I am solving a problem on LeetCode:
Given an unsorted array of integers nums, return the length of the longest consecutive elements sequence. You must write an algorithm that runs in O(n) time. So for nums = [100,4,200,1,3,2], the output is 4.
The Union Find solution to solve this is as below:
class Solution {
public:
vector<int> parent, sz;
int find(int i) {
if(parent[i]==i) return i;
return parent[i]=find(parent[i]);
}
void merge(int i, int j) {
int p1=find(i);
int p2=find(j);
if(p1==p2) return;
if(sz[p1]>sz[p2]) {
sz[p1]+=sz[p2];
parent[p2]=p1;
} else {
sz[p2]+=sz[p1];
parent[p1]=p2;
}
}
int longestConsecutive(vector<int>& nums) {
sz.resize(nums.size(),1);
parent.resize(nums.size(),0);
iota(begin(parent),end(parent),0);
unordered_map<int, int> m;
for(int i=0; i<nums.size(); i++) {
int n=nums[i];
if(m.count(n)) continue;
if(m.count(n-1)) merge(i,m[n-1]);
if(m.count(n+1)) merge(i,m[n+1]);
m[n]=i;
}
int res=0;
for(int i=0; i<parent.size(); i++) {
if(parent[i]==i && sz[i]>res) {
res=sz[i];
}
}
return res;
}
};
This gets accepted by the OJ (Runtime: 80 ms, faster than 76.03% of C++ online submissions for Longest Consecutive Sequence), but is this really O(n), as claimed by many answers, such as this one? My understanding is that Union Find is an O(NlogN) algorithm.
Are they right? Or, am I missing something?
They are right. A properly implemented Union Find with path compression and union by rank has linear run time complexity as a whole, while any individual operation has an amortized constant run time complexity. The exact complexity of m operations of any type is O(m * alpha(n)) where alpha is the inverse Ackerman function. For any possible n in the physical world, the inverse Ackerman function doesn't exceed 4. Thus, we can state that individual operations are constant and algorithm as a whole linear.
The key part for path compression in your code is here:
return parent[i]=find(parent[i])
vs the following that doesn't employ path compression:
return find(parent[i])
What this part of the code does is that it flattens the structure of the nodes in the hierarchy and links each node directly to the final root. Only in the first run of find will you traverse the whole structure. The next time you'll get a direct hit since you set the node's parent to its ultimate root. Notice that the second code snippet works perfectly fine, but it just does redundant work when you are not interested in the path itself and only in the final root.
Union by rank is evident here:
if(sz[p1]>sz[p2]) {...
It makes sure that the node with more children becomes the root of the node with less children. Therefore, less nodes need to be reassigned a new parent, hence less work.
Note: The above was updated and corrected based on feedback from #Matt-Timmermans and #kcsquared.
Related
The problem is: Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.
For example, given
s = "hithere",
dict = ["hi", "there"].
Return true because "hithere" can be segmented as "leet code".
My implementation is as below. This code is ok for normal cases. However, it suffers a lot for input like:
s = "aaaaaaaaaaaaaaaaaaaaaaab", dict = {"aa", "aaaaaa", "aaaaaaaa"}.
I want to memorize the processed substrings, however, I cannot done it right. Any suggestion on how to improve? Thanks a lot!
class Solution {
public:
bool wordBreak(string s, unordered_set<string>& wordDict) {
int len = s.size();
if(len<1) return true;
for(int i(0); i<len; i++) {
string tmp = s.substr(0, i+1);
if((wordDict.find(tmp)!=wordDict.end())
&& (wordBreak(s.substr(i+1), wordDict)) )
return true;
}
return false;
}
};
It's logically a two-step process. Find all dictionary words within the input, consider the found positions (begin/end pairs), and then see if those words cover the whole input.
So you'd get for your example
aa: {0,2}, {1,3}, {2,4}, ... {20,22}
aaaaaa: {0,6}, {1,7}, ... {16,22}
aaaaaaaa: {0,8}, {1,9} ... {14,22}
This is a graph, with nodes 0-23 and a bunch of edges. But node 23 b is entirely unreachable - no incoming edge. This is now a simple graph theory problem
Finding all places where dictionary words occur is pretty easy, if your dictionary is organized as a trie. But even an std::map is usable, thanks to its equal_range method. You have what appears to be an O(N*N) nested loop for begin and end positions, with O(log N) lookup of each word. But you can quickly determine if s.substr(begin,end) is a still a viable prefix, and what dictionary words remain with that prefix.
Also note that you can build the graph lazily. Staring at begin=0 you find edges {0,2}, {0,6} and {0,8}. (And no others). You can now search nodes 2, 6 and 8. You even have a good algorithm - A* - that suggests you try node 8 first (reachable in just 1 edge). Thus, you'll find nodes {8,10}, {8,14} and {8,16} etc. As you see, you'll never need to build the part of the graph that contains {1,3} as it's simply unreachable.
Using graph theory, it's easy to see why your brute-force method breaks down. You arrive at node 8 (aaaaaaaa.aaaaaaaaaaaaaab) repeatedly, and each time search the subgraph from there on.
A further optimization is to run bidirectional A*. This would give you a very fast solution. At the second half of the first step, you look for edges leading to 23, b. As none exist, you immediately know that node {23} is isolated.
In your code, you are not using dynamic programming because you are not remembering the subproblems that you have already solved.
You can enable this remembering, for example, by storing the results based on the starting position of the string s within the original string, or even based on its length (because anyway the strings you are working with are suffixes of the original string, and therefore its length uniquely identifies it). Then, in the beginning of your wordBreak function, just check whether such length has already been processed and, if it has, do not rerun the computations, just return the stored value. Otherwise, run computations and store the result.
Note also that your approach with unordered_set will not allow you to obtain the fastest solution. The fastest solution that I can think of is O(N^2) by storing all the words in a trie (not in a map!) and following this trie as you walk along the given string. This achieves O(1) per loop iteration not counting the recursion call.
Thanks for all the comments. I changed my previous solution to the implementation below. At this point, I didn't explore to optimize on the dictionary, but those insights are very valuable and are very much appreciated.
For the current implementation, do you think it can be further improved? Thanks!
class Solution {
public:
bool wordBreak(string s, unordered_set<string>& wordDict) {
int len = s.size();
if(len<1) return true;
if(wordDict.size()==0) return false;
vector<bool> dq (len+1,false);
dq[0] = true;
for(int i(0); i<len; i++) {// start point
if(dq[i]) {
for(int j(1); j<=len-i; j++) {// length of substring, 1:len
if(!dq[i+j]) {
auto pos = wordDict.find(s.substr(i, j));
dq[i+j] = dq[i+j] || (pos!=wordDict.end());
}
}
}
if(dq[len]) return true;
}
return false;
}
};
Try the following:
class Solution {
public:
bool wordBreak(string s, unordered_set<string>& wordDict)
{
for (auto w : wordDict)
{
auto pos = s.find(w);
if (pos != string::npos)
{
if (wordBreak(s.substr(0, pos), wordDict) &&
wordBreak(s.substr(pos + w.size()), wordDict))
return true;
}
}
return false;
}
};
Essentially one you find a match remove the matching part from the input string and so continue testing on a smaller input.
I created a program that finds the median of a list of numbers. The list of numbers is dynamic in that numbers can be removed and inserted (duplicate numbers can be entered) and during this time, the new median is re-evaluated and printed out.
I created this program using a multimap because
1) the benefit of it being already being sorted,
2) easy insertion, deletion, searching (since multimap implements binary search)
3) duplicate entries are allowed.
The constraints for the number of entries + deletions (represented as N) are: 0 < N <= 100,000.
The program I wrote works and prints out the correct median, but it isn't fast enough. I know that the unsorted_multimap is faster than multimap, but then the problem with unsorted_multimap is that I would have to sort it. I have to sort it because to find the median you need to have a sorted list. So my question is, would it be practical to use an unsorted_multimap and then quick sort the entries, or would that just be ridiculous? Would it be faster to just use a vector, quicksort the vector, and use a binary search? Or maybe I am forgetting some fabulous solution out there that I haven't even thought of.
Though I'm not new to C++, I will admit, that my skills with time-complexity are somewhat medicore.
The more I look at my own question, the more I'm beginning to think that just using a vector with quicksort and binary search would be better since the data structures basically already implement vectors.
the more I look at my own question, the more I'm beginning to think that just using vector with quicksort and binary search would be better since the data structures basically already implement vectors.
If you have only few updates - use unsorted std::vector + std::nth_element algorithm which is O(N). You don't need full sorting which is O(N*ln(N)).
live demo of nth_element:
#include <algorithm>
#include <iterator>
#include <iostream>
#include <ostream>
#include <vector>
using namespace std;
template<typename RandomAccessIterator>
RandomAccessIterator median(RandomAccessIterator first,RandomAccessIterator last)
{
RandomAccessIterator m = first + distance(first,last)/2; // handle even middle if needed
nth_element(first,m,last);
return m;
}
int main()
{
vector<int> values = {5,1,2,4,3};
cout << *median(begin(values),end(values)) << endl;
}
Output is:
3
If you have many updates and only removing from middle - use two heaps as comocomocomocomo suggests. If you would use fibonacci_heap - then you would also get O(N) removing from arbitary position (if don't have handle to it).
If you have many updates and need O(ln(N)) removing from arbitary places - then use two multisets as ipc suggests.
If your purpose is to keep track of the median on the fly, as elements are inserted/removed, you should use a min-heap and a max-heap. Each one would contain one half of the elements... There was a related question a couple of days ago: How to implement a Median-heap
Though, if you need to search for specific values in order to remove elements, you still need some kind of map.
You said that it is slow. Are you iterating from the beginning of the map to the (N/2)'th element every time you need the median? You don't need to. You can keep track of the median by maintaining an iterator pointing to it at all times and a counter of the number of elements less than that one. Every time you insert/remove, compare the new/old element with the median and update both iterator and counter.
Another way of seeing it is as two multimaps containing half the elements each. One holds the elements less than the median (or equal) and the other holds those greater. The heaps do this more efficiently, but they don't support searches.
If you only need the median a few times you can use the "select" algorithm. It is described in Sedgewick's book. It takes O(n) time on average. It is similar to quick sort but it does not sort completely. It just partitions the array with random pivots until, eventually, it gets to "select" on one side the smaller m elements (m=(n+1)/2). Then you search for the greatest of those m elements, and this is the median.
Here is how you could implement that in O(log N) per update:
template <typename T>
class median_set {
public:
std::multiset<T> below, above;
// O(log N)
void rebalance()
{
int diff = above.size() - below.size();
if (diff > 0) {
below.insert(*above.begin());
above.erase(above.begin());
} else if (diff < -1) {
above.insert(*below.rbegin());
below.erase(below.find(*below.rbegin()));
}
}
public:
// O(1)
bool empty() const { return below.empty() && above.empty(); }
// O(1)
T const& median() const
{
assert(!empty());
return *below.rbegin();
}
// O(log N)
void insert(T const& value)
{
if (!empty() && value > median())
above.insert(value);
else
below.insert(value);
rebalance();
}
// O(log N)
void erase(T const& value)
{
if (value > median())
above.erase(above.find(value));
else
below.erase(below.find(value));
rebalance();
}
};
(Work in action with tests)
The idea is the following:
Keep track of the values above and below the median in two sets
If a new value is added, add it to the corresponding set. Always ensure that the set below has exactly 0 or 1 more then the other
If a value is removed, remove it from the set and make sure that the condition still holds.
You can't use priority_queues because they won't let you remove one item.
Can any one help me what is Space and Time complexity of my following C# program with details.
//Passing Integer array to Find Extreme from that Integer Array
public int extreme(int[] A)
{
int N = A.Length;
if (N == 0)
{
return -1;
}
else
{
int average = CalculateAverage(A);
return FindExtremes(A, average);
}
}
// Calaculate Average of integerArray
private int CalculateAverage(int[] integerArray)
{
int sum = 0;
foreach (int value in integerArray)
{
sum += value;
}
return Convert.ToInt32(sum / integerArray.Length);
}
//Find Extreme from that Integer Array
private int FindExtremes(int[] integerArray, int average) {
int Index = -1; int ExtremeElement = integerArray[0];
for (int i = 0; i < integerArray.Length; i++)
{
int absolute = Math.Abs(integerArray[i] - average);
if (absolute > ExtremeElement)
{
ExtremeElement = integerArray[i];
Index = i;
}
}
return Index;
}
You are almost certainly better off using a vector. Possibly maintaining an auxiliary vector of indexes to be removed between median calculations so you can delete them in batches. New additions can also be put into an auxiliary vector, sorted, then merged in.
I was trying a problem on SPOJ,in which we have to simply find the Length of Longest Increasing Sub-sequence of the given Array A.
I had solved this problem using a dynamic programming O(n^2)algorithm and the solution got accepted..Here is the code,that got Accepted:
void LIS(int *A,int A_Length)
{
int Seq[MAX];
for(int i=0;i<A_Length;++i)
{
int maxima=0;
for(int j=0;j<i;++j)
{
if(A[i]>A[j])
{
maxima=max(Seq[j],maxima);
}
}
Seq[i]=maxima+1;
//cout<<Seq[i]<<endl;
}
cout<<*max_element(Seq,Seq+A_Length)<<endl;
}
But When i tried to solve it using the Second Method (LINK),which is ::
A simple way of finding the longest increasing subsequence is
to use the Longest Common Subsequence (Dynamic Programming) algorithm.
[1]Make a sorted copy of the sequence A, denoted as B. O(nlog(n)) time.
[2]Use Longest Common Subsequence on with A and B. O(n2) time.
,I got Wrong Answer .
This is my c++ code
//Global Variable
int A[100],B[100];
int DP[100][100];
//This function Finds the Longest common subsequce of Array A[1,2,3...,N] and B[1,2,3...,N]
void LIS(int N)
{
sort((B+1),(B+1)+N);//STL SORT sort from index 1 to N of Array B.
int i,j;
//Base Cases
for(i=0;i<=N;++i)
DP[i][0]=0;
for(j=0;j<=N;++j)
DP[0][j]=0;
for(i=1;i<=N;++i)
{
for(j=1;j<=N;++j)
{
if(A[i]==B[j])
DP[i][j]=DP[i-1][j-1]+1;
else
DP[i][j]=max(DP[i-1][j],DP[i][j-1]);
}
}
printf("%d\n",DP[N][N]);
}
int main()
{
int N,i;
scanf("%d",&N);
for(i=1;i<=N;++i)
{
scanf("%d",&A[i]);
B[i]=A[i];
}
LIS(N);
return 0;
}
I don't know why i am getting the Wrong Answer.Can You please Help me in Finding the Bug. Or the LIS by LCS Algorithm given in the site is incorrect??
The Second Method is correct, but can't be applied to this problem directly. That's because the numbers in the sequence are not guaranteed to be unique in this SPOJ problem, and the target is to find a strict increasing subsequence, while Your Second Method's output is non-decreasing subsequence here. Demonstrating on a simple test case [1,2,2,3] will help you find the difference.
This solution is also simple: just remove the duplicated elements after sorting.
I have coded DFS as the way it is on my mind and didn't referred any Text book or Pseudo-code for ideas. I think I have some lines of codes that are making unnecessary calculations. Any ideas on reducing the complexity of my algorithm ?
vector<int>visited;
bool isFound(vector<int>vec,int value)
{
if(std::find(vec.begin(),vec.end(),value)==vec.end())
return false;
else
return true;
}
void dfs(int **graph,int numOfNodes,int node)
{
if(isFound(visited,node)==false)
visited.push_back(node);
vector<int>neighbours;
for(int i=0;i<numOfNodes;i++)
if(graph[node][i]==1)
neighbours.push_back(i);
for(int i=0;i<neighbours.size();i++)
if(isFound(visited,neighbours[i])==false)
dfs(graph,numOfNodes,neighbours[i]);
}
void depthFirstSearch(int **graph,int numOfNodes)
{
for(int i=0;i<numOfNodes;i++)
dfs(graph,numOfNodes,i);
}
PS: Could somebody please sent me a link teaching me how can to insert C++ code with good quality. I've tried syntax highlighting but it didn't work out.
Your DFS has O(n^2) time complexity, which is really bad (it should run in O(n + m)).
This line ruins your implementation, because searching in vector takes time proportional to its length:
if(std::find(vec.begin(),vec.end(),value)==vec.end())
To avoid this, you can remember what was visited in an array of boolean values.
Second problem with your DFS is that for bigger graph it will probably cause stack overflow, because worst case recursion depth is equal to number of vertices in graph. Remedy to this problem is also simple: use std::list<int> as your own stack.
So, code that does DFS should look more or less like this:
// n is number of vertices in graph
bool visited[n]; // in this array we save visited vertices
std::list<int> stack;
std::list<int> order;
for(int i = 0; i < n; i++){
if(!visited[i]){
stack.push_back(i);
while(!stack.empty()){
int top = stack.back();
stack.pop_back();
if(visited[top])
continue;
visited[top] = true;
order.push_back(top);
for(all neighbours of top)
if(!visited[neighbour])
stack.push_back(neighbour);
}
}
}
This is not a homework.
I'm using a small "priority queue" (implemented as array at the moment) for storing last N items with smallest value. This is a bit slow - O(N) item insertion time. Current implementation keeps track of largest item in array and discards any items that wouldn't fit into array, but I still would like to reduce number of operations further.
looking for a priority queue algorithm that matches following requirements:
queue can be implemented as array, which has fixed size and _cannot_ grow. Dynamic memory allocation during any queue operation is strictly forbidden.
Anything that doesn't fit into array is discarded, but queue keeps all smallest elements ever encountered.
O(log(N)) insertion time (i.e. adding element into queue should take up to O(log(N))).
(optional) O(1) access for *largest* item in queue (queue stores *smallest* items, so the largest item will be discarded first and I'll need them to reduce number of operations)
Easy to implement/understand. Ideally - something similar to binary search - once you understand it, you remember it forever.
Elements need not to be sorted in any way. I just need to keep N smallest value ever encountered. When I'll need them, I'll access all of them at once. So technically it doesn't have to be a queue, I just need N last smallest values to be stored.
I initially thought about using binary heaps (they can be easily implemented via arrays), but apparently they don't behave well when array can't grow anymore. Linked lists and arrays will require extra time for moving things around. stl priority queue grows and uses dynamic allocation (I may be wrong about it, though).
So, any other ideas?
--EDIT--
I'm not interested in STL implementation. STL implementation (suggested by a few people) works a bit slower than currently used linear array due to high number of function calls.
I'm interested in priority queue algorithms, not implemnetations.
Array based heaps seem ideal for your purpose. I am not sure why you rejected them.
You use a max-heap.
Say you have an N element heap (implemented as an array) which contains the N smallest elements seen so far.
When an element comes in you check against the max (O(1) time), and reject if it is greater.
If the value coming in is lower, you modify the root to be the new value and sift-down this changed value - worst case O(log N) time.
The sift-down process is simple: Starting at root, at each step you exchange this value with it's larger child until the max-heap property is restored.
So, you will not have to do any deletes which you probably will have to, if you use std::priority_queue. Depending on the implementation of std::priority_queue, this could cause memory allocation/deallocation.
So you can have the code as follows:
Allocated Array of size N.
Fill it up with the first N elements you see.
heapify (you should find this in standard text books, it uses sift-down). This is O(N).
Now any new element you get, you either reject it in O(1) time or insert by sifting-down in worst case O(logN) time.
On an average, though, you probably will not have to sift-down the new value all the way down and might get better than O(logn) average insert time (though I haven't tried proving it).
You only allocate size N array once and any insertion is done by exchanging elements of the array, so there is no dynamic memory allocation after that.
Check out the wiki page which has pseudo code for heapify and sift-down: http://en.wikipedia.org/wiki/Heapsort
Use std::priority_queue with the largest item at the head. For each new item, discard it if it is >= the head item, otherwise pop the head item and insert the new item.
Side note: Standard containers will only grow if you make them grow. As long as you remove one item before inserting a new item (after it reaches its maximum size, of course), this won't happen.
Most priority queues I work are based on linked lists. If you have a pre-determined number of priority levels, you can easily create a priority queue with O(1) insertion by having an array of linked lists--one linked list per priority level. Items of the same priority will of course degenerate into either a FIFO, but that can be considered acceptable.
Adding and removal then becomes something like (your API may vary) ...
listItemAdd (&list[priLevel], &item); /* Add to tail */
pItem = listItemRemove (&list[priLevel]); /* Remove from head */
Getting the first item in the queue then becomes a problem of finding the non-empty linked-list with the highest priority. That may be O(N), but there are several tricks you can use to speed it up.
In your priority queue structure, keep a pointer or index or something to the linked list with the current highest priority. This would need to be updated each time an item is added or removed from the priority queue.
Use a bitmap to indicate which linked lists are not empty. Combined with a find most significant bit, or find least significant bit algorithm you can usually test up to 32 lists at once. Again, this would need to be updated on each add / remove.
Hope this helps.
If amount of priorities is small and fixed than you can use ring-buffer for each priority. That will lead to waste of the space if objects is big, but if their size is comparable with pointer/index than variants with storing additional pointers in objects may increase size of array in the same way.
Or you can use simple single-linked list inside array and store 2*M+1 pointers/indexes, one will point to first free node and other pairs will point to head and tail of each priority. In that cases you'll have to compare in avg. O(M) before taking out next node with O(1). And insertion will take O(1).
If you construct an STL priority queue at the maximum size (perhaps from a vector initialized with placeholders), and then check the size before inserting (removing an item if necessary beforehand) you'll never have dynamic allocation during insert operations. The STL implementation is quite efficient.
Matters Computational see page 158. The implementation itself is quite well, and you can even tweak it a little without making it less readable. For example, when you compute the left child like:
int left = i / 2;
You can compute the rightchild like so:
int right = left + 1;
Found a solution ("difference" means "priority" in the code, and maxRememberedResults is 255 (could be any (2^n - 1)):
template <typename T> inline void swap(T& a, T& b){
T c = a;
a = b;
b = c;
}
struct MinDifferenceArray{
enum{maxSize = maxRememberedResults};
int size;
DifferenceData data[maxSize];
void add(const DifferenceData& val){
if (size >= maxSize){
if(data[0].difference <= val.difference)
return;
data[0] = val;
for (int i = 0; (2*i+1) < maxSize; ){
int next = 2*i + 1;
if (data[next].difference < data[next+1].difference)
next++;
if (data[i].difference < data[next].difference)
swap(data[i], data[next]);
else
break;
i = next;
}
}
else{
data[size++] = val;
for (int i = size - 1; i > 0;){
int parent = (i-1)/2;
if (data[parent].difference < data[i].difference){
swap(data[parent], data[i]);
i = parent;
}
else
break;
}
}
}
void clear(){
size = 0;
}
MinDifferenceArray()
:size(0){
}
};
build max-based queue (root is largest)
until it is full, fill up normally
when it is full, for every new element
Check if new element is smaller than root.
if it is larger or equal than root, reject.
otherwise, replace root with new element and perform normal heap "sift-down".
And we get O(log(N)) insert as a worst case scenario.
It is the same solution as the one provided by user with nickname "Moron".
Thanks to everyone for replies.
P.S. Apparently programming without sleeping enough wasn't a good idea.
It's better to implement your own class using std::array and heap algorithms.
`template<class T, int fixed_size = 5>
class fixed_size_arr_pqueue_v2
{
std::array<T, fixed_size> _data;
int _size = 0;
int parent(int i)
{
return (i - 1)/2;
}
void heapify(int i, bool downward = false)
{
int l = 2*i + 1;
int r = 2*i + 2;
int largest = 0;
if (l < size() && _data[l] > _data[i])
largest = l;
else
largest = i;
if (r < size() && _data[r] > _data[largest])
largest = r;
if (largest != i)
{
std::swap(_data[largest], _data[i]);
if (!downward)
heapify(parent(i));
else
heapify(largest, true);
}
}
public:
void push(T &d)
{
if (_size == fixed_size)
{
//min elements in a max heap lies at leaves only.
auto minItr = std::min_element(begin(_data) + _size/2, end(_data));
auto minPos {minItr - _data.begin()};
auto min { *minItr};
if (d > min)
{
_data.at(minPos) = d;
if (_data[parent(minPos)] > d)
{
//this is unlikely to happen in our case? as this position is a leaf?
heapify(minPos, true);
}
else
heapify(parent(minPos));
}
return ;
}
_data.at(_size++) = d;
std::push_heap(_data.begin(), _data.begin() + _size);
}
T pop()
{
T d = _data.front();
std::pop_heap(_data.begin(), _data.begin() + _size);
_size--;
return d;
}
T top()
{
return _data.front();
}
int size() const
{
return _size;
}
};`