Python

1. Two Sum

Was able to do this from one list where, get the compliment and check from i+1 index forward within the list to see if the value exists in there or not

Python

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        for i, num in enumerate(nums):
            compliment = target - num 
            if compliment in nums[i+1:]:
                idx = nums.index(compliment, i+1)
                out = [i, idx]
                return out
        
        return None

C++

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) { 
        std::unordered_map<int, int> sumMap;
        for (int i = 0; i < nums.size(); i++) {
            int compliment = target - nums[i];
            if (sumMap.find(compliment) != sumMap.end()) {
                int compliment_index = sumMap[compliment];
                return {compliment_index, i};
            }
            else {
                sumMap[nums[i]] = i;
            }
        }
        return {};
    }
};

1. Palindrome Number

Set the default cases first. Then got the last digit, and started storing the reverse. Then removed the final digit.

Time complexity: O(log n) because it will run different amount of times based on the number of digits

Space complexity: O(1) because we used a fixed number of variables

class Solution {
public:
    bool isPalindrome(int x) {

        if (x == 0) {
            return true;
        }
 
        if (x < 0 || x % 10 == 0) {
            return false;
        }

        if (x < 10) {
            return true;
        }

        int copy = x;
        long long reversed = 0;

        while (x != 0) {
            int digit = x % 10; // last digit
            reversed = (reversed * 10) + digit; // store reverse
            x /= 10;
        }

        return reversed == copy;
    }
};

1. Longest Common Prefix