[LeetCode] Longest Consecutive Sequence, Solution

Given an unsorted array of integers, find the length of the longest consecutive elements sequence.

For example,
Given  [100, 4, 200, 1, 3, 2],
The longest consecutive elements sequence is  [1, 2, 3, 4]. Return its length:  4.

Your algorithm should run in O( n) complexity.

» Solve this problem

[Thoughts]
Since it requires O(n) solution, normal sort won’t be work here. Hash probably is the best choice.
3 Steps:
1. create the hashmap to hold <num, index>
2. scan the num vector from left to right, for each num
               i, check whether num -1 is in the map  (loop)
              ii, check whether num+1 is in the map  (loop)
3. track the sequence length during scanning.

[Code]

1:       int longestConsecutive(vector<int> &num) {  
2:            unordered_map<int, int> hashmap;            
3:            for(int i =0; i< num.size(); i++)  
4:            {  
5:                 hashmap[num[i]] = i;  
6:            }  
7:            vector<int> visited(num.size(), 0);  
8:            int maxV = INT_MIN;  
9:            for(int i =0; i< num.size(); i++)  
10:            {  
11:                 if(visited[i]==1) continue;  
12:                 visited[i]=1;  
13:                 int len = 1;  
14:                 int index = num[i]-1;  
15:                 while(hashmap.find(index) != hashmap.end())  
16:                 {  
17:                      visited[hashmap[index]] =1;  
18:                      index--;  
19:                      len++;  
20:                 }  
21:                 index = num[i]+1;  
22:                 while(hashmap.find(index) != hashmap.end())  
23:                 {  
24:                      visited[hashmap[index]] =1;  
25:                      index++;  
26:                      len++;  
27:                 }  
28:                 if(len > maxV)  
29:                      maxV = len;  
30:            }  
31:            return maxV;  
32:       }  

Update 08/23/2014
Interesting discussion in the comments:) I haven’t visited my blog for quite long time.

For each num, define D[num] as the longest consecutive sequence from k to num,  0<k<num

So D[num] = D[num-1] +1   if num-1 in the map
                  =0                      if num-1  not in the map

But unfortunately, the unordered_map doesn’t keep any order of sequence. It’s hard to do the DP via a loop.

Here can use Memorization to optimize the code. Since for each fresh node, it only visits once. It is O(n) code.

And in the code , the second ‘for’ loop and the third ‘for’ loop could be merged together, but here keep them separated for readability.

1:       int longestConsecutive(vector<int> &num) {  
2:            unordered_map<int, int> hashmap;  
3:            vector<int> length(num.size(),0);  
4:            for(int i =0; i< num.size(); i++)  
5:            {  
6:                 hashmap[num[i]]=i;  
7:            }  
8:            for(int i =0; i< num.size(); i++)  
9:            {  
10:                 // skip the node, which already calculate the length  
11:                 if(length[i] > 0) continue;                 
12:                 length[i] = consecutiveLength(num[i], hashmap, length);  
13:            }  
14:            int maxV = INT_MIN;  
15:            for(int i =0; i< num.size(); i++)  
16:            {  
17:                 maxV = length[i]>maxV?length[i]:maxV;  
18:            }  
19:            return maxV;  
20:       }  
21:       int consecutiveLength(int num, unordered_map<int, int>& hashmap, vector<int>& length)  
22:       {  
23:            if(hashmap.find(num) == hashmap.end()) return 0;  
24:            int index = hashmap[num];  
25:            // skip the node, which already calculate the length  
26:            if(length[index] > 0) return length[index];  
27:            else  
28:            {  
29:                 // hit each fresh node only once  
30:                 length[index] = consecutiveLength(num - 1, hashmap, length) + 1;  
31:                 return length[index];  
32:            }  
33:       }