I knew the answer was the median basically immediately, but the proof part is where I fumbled a bit.
Recognize that the optimal meeting point is the median of the positions, which minimizes the sum of absolute deviations. Prove optimality by showing that moving away from the median increases the total distance, using a pairwise exchange argument. Implement by sorting the array and selecting the middle element (or any point between the two middle elements for even N).
Pro tip: Mention that the median is robust to outliers and that the solution runs in O(N log N) due to sorting, but can be O(N) with quickselect. Also, note that for even N, any point between the two middle values yields the same minimum sum, which is a common edge case.
Restate the problem: minimize sum of absolute distances. Clarify that the point can be anywhere on the line, not necessarily at a person's position.
Show that the median minimizes the sum of absolute deviations. For odd N, it's the middle element; for even N, any point between the two middle elements.
Use a pairwise argument: if you move away from the median, the number of points on the side you move towards increases, increasing the total distance. Alternatively, use the derivative of the sum of absolute values.
Sort the array and return the middle element (or the average of the two middle elements for even N, but any point in between works). Discuss time complexity O(N log N) and space O(1) if sorting in place.
Mention alternative algorithms like quickselect for O(N) average time. Discuss if the problem changes with weighted points or different distance metrics.
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