Explain, with intuition and a brief derivation, why the median minimizes the sum of absolute deviations (L 1) while the mean minimizes the sum of squared deviations (L 2). How does this extend to selecting an optimal point in two dimensions under Manhattan versus Euclidean distance, and when is the average a poor choice due to outliers?

This question evaluates understanding of central tendency and loss functions—specifically the roles of median and mean under L1 (absolute) and L2 (squared) norms—and the geometric extension of these concepts to two-dimensional data.

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This is a medium difficulty Statistics & Math question, commonly asked during Technical Screen rounds at Snapchat.

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This question is commonly asked for Software Engineer candidates at Snapchat during technical interviews.

Explain median vs mean for L1/L2 | Snapchat Interview Question

Median vs Mean Under L1 and L2, and the 2D Extension

Task

Explain, with intuition and a brief derivation, why:

In 1D, the median minimizes the sum of absolute deviations (L1), while the mean minimizes the sum of squared deviations (L2).
In 2D, the optimal point under Manhattan distance (L1) versus Euclidean distance (L2) corresponds to which estimator.
When the average (mean) is a poor choice due to outliers.

Assumptions/Setup

1D data: x₁, x₂, …, xₙ ∈ ℝ.
2D points: pᵢ = (xᵢ, yᵢ) ∈ ℝ².
“L1” refers to minimizing sum of absolute deviations; “L2” refers to minimizing sum of squared deviations. For 2D Euclidean distance, clarify whether the loss is squared or unsquared; both are addressed.

Median vs Mean Under L1 and L2, and the 2D Extension

Task

Explain, with intuition and a brief derivation, why:

In 1D, the median minimizes the sum of absolute deviations (L1), while the mean minimizes the sum of squared deviations (L2).
In 2D, the optimal point under Manhattan distance (L1) versus Euclidean distance (L2) corresponds to which estimator.
When the average (mean) is a poor choice due to outliers.

Assumptions/Setup

1D data: x₁, x₂, …, xₙ ∈ ℝ.
2D points: pᵢ = (xᵢ, yᵢ) ∈ ℝ².
“L1” refers to minimizing sum of absolute deviations; “L2” refers to minimizing sum of squared deviations. For 2D Euclidean distance, clarify whether the loss is squared or unsquared; both are addressed.

Explain median vs mean for L1/L2

Quick Overview

Median vs Mean Under L1 and L2, and the 2D Extension

Task

Assumptions/Setup

Solution

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Explain median vs mean for L1/L2

Quick Overview

Median vs Mean Under L1 and L2, and the 2D Extension

Task

Assumptions/Setup

Solution

Comments (0)