Difficulty: Hard | Asked-in: Google, Amazon, LinkedIn, Walmart, Zoho
Key takeaway: This is an excellent problem to learn problem-solving and step-by-step optimization using sorting and a hash table.
Given an array X[] of n integers, write a program to find the length of the longest consecutive sequence. In other words, we need to find the length of the longest subsequence such that the elements in the subsequence are consecutive integers.
Input: X[] = [4, 7, 1, 2, 8, 10, 3], Output: 4
Explanation: [1, 2, 3, 4] is the longest subsequence of consecutive elements.
Input: X[] = [0, -3, 5, -1, 7, -2, -4, 1, 3], Output: 6
Explanation: There are two longest consecutive sequences of length 6: [-4, -3, -2, -1, 0, 1] and [-2, -1, 0, 1, 2, 3]. So, we return 6 as an output.
Input: X[] = [0, 3, 7, 2, 5, 8, 4, 6, 0, 2, 1], Output: 9
Explanation: Here 2 and 3 are repeated, but all the unique integers are part of the longest consecutive sequence i.e. 0, 1, 2, 3, 4, 5, 6, 7, 8.
The longest consecutive sequence must start from some element in the array. So the basic idea would be to explore each possibility: pick each element in the input and do a linear search to count the length of the longest consecutive sequence starting from that element.
We also keep track of the longest length of consecutive sequences seen so far during this process. The critical question is: how do we implement this? Let's think!
Step 1: We initialize a variable longestLength to track the length of the longest consecutive sequence. longestLength = 0.
Step 2: We run an outer loop from i = 0 to n - 1 to traverse the input array. Inside the loop, we initialize two variables: currLength to store the length of the longest consecutive sequence starting from any element X[i] and currElement to track the consecutive element of the sequence starting from element X[i]. currElement = X[i], currLength = 1.
Step 3: Now we run an inner loop to search the next element in the sequence (currElement + 1) using linear search. If the next element is present in the array, we increment the value of currLength by 1 and move to the next possibility of the consecutive element by incrementing currElement by 1.
This process continues in an inner loop until we find an element missing in the sequence. If the linear search returns false, we stop the inner loop.
while(searchNext(X, n, currElement + 1) == true)
{
currElement = currElement + 1
currLength = currLength + 1
}
Step 4: At the end of the inner loop, if (currLength > longestLength), we update the longestLength with currLength. In other words, the length of the consecutive sequence starting from element X[i] is larger than the length of the longest consecutive sequence calculated until that point.
if(longestLength < currLength)
longestLength = currLength
Step 5: Now we move to the next iteration of the outer loop to do a similar process and calculate the longest consecutive sequence starting from the element X[i+1]. At the end of the outer loop, we return the value stored in the variable longestLength.
bool searchNext(int X[], int n, int k)
{
for (int i = 0; i < n; i = i + 1)
{
if (X[i] == k)
return true;
}
return false;
}
int longestConsecutiveSequenceCpp(int X[], int n)
{
int longestLength = 0;
for (int i = 0; i < n; i = i + 1)
{
int currElement = X[i];
int currLength = 1;
while (searchNext(X, n, currElement + 1))
{
currElement = currElement + 1;
currLength = currLength + 1;
}
longestLength = max(longestLength, currLength);
}
return longestLength;
}
def searchNext(X, n, k):
for i in range(n):
if X[i] == k:
return True
return False
def longestConsecutiveSequencePython(X, n):
longestLength = 0
for i in range(n):
currElement = X[i]
currLength = 1
while searchNext(X, n, currElement + 1):
currElement = currElement + 1
currLength = currLength + 1
longestLength = max(longestLength, currLength)
return longestLength
For each element X[i], we are finding the length of the longest consecutive streak of integers using the inner loop. So overall time complexity = n * Time complexity of finding the longest consecutive streak starting from each element = n * Time complexity of inner while loop.
The time complexity of the inner while loop depends on two things: 1) The length of the longest consecutive streak starting from a given element (This could be n in the worst case) and 2) The time complexity of searching an element in the streak linearly (This is O(n) in the worst case).
The time complexity of the inner while loop in the worst case = n * O(n) = O(n²). This process is repeated for each element of the array. So the overall time Complexity = n * O(n²) = O(n³).
Space complexity = O(1), as we are using a constant number of variables.
Suppose we sort the input and iterate over each element. In that case, it will be easy to find sequences of consecutive numbers because consecutive elements will be lined up linearly next to each other.
Step 1: We initialize two variables, currLength and longestLength, to track the current length of the consecutive sequence and the length of the longest consecutive sequence.
Step 2: We sort the input in increasing order. We can use an efficient in-place sorting algorithm such as heap sort.
Step 3: Now we traverse the sorted array and compare each element X[i] to its previous element X[i - 1].
By the end of the loop, it is possible that the last element X[n - 1] may be part of the longest sequence. In other words, if (X[n - 1] == X[n - 2] + 1), then X[n - 1] is a part of the continuous sequence of X[n - 2], and currLength gets incremented to 1.
After this, the loop will end due to the loop condition in the next iteration, and the updated value of currLength will not be considered for the calculation of longestLength. To handle this, we need to return the maximum of currLength and longestLength by the end of the loop, i.e., return max(currLength, longestLength).
int longestConsecutiveSequenceCpp(int X[], int n)
{
sort(X, X + n);
int longestLength = 1;
int currLength = 1;
for (int i = 1; i < n; i = i + 1)
{
if (X[i] != X[i - 1])
{
if (X[i] == X[i - 1] + 1)
currLength = currLength + 1;
else
{
longestLength = max(longestLength, currLength);
currLength = 1;
}
}
}
return max(longestLength, currLength);
}
def longestConsecutiveSequencePython(X, n):
X = sorted(X)
longestLength = 1
currLength = 1
for i in range(1, n):
if X[i] != X[i - 1]:
if X[i] == X[i - 1] + 1:
currLength = currLength + 1
else:
longestLength = max(longestLength, currLength)
currLength = 1
return max(longestLength, currLength)
Suppose we are using some efficient O(nlogn) sorting algorithm like merge sort or heap sort or quicksort. So time complexity = Time complexity of sorting + Linear traversal of the array = O(nlogn) + O(n) = O(nlogn).
Space complexity: If we use heap sort, O(1), else if we use merge sort, O(n).
In the previous solution, sorting helped us calculate the longest sequence in O(n), but the sorting algorithm still dominates the overall time complexity. The critical question is: how can we optimize the time complexity further? Let’s think.
The solution idea is inspired by the brute force approach. Instead of using linear search to find the next element in the sequence, can we think to use a hash table? As we know, the hash table does fast searching in O(1) time complexity on average. If we observe the problem clearly, there will be two types of elements in the array:
If we know the starting element of any consecutive sequence (Type 1), we can easily calculate the length of the sequence by searching all the next successive elements. So, one solution idea would be to identify all elements of type 1, calculate the consecutive sequence length starting from any such element, and return the max among them.
If we observe the sorted array approach, we are doing a similar process. When we encounter a different starting element, we reset the sequence length and update the max sequence length seen so far. But how do we implement this idea using a hash table? Let's think!
Now, we traverse each element X[i] using a loop:
int longestConsecutiveSequence(int X[], int n)
{
HashTable H
int longestLength = 0
for(int i = 0; i < n; i = i + 1)
H.insert(X[i])
for(int i = 0; i < n; i = i + 1)
{
if (H.search(X[i] - 1) == false)
{
int currLength = 1
int currElement = X[i]
while(H.search(X, currElement + 1) == true)
{
currLength = currLength + 1
currElement = currElement + 1
}
longestLength = max(longestLength, currLength)
}
}
return longestLength
}
int longestConsecutiveSequenceCpp(int X[], int n)
{
unordered_set<int> H;
for (int i = 0; i < n; i = i + 1)
H.insert(X[i]);
int longestLength = 0;
for (int i = 0; i < n; i = i + 1)
{
if (H.find(X[i] - 1) == H.end())
{
int currLength = 1;
int currElement = X[i];
while (H.find(currElement + 1) != H.end())
{
currLength = currLength + 1;
currElement = currElement + 1;
}
longestLength = max(longestLength, currLength);
}
}
return longestLength;
}
int longestConsecutiveSequenceJava(int[] X, int n)
{
HashSet<Integer> H = new HashSet<>();
for (int i = 0; i < n; i = i + 1)
H.add(X[i]);
int longestLength = 0;
for (int i = 0; i < n; i = i + 1)
{
if (H.contains(X[i] - 1) == false)
{
int currLength = 1;
int currElement = X[i];
while (H.contains(currElement + 1))
{
currLength = currLength + 1;
currElement = currElement + 1;
}
longestLength = Math.max(longestLength, currLength);
}
}
return longestLength;
}
def longestConsecutiveSequencePython(X, n):
H = set(X)
longestLength = 0
for i in range(n):
if X[i] - 1 not in H:
currLength = 1
currElement = X[i]
while currElement + 1 in H:
currLength = currLength + 1
currElement = currElement + 1
longestLength = max(longestLength, currLength)
return longestLength
At first sight, the time complexity appears to be quadratic due to the two nested loops. But it requires a closer look because while loop is running only when any element X[i] is the beginning of a sequence. A better idea would be to calculate the count of the critical operations inside the loop for better analysis.
Overall time complexity = Time complexity of inserting n elements into hash table + Time complexity of searching n elements twice = n*O(1) + 2*n*O(1)= O(n).
Space Complexity = O(n), for the hash table.
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