Visa Interview Questions (12+ Questions)

Last Updated: June 8, 2026 • 12 QuestionsReal Company Interviews

Prepare for your Visa interview with our comprehensive collection of 12+ real interview questions and detailed answers. These questions have been curated from actual Visa technical interviews across various roles including DevOps Engineer, Data Engineer, QA Engineer, and more.

12
Interview Questions
1
Categories
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Difficulty Levels

Table of Contents

Our Visa interview questions cover a wide range of technical topics and difficulty levels, from entry-level positions to senior roles. Each question includes detailed explanations and answers to help you understand the concepts and prepare effectively for your interview.

💡 Pro Tips for Visa Interviews

  • Practice each question and understand the underlying concepts
  • Review Visa's specific technologies and methodologies
  • Prepare follow-up questions and edge cases
  • Practice explaining your solutions clearly and concisely

Interview Questions & Answers

1. Allow Outbound SMTP Traffic

Company: Visa Difficulty: medium 🔒 Premium Categories: Devops

Learn how to diagnose and restore outbound SMTP connectivity using Linux Bash commands and firewall management. This guide covers identifying active firewalls, inspecting outbound rules, allowing SMTP port 25, testing connectivity, and verifying fixes, essential for troubleshooting email delivery issues and managing host-level firewall policies in production systems.

2. Traffic Management with Ingress

Company: Visa Difficulty: hard 🔒 Premium Categories: Devops

Kubernetes NGINX Ingress Canary: 10% app-v2 Auto-Rollback hashicorp/http-echo. Production-grade canary deployment in canary namespace with NGINX Ingress traffic splitting: 90% app-v1 (3 replicas) vs 10% app-v2 (1 replica) using hashicorp/http-echo containers. Dual Ingress app-ingress + app-ingress-canary with canary-weight: 10, header routing X-Canary: true, and auto-rollback script on >5 v2 pod restarts. Perfect for progressive delivery, A/B testing infrastructure, golden path releases, SRE automation, and zero-risk production deployments.

3. Immutable Artifact Promotion

Company: Visa Difficulty: medium 🔒 Premium Categories: Devops

Implement immutable artifact promotion patterns in GitHub Actions. Learn to build Docker images once, pass their unique digests between jobs using outputs and needs, and deploy to production without rebuilding.

4. Using Service Containers (PostgreSQL)

Company: Visa Difficulty: medium 🔒 Premium Categories: Devops

Learn to use GitHub Actions Service Containers. Spin up ephemeral databases (like PostgreSQL) inside your workflow for robust integration testing using the services keyword.

5. Subsets

Company: Visa Difficulty: medium Categories: Devops, Data engineering

def subsets(nums: list[int]) -> list[list[int]]:
res = []

def dfs(i, current_subset):
    res.append(current_subset.copy())
    
    for j in range(i, len(nums)):
        current_subset.append(nums[j])
        dfs(j + 1, current_subset)
        current_subset.pop()
        
dfs(0, [])
return res

6. Insert New Records into SQLite Database from CSV

Company: Visa Difficulty: medium Categories: Data analysis, Data engineering

Read customer records from a CSV file, check for existing records in a SQLite database table based on ID field, and insert only new records to avoid duplicates.

7. Combine Firms, Funds, and Investments Data

Company: Visa Difficulty: hard Categories: Data analysis, Data engineering

WITH combined AS (
SELECT
i.investment_id,
COALESCE(fd.fund_id, i.fund_id) AS fund_id,
COALESCE(f.firm_id, fd.firm_id) AS firm_id,
f.firm_name,
f.founded_year,
f.location,
fd.fund_name,
fd.fund_size,
fd.fund_start_year,
fd.fund_end_year,
i.company_name,
i.investment_amount,
i.investment_date
FROM {{ ref("firms") }} f
FULL OUTER JOIN {{ ref("funds") }} fd
ON f.firm_id = fd.firm_id
FULL OUTER JOIN {{ ref("investments") }} i
ON fd.fund_id = i.fund_id
)
SELECT *
FROM combined
WHERE NOT (
investment_id IS NULL
AND fund_id IS NULL
AND firm_id IS NULL
AND firm_name IS NULL
AND founded_year IS NULL
AND location IS NULL
AND fund_name IS NULL
AND fund_size IS NULL
AND fund_start_year IS NULL
AND fund_end_year IS NULL
AND company_name IS NULL
AND investment_amount IS NULL
AND investment_date IS NULL
)

8. Mining Operations Aggregation

Company: Visa Difficulty: easy Categories: Data analysis, Data engineering

Master data aggregation in PySpark. Learn how to join tables, perform multi-column GroupBy operations, and cast aggregated sums to specific data types.

9. Employee Bonus Eligibility Checker

Company: Visa Difficulty: medium 🔒 Premium Categories: Data engineering

Interview Question: SQL Query for Employee Performance and Bonus Status

Objective:

Crafting an SQL query to showcase employees' names, performance scores, annual salaries, and bonus statuses is an essential exercise in database management. This query will help in evaluating and rewarding ...


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10. Trapping Rain Water

Company: Visa Difficulty: hard Categories: Data engineering

def trap(height: list[int]) -> int:
if not height:
return 0

l, r = 0, len(height) - 1
left_max, right_max = height[l], height[r]
res = 0

while l < r:
    if left_max < right_max:
        l += 1
        left_max = max(left_max, height[l])
        res += left_max - height[l]
    else:
        r -= 1
        right_max = max(right_max, height[r])
        res += right_max - height[r]
        
return res

11. Reconstruct Itinerary

Company: Visa Difficulty: hard Categories: Data engineering

def find_itinerary(tickets: list[list[str]]) -> list[str]:
adj = defaultdict(list)

tickets.sort(reverse=True)

for src, dst in tickets:
    adj[src].append(dst)
    
res = []

def dfs(src):
    while adj[src]:
        dst = adj[src].pop()
        dfs(dst)
    res.append(src)
    
dfs("JFK")
res.reverse()

return res

12. Self-Join to Compare Order Dates

Company: Visa Difficulty: medium 🔒 Premium Categories: Data engineering

How to Identify Customers with Multiple Orders on the Same Date in SQL

Interview Question Objective

Write an SQL query to identify customers who have placed multiple orders on the same date. The query should return distinct pairs of customer names and order dates, sorted by the order date...


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