SWE(C++)

Count Character Pairs: Count the frequencies of all characters in the array. Since a palindrome is built using pairs of identical characters (with at most one middle character), count how many pairs you have in total and how many “leftover” odd characters remain.

Sort by Length: Sort the original string lengths in ascending order. It is always more efficient to complete a short palindrome first because it requires fewer character pairs, leaving more resources to try and complete longer strings.

Greedy Allocation: Iterate through the sorted lengths. For each string of length L, you need ⌊L/2⌋ pairs to fill it. If you have enough pairs, subtract them and increment your count. Don’t worry about the middle character for odd-length strings; the leftover characters from Step 1 will naturally fill those spots.

Bit-by-Bit Evaluation (LSB to MSB): Since the sum S is fixed, the total contribution of all ai​⊕x must equal S. Iterate through bits from k=0 to 61. At each step k, the k-th bit of the sum is determined by the k-th bits of the modified elements (ai​⊕x) and the carry from the (k−1) position.

Greedy Choice with Carry Management: For each bit k, you have two choices: set the k-th bit of x to 0 or 1.

If you set xk​=0, the k-th bit of the i-th term is ai,k​.

If you set xk​=1, the k-th bit of the i-th term is 1−ai,k​.
Calculate how many 1s are contributed to the k-th position for both choices. Choose the value for xk​ that results in a k-th bit of the total sum that matches Sk​ (the k-th bit of S).

Recursive Backtracking (or DP): Because a choice at bit k affects the carry to bit k+1, a pure greedy approach might fail. You must use a recursive function solve(bit, carry) to explore whether a valid x exists. Since the carry can be large, you must observe that the carry eventually stabilizes or can be capped, allowing for a memoized search.

Identify the Greedy Choice: To minimize the total sum, each operation should provide the maximum possible reduction. The reduction gained by dividing x is x−⌈x/2⌉=⌊x/2⌋. This reduction is always non-decreasing with respect to x, meaning you should always pick the largest current element in the array for each of the k operations.

Use a Max-Heap (Priority Queue): Repeatedly finding the maximum in a raw array would take O(n) per operation, leading to O(nk), which is too slow. Instead, push all elements into a Max-Priority Queue. This allows you to extract the maximum and re-insert the halved value in O(logn) time.

Iterate and Re-sum: Perform the operation exactly k times (or until the maximum element becomes 0). In each iteration:

Extract the top element max_val.

Calculate new_val = ceil(max_val / 2.0).

Push new_val back into the heap.
After k operations, the sum of all elements currently in the heap is your answer.

Tip 1: Read the OSTEP book for OS 
Tip 2: Study the Multithreading in Action (C++) book for multithreading

Tip 1: Practice implementing std::vector, thread pool, memory pool, or have a rough idea of them.

Tip 1: Practice implementing std::vector, thread pool, and memory pool, or have a rough idea of them.

Tip 1: Again, read any C++ book to understand the underlying concepts of how memory interacts with C++.

Tip 1: Practice some basic HR questions from the internet.
Tip 2: Be confident.