Longest Consecutive Sequence LeetCode problem

data_structures/arrays/longest_consecutive_sequence.py

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+"""
+author: Sarthak Sharma https://github.com/Sarthak950  https://sarthak950.netlify.app
+date:   4 OCT 2023
+Longest Consecutive Sequence Problem from LeetCode
+
+"""
+
+
+    def longestConsecutiveSequence(self, nums: List[int]) -> int:
+        # Create a map of all numbers in the array to their index
+        nmap = defaultdict(int)
+        for i in range(len(nums)):
+            if nums[i] not in nmap:
+                nmap[nums[i]] = i
+
+        # Create a seen array to keep track of whether a number has been seen before
+        seen, ans = [0] * len(nums), 0
+
+        # Iterate through each number in the array
+        for n in nums:
+            # Set curr to the current number, and count to 1
+            curr, count = n, 1
+
+            # If we've already seen this number, skip it
+            if seen[nmap[n]]:
+                continue
+
+            # Otherwise, iterate through all consecutive numbers after curr
+            while curr + 1 in nmap:
+                # Increment curr
+                curr += 1
+
+                # Check if we've seen this number before
+                ix = nmap[curr]
+                if seen[ix]:
+                    # If we have, add it to the count and break out of the loop
+                    count += seen[ix]
+                    break
+                else:
+                    # Otherwise, add it to the seen array and increment the count
+                    seen[ix] = 1
+                    count += 1
+
+            # Add the count to the seen array and update the answer
+            seen[nmap[n]], ans = count, max(ans, count)
+
+        # Return the answer
+        return ans
+
+
+"""
+Idea
+1.  First, we put all the numbers into a dictionary, and note the index of each number.
+    This is to make sure the lookup time when we check if a number is in the list is O(1).
+    We also initialize a seen list for later use.
+
+2.  Then, we loop through the numbers, and for each number n, we check if it's already in seen list.
+    If so, we skip it. Otherwise, we start counting the length of the consecutive sequence starting from n.
+    This is done by checking if n+1 is in the dictionary. If so, we increment the counter, and set the seen status of n to 1.
+    Otherwise, we set the seen status of n to 1, and break the loop.
+    This is because if n+1 is not in the dictionary, then the consecutive sequence starting from n is over.
+    And we don't need to count the length of the consecutive sequence starting from n+1, since we will eventually count it when we reach n+1.
+
+"""