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Maximize Minimum Scaled Value

Last updated: May 14, 2026

Quick Overview

This question evaluates algorithmic optimization and resource-allocation skills, focusing on integer scaling under a global budget to maximize the minimum final value across elements. It is asked in the Coding & Algorithms domain to probe trade-off reasoning and problem-solving under constraints, and the level of abstraction is practical application (algorithm design and implementation).

  • hard
  • Uber
  • Coding & Algorithms
  • Software Engineer

Maximize Minimum Scaled Value

Company: Uber

Role: Software Engineer

Category: Coding & Algorithms

Difficulty: hard

Interview Round: Take-home Project

You are given two integer arrays `values` and `costs` of the same length `n`, and a non-negative integer `totalCost`. For each index `i`, you may choose an integer multiplier `times[i] >= 1`. - The final value at index `i` becomes `values[i] * times[i]`. - Increasing index `i` from multiplier `1` to `times[i]` costs `costs[i] * (times[i] - 1)`. Your total spending across all indices must satisfy: `sum(costs[i] * (times[i] - 1)) <= totalCost` Your goal is to maximize the minimum final value among all indices, that is, maximize: `min(values[i] * times[i])` Return the largest possible value of this minimum.

Quick Answer: This question evaluates algorithmic optimization and resource-allocation skills, focusing on integer scaling under a global budget to maximize the minimum final value across elements. It is asked in the Coding & Algorithms domain to probe trade-off reasoning and problem-solving under constraints, and the level of abstraction is practical application (algorithm design and implementation).

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Uber logo
Uber
Jan 16, 2026, 12:00 AM
Software Engineer
Take-home Project
Coding & Algorithms
1
0
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You are given two integer arrays values and costs of the same length n, and a non-negative integer totalCost.

For each index i, you may choose an integer multiplier times[i] >= 1.

  • The final value at index i becomes values[i] * times[i] .
  • Increasing index i from multiplier 1 to times[i] costs costs[i] * (times[i] - 1) .

Your total spending across all indices must satisfy:

sum(costs[i] * (times[i] - 1)) <= totalCost

Your goal is to maximize the minimum final value among all indices, that is, maximize:

min(values[i] * times[i])

Return the largest possible value of this minimum.

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