Adding solutions of 200, 203, 206, 208, 210 and top interview questions

Author: Garvit244

100-200q/125.py

@@ -25,4 +25,7 @@ def numDistinct(self, s, t):
         			dp[r][c] = dp[r-1][c]
         			if s[r-1] == t[c-1]:
         				dp[r][c] += dp[r-1][c-1]
-        return dp[row][col] 
+        return dp[row][col] 
+
+# Time: O(N^2)
+# Space: O(N^2)

100-200q/127.py

@@ -0,0 +1,62 @@
+'''
+	Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that:
+
+	Only one letter can be changed at a time.
+	Each transformed word must exist in the word list. Note that beginWord is not a transformed word.
+	Note:
+
+	Return 0 if there is no such transformation sequence.
+	All words have the same length.
+	All words contain only lowercase alphabetic characters.
+	You may assume no duplicates in the word list.
+	You may assume beginWord and endWord are non-empty and are not the same.
+	Example 1:
+
+	Input:
+	beginWord = "hit",
+	endWord = "cog",
+	wordList = ["hot","dot","dog","lot","log","cog"]
+
+	Output: 5
+
+	Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
+	return its length 5.
+'''
+
+class Solution(object):
+    def ladderLength(self, beginWord, endWord, wordList):
+        """
+        :type beginWord: str
+        :type endWord: str
+        :type wordList: List[str]
+        :rtype: int
+        """
+        d = {}
+        for word in wordList:
+        	for i in range(len(word)):
+        		s = word[:i] + "_" + word[i+1:]
+        		if s in d:
+        			d[s].append(word)
+        		else:
+        			d[s] = [word]
+
+        queue, visited = [], set()
+        queue.append((beginWord, 1))
+        while queue:
+        	word, steps = queue.pop(0)
+        	if word not in visited:
+        		visited.add(word)
+
+        		if word == endWord:
+        			return steps
+        		else:
+	        		for index in range(len(word)):
+	        			s = word[:index] + "_" + word[index+1:]
+	        			neigh_words = []
+	        			if s in d:
+	        				neigh_words = d[s]
+
+	        			for neigh in neigh_words:
+	        				if neigh not in visited:
+	        					queue.append((neigh, steps+1))
+        return 0

100-200q/128.py

@@ -0,0 +1,31 @@
+'''
+	Given an unsorted array of integers, find the length of the longest consecutive elements sequence.
+
+	Your algorithm should run in O(n) complexity.
+
+	Example:
+
+	Input: [100, 4, 200, 1, 3, 2]
+	Output: 4
+	Explanation: The longest consecutive elements sequence is [1, 2, 3, 4]. Therefore its length is 4.
+'''
+
+class Solution(object):
+    def longestConsecutive(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        result = 0
+        nums = set(nums)
+
+        for num in nums:
+        	if num-1 not in nums:
+        		curr = num
+        		length = 1
+
+        		while curr+1 in nums:
+        			curr += 1
+        			length += 1
+        		result = max(result, length)
+        return result

100-200q/129.py

@@ -0,0 +1,49 @@
+'''
+Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.
+
+An example is the root-to-leaf path 1->2->3 which represents the number 123.
+
+Find the total sum of all root-to-leaf numbers.
+
+Note: A leaf is a node with no children.
+
+Example:
+
+Input: [1,2,3]
+    1
+   / \
+  2   3
+Output: 25
+Explanation:
+The root-to-leaf path 1->2 represents the number 12.
+The root-to-leaf path 1->3 represents the number 13.
+Therefore, sum = 12 + 13 = 25.
+'''
+
+# Definition for a binary tree node.
+# class TreeNode(object):
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution(object):
+    def sumNumbers(self, root):
+        """
+        :type root: TreeNode
+        :rtype: int
+        """
+        if not root:
+        	return 0
+
+        def dfs(root, num, total):
+        	if not root:
+        		return total
+        	num = num*10 + root.val
+        	if not root.left and not root.right:
+        		total += num
+        		return total
+
+        	return dfs(root.left, num) + dfs(root.right, num)
+
+        return dfs(root, 0, 0)

100-200q/130.py

@@ -0,0 +1,26 @@
+'''
+	Given a 2D board containing 'X' and 'O' (the letter O), capture all regions surrounded by 'X'.
+
+	A region is captured by flipping all 'O's into 'X's in that surrounded region.
+
+	Example:
+
+	X X X X
+	X O O X
+	X X O X
+	X O X X
+	After running your function, the board should be:
+
+	X X X X
+	X X X X
+	X X X X
+	X O X X
+'''
+
+class Solution(object):
+    def solve(self, board):
+        """
+        :type board: List[List[str]]
+        :rtype: void Do not return anything, modify board in-place instead.
+        """
+        

100-200q/131.py

@@ -0,0 +1,25 @@
+'''
+	Given a string s, partition s such that every substring of the partition is a palindrome.
+
+	Return all possible palindrome partitioning of s.
+'''
+
+class Solution(object):
+    def partition(self, s):
+    	result = []
+    	def valid(s):
+    		for i in range(len(s)/2):
+    			if s[i] != s[-(i+1)]:
+    				return False
+    		return True
+
+    	def partitionRec(curr, s, i):
+    		if i == len(s):
+    			result.append(curr)
+    		else:
+    			for j in range(i, len(s)):
+    				if valid(s[i:j+1]):
+    					partitionRec(curr + [s[i:j+1]], s, j+1)
+
+    	partitionRec([], s, 0)
+    	return result

100-200q/134.py

@@ -0,0 +1,28 @@
+'''
+	There are N gas stations along a circular route, where the amount of gas at station i is gas[i].
+
+	You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to its next station (i+1). You begin the journey with an empty tank at one of the gas stations.
+
+	Return the starting gas station's index if you can travel around the circuit once in the clockwise direction, otherwise return -1.
+'''
+
+class Solution(object):
+    def canCompleteCircuit(self, gas, cost):
+        """
+        :type gas: List[int]
+        :type cost: List[int]
+        :rtype: int
+        """
+        start, curr_sum, total_sum =0, 0, 0
+        for index in range(len(gas)):
+            diff = gas[index] - cost[index]
+            total_sum += diff
+            curr_sum += diff
+            
+            if curr_sum < 0:
+                start = index + 1
+                curr_sum = 0
+                
+        if total_sum >= 0:
+            return start
+        return -1

100-200q/139.py

@@ -25,5 +25,5 @@ def wordBreak(self, s, wordDict):
         	for j in range(i, -1, -1):
         		if dp[j] and s[j:i+1] in wordDict:
         			dp[i+1] = True
-        			wordBreak
+        			break
         return dp[len(s)]

100-200q/141.py

@@ -0,0 +1,30 @@
+'''
+	Given a linked list, determine if it has a cycle in it.
+
+	Follow up:
+	Can you solve it without using extra space?
+'''
+
+# Definition for singly-linked list.
+# class ListNode(object):
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution(object):
+    def hasCycle(self, head):
+        """
+        :type head: ListNode
+        :rtype: bool
+        """
+        
+        if not head:
+        	return False
+
+        slow, fast = head,head
+        while fast and fast.next:
+        	slow = slow.next
+        	fast = fast.next.next
+        	if slow == fast:
+        		return True
+        return False

100-200q/148.py

@@ -0,0 +1,62 @@
+'''
+	Sort a linked list in O(n log n) time using constant space complexity.
+
+	Example 1:
+
+	Input: 4->2->1->3
+	Output: 1->2->3->4
+'''
+
+# Definition for singly-linked list.
+# class ListNode(object):
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution(object):
+    def sortList(self, head):
+        """
+        :type head: ListNode
+        :rtype: ListNode
+        """
+
+        if not head or not head.next:
+        	return head
+
+        slow, fast = head, head.next
+
+        while fast.next and fast.next.next:
+        	slow = slow.next
+        	fast = fast.next.next
+
+        head1, head2 = head, slow.next
+        slow.next = None
+        head1 = self.sortList(head1)
+        head2 = self.sortList(head2)
+        head = self.merge(head1, head2)
+        return head
+
+    def merge(self, head1, head2):
+    	if not head1:
+    		return head2
+    	if not head2:
+    		return head1
+
+    	result = ListNode(0)
+    	p = result
+
+    	while head1 and head2:
+    		if head1.val <= head2.val:
+    			p.next = ListNode(head1.val)
+    			head1 = head1.next
+    			p = p.next
+    		else:
+    			p.next = ListNode(head2.val)
+    			head2 = head2.next
+    			p = p.next
+
+    	if head1:
+    		p.next = head1
+    	if head2:
+    		p.next = head2
+    	return result.next