Associate Software Engineer

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The Two Sum problem asks us to find two numbers in an array that sum up to a given target value. We need to return the indices of these two numbers.

Approach
One brute force approach is to consider every pair of elements and check if their sum equals the target. This can be done using nested loops, where the outer loop iterates from the first element to the second-to-last element, and the inner loop iterates from the next element to the last element. However, this approach has a time complexity of O(n^2).
A more efficient approach is to use a hash table (unordered_map in C++). We can iterate through the array once, and for each element, check if the target minus the current element exists in the hash table. If it does, we have found a valid pair of numbers. If not, we add the current element to the hash table.
An approach using a hash table:

Create an empty hash table to store elements and their indices.
Iterate through the array from left to right.
For each element nums[i], calculate the complement by subtracting it from the target: complement = target - nums[i].
Check if the complement exists in the hash table. If it does, we have found a solution.
If the complement does not exist in the hash table, add the current element nums[i] to the hash table with its index as the value.
Repeat steps 3-5 until we find a solution or reach the end of the array.
If no solution is found, return an empty array or an appropriate indicator.
This approach has a time complexity of O(n) since hash table lookups take constant time on average. It allows us to solve the Two Sum problem efficiently by making just one pass through the array.

Initialize an empty string ans to store the common prefix.
Sort the input list v lexicographically. This step is necessary because the common prefix should be common to all the strings, so we need to find the common prefix of the first and last string in the sorted list.
Iterate through the characters of the first and last string in the sorted list, stopping at the length of the shorter string.
If the current character of the first string is not equal to the current character of the last string, return the common prefix found so far.
Otherwise, append the current character to the ans string.
Return the ans string containing the longest common prefix.

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