Assistant System Engineer

To solve it, I aimed to find the target value that minimizes the effort to make all locks the same. First, I defined a helper that calculates the cost to adjust each lock to a specific target, considering both direct adjustments and the option to "wrap around" to minimize changes. Using binary search, I then explored potential target values, starting with edge cases and iteratively refining the target to find the one that yields the lowest overall cost. This efficient approach narrows down possibilities quickly, leading to the minimum operations needed to align all locks.

To solve this problem, I used depth-first search (DFS) to calculate the minimum and maximum sum of distances when values in a tree are shuffled. For the minimum sum, I aimed to minimize the distances by assigning values to the closest possible nodes without repeating any initial values. This involves identifying and traversing only necessary paths to reduce redundant moves, particularly focusing on nodes that aren't leaves. For the maximum sum, I maximized the distances by assigning values as far apart as possible within the tree's structure. By carefully counting and adding the distances for both configurations, I obtained the required minimum and maximum sums of distances in an efficient way.

To solve this problem, the approach involves identifying the nodes that need to be swapped by their positions from the beginning and the end of the linked list. First, traverse the list to determine its length, which helps locate the `k`th node from the beginning and the `k`th node from the end. Specifically, the `k`th node from the end can be found at position `(length - k + 1)`. After determining these positions, traverse the list again to reach these nodes. Once both nodes are identified, simply swap their values rather than their positions, which avoids restructuring the list. Finally, return the head of the linked list, as the list structure remains the same, with only the values swapped. This method ensures efficiency by only requiring two traversals, making it an optimal solution.

To solve this problem, I first mapped each Roman numeral character to its respective integer value for easy reference. Then, I iterated through the characters of the Roman numeral string to apply the conversion rules. For each character, I checked if its value was smaller than the value of the next character, indicating a subtractive combination (like "IV" representing 4). In such cases, I subtracted its value from the running total; otherwise, I added its value to the total. By the end of the traversal, the accumulated result represented the integer value of the Roman numeral. This approach allowed for efficient handling of both additive and subtractive cases in a single pass.