Sung’s Dev Blog

Subtree of Another Tree

2023-09-12T00:00:00+00:00

572. Subtree of Another Tree

Difficulty: Easy

Related Topics: Tree, DFS, String Matching, Binary Tree, Hash Function

Given the roots of two binary trees root and subRoot, return true if there is a subtree of root with the same structure and node values of subRoot and false otherwise.

A subtree of a binary tree tree is a tree that consists of a node in tree and all of this node’s descendants. The tree tree could also be considered as a subtree of itself.

Example 1:

Input: root = [3,4,5,1,2], subRoot = [4,1,2] Output: true

Example 2:

Input: root = [3,4,5,1,2,null,null,null,null,0], subRoot = [4,1,2] Output: false

Code:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isSubtree(self, root: Optional[TreeNode], subRoot: Optional[TreeNode]) -> bool:
        if not root: return False
        if self.isSameTree(root, subRoot):
            return True
        return self.isSubtree(root.left, subRoot) or self.isSubtree(root.right, subRoot)

    def isSameTree(self, root, subRoot) -> bool:
        if root and subRoot:
            return root.val == subRoot.val and self.isSameTree(root.left, subRoot.left) and self.isSameTree(root.right, subRoot.right)
        return root is subRoot

House Robber II

2023-09-11T00:00:00+00:00

213. House Robber II

Difficulty: Medium

Related Topics: Array, Dynamic Programming

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have a security system connected, and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.

Example 1:

Input: nums = [2,3,2] Output: 3 Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2), because they are adjacent houses.

Example 2:

Input: nums = [1,2,3,1] Output: 4 Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3). Total amount you can rob = 1 + 3 = 4.

Example 3:

Input: nums = [1,2,3] Output: 3

Code:

class Solution:
    def rob(self, nums: List[int]) -> int:
        if len(nums) == 0: return 0
        elif len(nums) == 1: return nums[0]
        elif len(nums) == 2: return max(nums)
        def houserob(house):
            length = len(house)
            dp = [0] * length
            for i in range(length):
                if i < 2:
                    dp[i] = house[i]
                else:
                    dp[i] = max(dp[:i-1]) + house[i]
            
            return max(dp)
        
        return max(houserob(nums[1:]), houserob(nums[:-1]))

Longest Repeating Character Replacement

2023-09-10T00:00:00+00:00

746. Longest Repeating Character Replacement

Difficulty: Medium

Related Topics: Hash Table, String, Sliding Window

You are given a string s and an integer k. You can choose any character of the string and change it to any other uppercase English character. You can perform this operation at most k times.

Return the length of the longest substring containing the same letter you can get after performing the above operations.

Example 1:

Input: s = “ABAB”, k = 2 Output: 4 Explanation: Replace the two ‘A’s with two ‘B’s or vice versa. Example 2:

Input: s = “AABABBA”, k = 1 Output: 4 Explanation: Replace the one ‘A’ in the middle with ‘B’ and form “AABBBBA”. The substring “BBBB” has the longest repeating letters, which is 4. There may exists other ways to achive this answer too.

Code:

from collections import Counter
class Solution:
    def characterReplacement(self, s: str, k: int) -> int:
        ptr1, ptr2 = 0, 0
        ans = 0
        while ptr2 < len(s):
            word = s[ptr1:ptr2+1]
            count = Counter(word)
            if ptr2-ptr1+1 - count[max(count, key=count.get)] <= k:
                ans = max(ans, ptr2-ptr1+1)
                ptr2 += 1
            else:
                ptr1 += 1
                ptr2 = ptr1 + ans - 1
        
        return ans

Min Cost Climbing Stairs

2023-09-09T00:00:00+00:00

746. Min Cost Climbing Stairs

Difficulty: Medium

Related Topics: Array, Dynamic Programming

You are given an integer array cost where cost[i] is the cost of ith step on a staircase. Once you pay the cost, you can either climb one or two steps.

You can either start from the step with index 0, or the step with index 1.

Return the minimum cost to reach the top of the floor.

Example 1:

Input: cost = [10,15,20] Output: 15 Explanation: You will start at index 1.

Pay 15 and climb two steps to reach the top. The total cost is 15.

Example 2:

Input: cost = [1,100,1,1,1,100,1,1,100,1] Output: 6 Explanation: You will start at index 0.

Pay 1 and climb two steps to reach index 2.
Pay 1 and climb two steps to reach index 4.
Pay 1 and climb two steps to reach index 6.
Pay 1 and climb one step to reach index 7.
Pay 1 and climb two steps to reach index 9.
Pay 1 and climb one step to reach the top. The total cost is 6.

Code:

class Solution:
    def minCostClimbingStairs(self, cost: List[int]) -> int:
        res = [0] * len(cost)

        for i in range(len(cost)):
            if i == 0 or i == 1:
                res[i] = cost[i]
            else:
                res[i] = min(res[i-1], res[i-2]) + cost[i]
        
        return min(res[-1], res[-2])

Subset II

2023-09-08T00:00:00+00:00

90. Subset II

Difficulty: Medium

Related Topics: Array, Backtracking, Bit Manipulation

Given an integer array nums that may contain duplicates, return all possible subsets (the power set).

The solution set must not contain duplicate subsets. Return the solution in any order.

Example 1:

Input: nums = [1,2,2] Output: [[],[1],[1,2],[1,2,2],[2],[2,2]]

Example 2:

Input: nums = [0] Output: [[],[0]]

Code 1:

class Solution:
    def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
        res = []
        nums.sort()

        def backtrack(num, tmp):
            if num == k:
                res.append(tmp[:])
                return
            
            for i in range(num, len(nums)):
                tmp.append(nums[i])
                backtrack(i+1, tmp)
                tmp.pop()
        
        for k in range(len(nums)+1):
            backtrack(0, [])

        ans = []
        for elem in res:
            if elem not in ans:
                ans.append(elem)
        return ans

Code 2:

class Solution:
    def subsetsWithDup(self, nums: List[int]) -> List[List[int]]:
        res = []
        nums.sort()

        def backtrack(ind, tmp):
            if ind == len(nums):
                res.append(tmp[:])
                return
            
            # All subsets that include nums[ind]
            tmp.append(nums[ind])
            backtrack(ind + 1, tmp)
            tmp.pop()
            # All subsets that don't include nums[ind]
            while ind + 1 < len(nums) and nums[ind] == nums[ind+1]:
                ind += 1
            backtrack(ind + 1, tmp)
        
        backtrack(0, [])
        return res

Combination sum

2023-09-07T00:00:00+00:00

39. Combination Sum

Difficulty: Medium

Related Topics: Array, Backtracking

Given an array of distinct integers candidates and a target integer target, return a list of all unique combinations of candidates where the chosen numbers sum to target. You may return the combinations in any order. The same number may be chosen from candidates an unlimited number of times. Two combinations are unique if the frequency of at least one of the chosen numbers is different. The test cases are generated such that the number of unique combinations that sum up to target is less than 150 combinations for the given input.

Example 1:

Input: candidates = [2,3,6,7], target = 7 Output: [[2,2,3],[7]] Explanation: 2 and 3 are candidates, and 2 + 2 + 3 = 7. Note that 2 can be used multiple times. 7 is a candidate, and 7 = 7. These are the only two combinations.

Example 2:

Input: candidates = [2,3,5], target = 8 Output: [[2,2,2,2],[2,3,3],[3,5]]

Example 3:

Input: candidates = [2], target = 1 Output: []

Code 1:

class Solution:
    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
        res = []
        n = len(candidates)

        def backTracking(num, tmp):
            if num == target:
                res.append(sorted(tmp[:]))
                return
            elif num > target:
                return

            for i in range(n):
                if num + candidates[i] <= target:
                    tmp.append(candidates[i])
                    backTracking(num + candidates[i], tmp)
                    tmp.pop()
        backTracking(0, [])

        ans = []
        for elem in res:
            if elem not in ans:
                ans.append(elem)
        return ans

Code 2:

class Solution:
    def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
        res = []
        n = len(candidates)
        """
        Decison Tree:
        Divide into two: Containing candidates[i], Not Containing candidates[i]
        """

        ans = []
        for elem in res:
            if elem not in ans:
                ans.append(elem)
        return ans

        def backTracking(i, cur, tmp):
            if cur == target:
                res.append(tmp[:])
                return
            if i >= n or cur > target:
                return
            
            # Divide into two branches
            tmp.append(candidates[i])
            backTracking(i, cur + candidates[i], tmp)
            tmp.pop()
            backTracking(i+1, cur, tmp)
        
        backTracking(0, 0, [])
        return res

Permutations

2023-09-06T00:00:00+00:00

46. Permutations

Difficulty: Medium

Related Topics: Array, Backtracking

Given an array nums of distinct integers, return all the possible permutations. You can return the answer in any order.

Example 1:

Input: nums = [1,2,3] Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]

Example 2:

Input: nums = [0,1] Output: [[0,1],[1,0]]

Example 3:

Input: nums = [1] Output: [[1]]

Code:

class Solution:
    def permute(self, nums: List[int]) -> List[List[int]]:
        res = []
        n = len(nums)
        check = [False] * n

        def recur(num, tmp):
            if num == n:
                res.append(tmp.copy())
                return
                
            for i in range(n):
                if check[i] == False:
                    check[i] = True
                    tmp.append(nums[i])
                    recur(num + 1, tmp)
                    check[i] = False
                    tmp.pop()
        
        recur(0, [])
        return res

Car Fleet

2023-09-05T00:00:00+00:00

853. Car Fleet

Difficulty: Medium

Related Topics: Array, Stack, Monotonic Stack, Sorting

There are n cars going to the same destination along a one-lane road. The destination is target miles away. You are given two integer array position and speed, both of length n, where position[i] is the position of the ith car and speed[i] is the speed of the ith car (in miles per hour). A car can never pass another car ahead of it, but it can catch up to it and drive bumper to bumper at the same speed. The faster car will slow down to match the slower car’s speed. The distance between these two cars is ignored (i.e., they are assumed to have the same position). A car fleet is some non-empty set of cars driving at the same position and same speed. Note that a single car is also a car fleet. If a car catches up to a car fleet right at the destination point, it will still be considered as one car fleet. Return the number of car fleets that will arrive at the destination.

Example 1:

Input: target = 12, position = [10,8,0,5,3], speed = [2,4,1,1,3] Output: 3 Explanation: The cars starting at 10 (speed 2) and 8 (speed 4) become a fleet, meeting each other at 12. The car starting at 0 does not catch up to any other car, so it is a fleet by itself. The cars starting at 5 (speed 1) and 3 (speed 3) become a fleet, meeting each other at 6. The fleet moves at speed 1 until it reaches target. Note that no other cars meet these fleets before the destination, so the answer is 3.

Example 2:

Input: target = 10, position = [3], speed = [3] Output: 1 Explanation: There is only one car, hence there is only one fleet.

Example 3:

Input: target = 100, position = [0,2,4], speed = [4,2,1] Output: 1 Explanation: The cars starting at 0 (speed 4) and 2 (speed 2) become a fleet, meeting each other at 4. The fleet moves at speed 2. Then, the fleet (speed 2) and the car starting at 4 (speed 1) become one fleet, meeting each other at 6. The fleet moves at speed 1 until it reaches target.

Code:

class Solution:
    def carFleet(self, target: int, position: List[int], speed: List[int]) -> int:
        stack = []
        for pos, v in sorted(zip(position, speed), reverse = True):
            dist = target - pos
            time = dist / v

            if not stack:
                stack.append(time)
            elif time > stack[-1]:
                stack.append(time)
        
        return len(stack)

Subsets

2023-09-04T00:00:00+00:00

78. Subsets

Difficulty: Medium

Related Topics: Array, Backtracking, Bit Manipulation

Given an integer array nums of unique elements, return all possible subsets (the power set).

The solution set must not contain duplicate subsets. Return the solution in any order.

Example 1:

Input: nums = [1,2,3] Output: [[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]]

Example 2:

Input: nums = [0] Output: [[],[0]]

Code:

class Solution:
    def subsets(self, nums: List[int]) -> List[List[int]]:
        def backtrack(ind, tmp):
            if len(tmp) == k:
                res.append(tmp[:])
                return
            
            for i in range(ind, n):
                tmp.append(nums[i])
                backtrack(i + 1, tmp)
                tmp.pop()

        res = []
        n = len(nums)
        for k in range(n+1):
            backtrack(0, [])
        return res

Letter Combinations of a Phone Number

2023-09-03T00:00:00+00:00

17. Letter Combinations of a Phone Number

Difficulty: Medium

Related Topics: Hash Table, String, Backtracking

Given a string containing digits from 2-9 inclusive, return all possible letter combinations that the number could represent. Return the answer in any order.

A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.

Example 1:

Input: digits = “23” Output: [“ad”,”ae”,”af”,”bd”,”be”,”bf”,”cd”,”ce”,”cf”]

Example 2:

Input: digits = “” Output: []

Example 3:

Input: digits = “2” Output: [“a”,”b”,”c”]

Code:

class Solution:
    def letterCombinations(self, digits: str) -> List[str]:
        res = []
        if not digits:
            return res
        hashmap = {
            "2": "abc",
            "3": "def",
            "4": "ghi",
            "5": "jkl",
            "6": "mno",
            "7": "pqrs",
            "8": "tuv",
            "9": "wxyz"
        }
        res = []
        
        def backtracking(ind, letter):
            nonlocal res
            if ind == len(digits):
                res.append(letter)
                return      ## Return to break out from recursion

            num = digits[ind]
            chars = hashmap[num]
            for i in chars:
                backtracking(ind+1, letter+i)
        
        backtracking(0, "")
        return res