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This question evaluates algorithmic problem-solving and data-processing competencies, focusing on numeric optimization for making array elements distinct with minimal cost and on traversal and aggregation over a social graph to rank movies by friend and friends-of-friends frequency.

  • medium
  • Amazon
  • Coding & Algorithms
  • Software Engineer

Minimize cost & recommend movies

Company: Amazon

Role: Software Engineer

Category: Coding & Algorithms

Difficulty: medium

Interview Round: Technical Screen

##### Question Given arrays size[] and cost[] where size[i] is the size of the i-th product and cost[i] is the cost to increase size[i] by one unit, compute the minimal total cost required to increment elements so that all sizes become distinct. Implement recommendMovies(userName) that returns a list of movie titles ordered by how frequently they were watched by the user’s friends and friends-of-friends, using Helper.getFriends and Helper.getMoviesWatched.

Quick Answer: This question evaluates algorithmic problem-solving and data-processing competencies, focusing on numeric optimization for making array elements distinct with minimal cost and on traversal and aggregation over a social graph to rank movies by friend and friends-of-friends frequency.

You are given two arrays size and cost of equal length n. size[i] is the initial size of the i-th product, and cost[i] is the cost to increase size[i] by one unit. In one operation you may increase size[i] by 1, paying cost[i]. You cannot decrease any value. Compute the minimal total cost required to transform the array so that all final sizes are pairwise distinct integers.

Constraints

  • 1 <= n <= 200000
  • 0 <= size[i] <= 10^9
  • 1 <= cost[i] <= 10^9
  • Only increments are allowed
  • Output may exceed 32-bit; use 64-bit integer arithmetic

Hints

  1. Sort items by their initial size.
  2. Sweep positions from left to right; at each integer position, at most one item can be assigned.
  3. Maintain a max-heap (priority queue) of costs for items whose size is <= current position.
  4. Greedily assign the current position to the item with the highest per-unit cost to avoid paying that cost in future increments.
Last updated: Mar 29, 2026

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