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Problem Solution We can easily see that this problem is just elementary school math, except that the digits are divided into nodes in the linked lists. Turn each list into an integer, then add the two numbers up. Finally, turn the sum into a linked list by iterating through the digits. Time Complexity O(N1+N2), where N1 and N2 are the numbers of elements in each input list. Space Complexity O(1)..

Problem Solution Use two pointers that move with different speeds. One moves forward by steps of 1, and the other moves by steps of 2. If the faster pointer reaches the end of the list, there is no loop in the list. If there is a loop in the list, the two pointers will intersect eventually. Return true when that happens. Check for edge cases where the list is empty or there are less than 3 eleme..

Problem Solution The base idea is to go through both lists at the same time, add each nodes, and append it to a result linked list. Make a variable to keep track of the carry for each sum. When making a result linked list, create two pointers for it. One pointer will always point to the head of the list. Another will go forward through the list as the code add elements to the linked list. In eac..

Problem Solution Use 2 pointers from left and right ends. Keep updating the maximum height when iterating from both ends. The tapped water depends on the left pointer's height as long as the maximum height on the right so far is greater than the left. Then the water depends on the right pointer's height when maximum height on the left becomes greater. class Solution(object): def trap(self, heigh..

Problem Solution Two strings are anagrams if the frequencies of appearing alphabets are the same. Make a dictionary, where the keys are tuples of alphabet(a-z) frequencies. The values are the strings in the given list. When I initiated an dictionary and used append function with a new key, it gave a KeyError. To avoid this, use defaultdict function and pass list as parameter. Then a defaultdict ..

Problem Solution Use deque (double-ended queue) to store indices and a result array to keep track of the maximum integers for each window. When pushing an index into the deque, pop every index that contains smaller integer than the current integer. If current integer is less than the integer held by the index at the head, just append current index to the deque. class Solution(object): def maxSli..

Problem Solution class Solution(object): def search(self, nums, target): """ :type nums: List[int] :type target: int :rtype: int """ left = 0 right = len(nums)-1 while left nums[mid]: left = mid + 1 return -1

Problem Solution Start from the center of the string, and expand to left and right until the characters at those positions are not the same. class Solution(object): def longestPalindrome(self, s): """ :type s: str :rtype: str """ if s is None or len(s) < 1: return "" start = 0 end = 0 for i in range(len(s)): len1 = self.expandCenter(s, i, i) len2 = self.expandCenter(s, i, i+1) length = max(len1,..

Problem Solution Iterative: # Definition for singly-linked list. # class ListNode(object): # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution(object): def reverseList(self, head): """ :type head: ListNode :rtype: ListNode """ prev = None curr = head while curr is not None: nextNode = curr.next curr.next = prev prev = curr curr = nextNode return prev Recurs..

Problem Solution Make a dictionary of all appearing alphabets and their frequencies. Make a set of unique frequencies. Go through the dictionary and check if the frequency for this character already exists in the set. If it is, decrement the frequency until it is unique or reaches zero. Increment the counter at the same time. Add the frequency to the set. My code exceeded the time limit at first..