithms (3rd ed.). MIT Press and McGraw-Hill. ISBN 0-262-03384-4., pp.220-223
I have to find an algorithm for the following problem:
input are two numbers S and k of natural numbers and an unsorted set of n pair-wise different numbers.
decide in O(n) if there is a subset of k numbers that sum up to <=S. Note: k should not be part of the time complexity.
algorithm({x_1, ..., x_n}, k, S):
if exists |{x_i, ..., x_j}| = k and x_i + ... x_j <= S return true
I don't find a solution with time complexity O(n).
What I was able to get is in O(kn), as we search k times the minimum and sum is up:
algorithm(a={x_1, ..., x_n}, k, S):
sum = 0
for i=1,...,k:
min = a.popFirst()
for i=2,...,len(a):
if(a[i] < min):
t = a[i]
a[i] = min
min = t
sum += min
if sum <= S:
return true
else:
return false
this is in O(n) and return the right result. How can i loose the k?
Thanks for helping me, im really struggeling on this one!
Quickselect can be used to find the k smallest elements: https://en.wikipedia.org/wiki/Quickselect
It's basically quicksort, except that you only recurse on the interesting side of the pivot.
A simple implementation runs in O(N) expected time, but using median-of-medians to select a pivot, you can make that a real worst-case bound: https://en.wikipedia.org/wiki/Median_of_medians
You could build a min-heap of size k from the set. Time complexity of building this is O(n) expected time and O(n log k) worst case.
The heap should contain first k minimum elements from the set.
Then it is straightforward to see the sum of the elements in the heap is <= S. You don't need to remove the elements from the heap to calculate the sum. Just traverse the heap to calculate sum. Removing all elements entails k log k complexity.
You don't even need to consider the next higher elements, because adding them would result in sum greater than S
I had a job interview a few weeks ago and I was asked to design a divide and conquer algorithm. I could not solve the problem, but they just called me for a second interview! Here is the question:
we are giving as input two n-element arrays A[0..n − 1] and B[0..n − 1] (which
are not necessarily sorted) of integers, and an integer value. Give an O(nlogn) divide and conquer algorithm that determines if there exist distinct values i, j (that is, i != j) such that A[i] + B[j] = value. Your algorithm should return True if i, j exists, and return False otherwise. You may assume that the elements in A are distinct, and the elements in B are distinct.
can anybody solve the problem? Thanks
My approach is..
Sort any of the array. Here we sort array A. Sort it with the Merge Sort algorithm which is a Divide and Conquer algorithm.
Then for each element of B, Search for Required Value- Element of B in array A by Binary Search. Again this is a Divide and Conquer algorithm.
If you find the element Required Value - Element of B from an Array A then Both element makes pair such that Element of A + Element of B = Required Value.
So here for Time Complexity, A has N elements so Merge Sort will take O(N log N) and We do Binary Search for each element of B(Total N elements) Which takes O(N log N). So total time complexity would be O(N log N).
As you have mentioned you require to check for i != j if A[i] + B[j] = value then here you can take 2D array of size N * 2. Each element is paired with its original index as second element of the each row. Sorting would be done according the the data stored in the first element. Then when you find the element, You can compare both elements original indexes and return the value accordingly.
The following algorithm does not use Divide and Conquer but it is one of the solutions.
You need to sort both the arrays, maintaining the indexes of the elements maybe sorting an array of pairs (elem, index). This takes O(n log n) time.
Then you can apply the merge algorithm to check if there two elements such that A[i]+B[j] = value. This would O(n)
Overall time complexity will be O(n log n)
I suggest using hashing. Even if it's not the way you are supposed to solve the problem, it's worth mentioning since hashing has a better time complexity O(n) v. O(n*log(n)) and that's why more efficient.
Turn A into a hashset (or dictionary if we want i index) - O(n)
Scan B and check if there's value - B[j] in the hashset (dictionary) - O(n)
So you have an O(n) + O(n) = O(n) algorithm (which is better that required (O n * log(n)), however the solution is NOT Divide and Conquer):
Sample C# implementation
int[] A = new int[] { 7, 9, 5, 3, 47, 89, 1 };
int[] B = new int[] { 5, 7, 3, 4, 21, 59, 0 };
int value = 106; // 47 + 59 = A[4] + B[5]
// Turn A into a dictionary: key = item's value; value = item's key
var dict = A
.Select((val, index) => new {
v = val,
i = index, })
.ToDictionary(item => item.v, item => item.i);
int i = -1;
int j = -1;
// Scan B array
for (int k = 0; k < B.Length; ++k) {
if (dict.TryGetValue(value - B[k], out i)) {
// Solution found: {i, j}
j = k;
// if you want any solution then break
break;
// scan further (comment out "break") if you want all pairs
}
}
Console.Write(j >= 0 ? $"{i} {j}" : "No solution");
Seems hard to achieve without sorting.
If you leave the arrays unsorted, checking for existence of A[i]+B[j] = Value takes time Ω(n) for fixed i, then checking for all i takes Θ(n²), unless you find a trick to put some order in B.
Balanced Divide & Conquer on the unsorted arrays doesn't seem any better: if you divide A and B in two halves, the solution can lie in one of Al/Bl, Al/Br, Ar/Bl, Ar/Br and this yields a recurrence T(n) = 4 T(n/2), which has a quadratic solution.
If sorting is allowed, the solution by Sanket Makani is a possibility but you do better in terms of time complexity for the search phase.
Indeed, assume A and B now sorted and consider the 2D function A[i]+B[j], which is monotonic in both directions i and j. Then the domain A[i]+B[j] ≤ Value is limited by a monotonic curve j = f(i) or equivalently i = g(j). But strict equality A[i]+B[j] = Value must be checked exhaustively for all points of the curve and one cannot avoid to evaluate f everywhere in the worst case.
Starting from i = 0, you obtain f(i) by dichotomic search. Then you can follow the border curve incrementally. You will perform n step in the i direction, and at most n steps in the j direction, so that the complexity remains bounded by O(n), which is optimal.
Below, an example showing the areas with a sum below and above the target value (there are two matches).
This optimal solution has little to do with Divide & Conquer. It is maybe possible to design a variant based on the evaluation of the sum at a central point, which allows to discard a whole quadrant, but that would be pretty artificial.
Given an array A with N elements I need to find pair (i,j) such that i is not equal to j and if we write the sum A[i]+A[j] for all pairs of (i,j) then it comes at the kth position.
Example : Let N=4 and arrays A=[1 2 3 4] and if K=3 then answer is 5 as we can see it clearly that sum array becomes like this : [3,4,5,5,6,7]
I can't go for all pair of i and j as N can go up to 100000. Please help how to solve this problem
I mean something like this :
int len=N*(N+1)/2;
int sum[len];
int count=0;
for(int i=0;i<N;i++){
for(int j=i+1;j<N;j++){
sum[count]=A[i]+A[j];
count++;
}
}
//Then just find kth element.
We can't go with this approach
A solution that is based on a fact that K <= 50: Let's take the first K + 1 elements of the array in a sorted order. Now we can just try all their combinations. Proof of correctness: let's assume that a pair (i, j) is the answer, where j > K + 1. But there are K pairs with the same or smaller sum: (1, 2), (1, 3), ..., (1, K + 1). Thus, it cannot be the K-th pair.
It is possible to achieve an O(N + K ^ 2) time complexity by choosing the K + 1 smallest numbers using a quickselect algorithm(it is possible to do even better, but it is not required). You can also just the array and get an O(N * log N + K ^ 2 * log K) complexity.
I assume that you got this question from http://www.careercup.com/question?id=7457663.
If k is close to 0 then the accepted answer to How to find kth largest number in pairwise sums like setA + setB? can be adapted quite easily to this problem and be quite efficient. You need O(n log(n)) to sort the array, O(n) to set up a priority queue, and then O(k log(k)) to iterate through the elements. The reversed solution is also efficient if k is near n*n - n.
If k is close to n*n/2 then that won't be very good. But you can adapt the pivot approach of http://en.wikipedia.org/wiki/Quickselect to this problem. First in time O(n log(n)) you can sort the array. In time O(n) you can set up a data structure representing the various contiguous ranges of columns. Then you'll need to select pivots O(log(n)) times. (Remember, log(n*n) = O(log(n)).) For each pivot, you can do a binary search of each column to figure out where it split it in time O(log(n)) per column, and total cost of O(n log(n)) for all columns.
The resulting algorithm will be O(n log(n) log(n)).
Update: I do not have time to do the finger exercise of supplying code. But I can outline some of the classes you might have in an implementation.
The implementation will be a bit verbose, but that is sometimes the cost of a good general-purpose algorithm.
ArrayRangeWithAddend. This represents a range of an array, summed with one value.with has an array (reference or pointer so the underlying data can be shared between objects), a start and an end to the range, and a shiftValue for the value to add to every element in the range.
It should have a constructor. A method to give the size. A method to partition(n) it into a range less than n, the count equal to n, and a range greater than n. And value(i) to give the i'th value.
ArrayRangeCollection. This is a collection of ArrayRangeWithAddend objects. It should have methods to give its size, pick a random element, and a method to partition(n) it into an ArrayRangeCollection that is below n, count of those equal to n, and an ArrayRangeCollection that is larger than n. In the partition method it will be good to not include ArrayRangeWithAddend objects that have size 0.
Now your main program can sort the array, and create an ArrayRangeCollection covering all pairs of sums that you are interested in. Then the random and partition method can be used to implement the standard quickselect algorithm that you will find in the link I provided.
Here is how to do it (in pseudo-code). I have now confirmed that it works correctly.
//A is the original array, such as A=[1,2,3,4]
//k (an integer) is the element in the 'sum' array to find
N = A.length
//first we find i
i = -1
nl = N
k2 = k
while (k2 >= 0) {
i++
nl--
k2 -= nl
}
//then we find j
j = k2 + nl + i + 1
//now compute the sum at index position k
kSum = A[i] + A[j]
EDIT:
I have now tested this works. I had to fix some parts... basically the k input argument should use 0-based indexing. (The OP seems to use 1-based indexing.)
EDIT 2:
I'll try to explain my theory then. I began with the concept that the sum array should be visualised as a 2D jagged array (diminishing in width as the height increases), with the coordinates (as mentioned in the OP) being i and j. So for an array such as [1,2,3,4,5] the sum array would be conceived as this:
3,4,5,6,
5,6,7,
7,8,
9.
The top row are all values where i would equal 0. The second row is where i equals 1. To find the value of 'j' we do the same but in the column direction.
... Sorry I cannot explain this any better!