LeetCode

1. Two Sum easy

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

You can return the answer in any order.

 

Example 1:

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].

Example 2:

Input: nums = [3,2,4], target = 6
Output: [1,2]

Example 3:

Input: nums = [3,3], target = 6
Output: [0,1]

 

Constraints:

• 2 <= nums.length <= 104
• -109 <= nums[i] <= 109
• -109 <= target <= 109
• Only one valid answer exists.

 

Follow-up: Can you come up with an algorithm that is less than O(n2) time complexity?

Solutions:

a.Brute force solution:

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        for i in range(len(nums)):
            for j in range(i+1,len(nums)):
                if (nums[i] + nums[j]) == target:
                    return [i,j] 
        

b. Using HashMap

class Solution:
    # This function finds two numbers in the 'nums' list that add up to the 'target' value.
    # It returns a list containing the indices of the two numbers.
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        # Initialize an empty hashmap to store the values and their corresponding indices
        hashmap = {}  
    
        # Iterate through the 'nums' list using indices
        for i in range(len(nums)):  
            # Calculate the complement of the current number
            complement = target - nums[i]  
            
            # If the complement exists in the hashmap
            if complement in hashmap:  
                # Return the indices of the complement and the current number
                return [hashmap[complement], i]  
            
            # Add the current number and its index to the hashmap
            hashmap[nums[i]] = i  

9. Palindrome Number easy

Given an integer x, return true if x is a 

palindrome

 

, and false otherwise.

 

Example 1:

Input: x = 121
Output: true
Explanation: 121 reads as 121 from left to right and from right to left.

Example 2:

Input: x = -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.

Example 3:

Input: x = 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.

 

Constraints:

• -231 <= x <= 231 - 1

 

Follow up: Could you solve it without converting the integer to a string?

Solutions:

a. Convert  Integer to String:

class Solution:
    def isPalindrome(self, x: int) -> bool: 
        str_x = str(x)
        len_x = len(str_x)
        for i in range(len_x):
            if str_x[i] != str_x[len_x-i-1]:
                return False
        return True

or

class Solution:
    def isPalindrome(self, x: int) -> bool:
        return str(x) == str(x)[::-1]

b. Dont't Convert  Integer to String:

class Solution:
    # This function checks if a given integer is a palindrome.
    def isPalindrome(self, x: int) -> bool:
        # Check if the number is negative or a multiple of 10 (excluding 0)
        if x < 0 or (x != 0 and x % 10 == 0):
            return False

        reversed_num = 0
        original = x

        # Reverse the second half of the number
        while x > reversed_num:
            reversed_num = reversed_num * 10 + x % 10  # Add the last digit of 'x' to 'reversed_num'
            x //= 10  # Remove the last digit of 'x'

        # Check if the reversed number matches the original number, x == reversed_num // 10  is when X have odd width
        return x == reversed_num or x == reversed_num // 10

21. Merge Two Sorted Lists  Easy

You are given the heads of two sorted linked lists list1 and list2.

Merge the two lists into one sorted list. The list should be made by splicing together the nodes of the first two lists.

Return the head of the merged linked list.

 

Example 1:

Input: list1 = [1,2,4], list2 = [1,3,4]
Output: [1,1,2,3,4,4]

Example 2:

Input: list1 = [], list2 = []
Output: []

Example 3:

Input: list1 = [], list2 = [0]
Output: [0]

 

Constraints:

• The number of nodes in both lists is in the range [0, 50].
• -100 <= Node.val <= 100
• Both list1 and list2 are sorted in non-decreasing order.

---

Solutions:

class Solution(object):
    # This function merges two sorted linked lists into a single sorted linked list.
    # It takes two non-reduced sequences 'list1' and 'list2' as input.
    def mergeTwoLists(self, list1, list2):
        # Initialize a new ListNode as the head of the merged list
        head = ListNode()
        # Initialize a pointer 'current' to traverse the merged list
        current = head
        
        # Continue until either list1 or list2 becomes None
        while list1 and list2:
            # If the value of the current node in list1 is smaller than list2
            if list1.val < list2.val:
                # Append the current node of list1 to the merged list
                current.next = list1
                # Move to the next node in list1
                list1 = list1.next
            else:
                # Append the current node of list2 to the merged list
                current.next = list2
                # Move to the next node in list2
                list2 = list2.next
            # Move the 'current' pointer to the last appended node
            current = current.next
        
        # Connect the remaining nodes of list1 or list2 to the merged list
        current.next = list1 or list2
        
        # Return the head of the merged list (excluding the dummy node)
        return head.next

26. Remove Duplicates from Sorted Array  easy

Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place such that each unique element appears only once. The relative order of the elements should be kept the same. Then return the number of unique elements in nums.

Consider the number of unique elements of nums to be k, to get accepted, you need to do the following things:

• Change the array nums such that the first k elements of nums contain the unique elements in the order they were present in nums initially. The remaining elements of nums are not important as well as the size of nums.
• Return k.

Custom Judge:

The judge will test your solution with the following code:

int[] nums = [...]; // Input array
int[] expectedNums = [...]; // The expected answer with correct length

int k = removeDuplicates(nums); // Calls your implementation

assert k == expectedNums.length;
for (int i = 0; i < k; i++) {
    assert nums[i] == expectedNums[i];
}

If all assertions pass, then your solution will be accepted.

 

Example 1:

Input: nums = [1,1,2]
Output: 2, nums = [1,2,_]
Explanation: Your function should return k = 2, with the first two elements of nums being 1 and 2 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

Example 2:

Input: nums = [0,0,1,1,1,2,2,3,3,4]
Output: 5, nums = [0,1,2,3,4,_,_,_,_,_]
Explanation: Your function should return k = 5, with the first five elements of nums being 0, 1, 2, 3, and 4 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

 

Constraints:

• 1 <= nums.length <= 3 * 104
• -100 <= nums[i] <= 100
• nums is sorted in non-decreasing order.

Solutions:

class Solution(object):
    def removeDuplicates(self, nums):
        replace = 1
        for i in range(1, len(nums)):
            if nums[i-1] != nums[i]:
                nums[replace] = nums[i]
                replace += 1
        return replace

89. Gray Code medium

An n-bit gray code sequence is a sequence of 2n integers where:

• Every integer is in the inclusive range [0, 2n - 1],
• The first integer is 0,
• An integer appears no more than once in the sequence,
• The binary representation of every pair of adjacent integers differs by exactly one bit, and
• The binary representation of the first and last integers differs by exactly one bit.

Given an integer n, return any valid n-bit gray code sequence.

 

Example 1:

Input: n = 2
Output: [0,1,3,2]
Explanation:
The binary representation of [0,1,3,2] is [00,01,11,10].
- 00 and 01 differ by one bit
- 01 and 11 differ by one bit
- 11 and 10 differ by one bit
- 10 and 00 differ by one bit
[0,2,3,1] is also a valid gray code sequence, whose binary representation is [00,10,11,01].
- 00 and 10 differ by one bit
- 10 and 11 differ by one bit
- 11 and 01 differ by one bit
- 01 and 00 differ by one bit

Example 2:

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

 

Constraints:

• 1 <= n <= 16

Solution：

class Solution:
    def grayCode(self, n: int) -> List[int]:
        L = [0]
        for i in range(1,2**n):
            L.append(i^(i>>1))
        return L

231. Power of Two easy

Given an integer n, return true if it is a power of two. Otherwise, return false.

An integer n is a power of two, if there exists an integer x such that n == 2x.

 

Example 1:

Input: n = 1
Output: true
Explanation: 20 = 1

Example 2:

Input: n = 16
Output: true
Explanation: 24 = 16

Example 3:

Input: n = 3
Output: false

 

Constraints:

• -231 <= n <= 231 - 1

 

Follow up: Could you solve it without loops/recursion?

Solutions:

a.loop:

class Solution:
    def isPowerOfTwo(self, n: int) -> bool:
        for i in range(32):
            if 2 ** i == n:
                return True
        return False

b. bit information   pow_n:  100/ 1000/ 10000/ 100000    pow_n - 1 = 011/ 0111/ 01111/ 011111                  pow_n & pow_n-1 = 0

class Solution:
    def isPowerOfTwo(self, n: int) -> bool:
        return n > 0 and n & (n-1) == 0

344. Reverse String easy

Write a function that reverses a string. The input string is given as an array of characters s.

You must do this by modifying the input array in-place with O(1) extra memory.

 

Example 1:

Input: s = ["h","e","l","l","o"]
Output: ["o","l","l","e","h"]

Example 2:

Input: s = ["H","a","n","n","a","h"]
Output: ["h","a","n","n","a","H"]

 

Constraints:

• 1 <= s.length <= 105
• s[i] is a printable ascii character.

Solutions:

a.

class Solution:
    def reverseString(self, s: List[str]) -> None:
        """
        Do not return anything, modify s in-place instead.
        """
        str_len = len(s)
        for i in range(str_len>>1):
            s[i],s[str_len-i-1] = s[str_len-i-1],s[i]
        

b. we can't directly write s = s[::-1],becase it will create a new list,and let s point to zhe new list, but we can write s[:] = s[::-1],becasue it guarantees that opreate in original list. or we can write s.revrese()

class Solution:
    def reverseString(self, s: List[str]) -> None:
        """
        Do not return anything, modify s in-place instead.
        """
        s[:]=s[::-1]