Breaking the Illusion: Manufacturing Anxiety
Recently, NVIDIA's algorithm interviews are being conducted intensively.
Many candidates' first reaction upon seeing the question: "Simple! This is just one traversal + greedy approach, right?"
Dead wrong.
If you simply find the violation and modify it, even with correct logic, in the interviewer's eyes it's still procedural programming thinking.
Our oavoservice team conducted a deep review of this problem and found that 90% of candidates fail on boundary judgment and modification strategy selection.
Question Breakdown: Demonstrating Expertise
🔍 Core Question: Check if Array Can Become Non-decreasing by Modifying At Most One Element
Problem Description:
Given an array nums with n integers, check if it could become non-decreasing by modifying at most one element.
Non-decreasing array definition: For every i (0-based), nums[i] <= nums[i+1] holds, where 0 <= i <= n-2.
Key Challenges:
- Precise location of violation positions
- Optimal selection of modification strategies
- Complete coverage of boundary conditions
Input/Output Example:
Input: nums = [4,2,3]
Output: true
Explanation: Can modify 4 to 1 or 2, or modify 2 to 4
oavoservice Exclusive Analysis: This problem disguises itself as simple greedy while actually testing algorithm design patterns and Engineering Judgment.
Fatal Pitfalls: Why Others Fail
🤯 Hidden Killer: Trade-off Selection in Modification Strategy
This is where most candidates fail.
Many students are accustomed to hard coding thinking and modify immediately upon finding violations:
# Typical Toy Code approach
def checkPossibility(nums):
count = 0
for i in range(len(nums) - 1):
if nums[i] > nums[i + 1]:
count += 1
if count > 1:
return False
# Error: Blind modification without considering global impact
nums[i + 1] = nums[i]
return True
But in NVIDIA interviewers' eyes, this means:
- Shallow algorithmic understanding: Didn't consider applicable scenarios for two modification strategies
- Missing boundary handling: Improper handling of special cases when
i==0 - Poor code maintainability: Direct modification of input array violates Idempotency principles
The interviewer's deep follow-up questions are often not because you got it wrong, but because you lack Scalability thinking.
Correct Industrial-Grade Solution
def checkPossibility(nums):
n = len(nums)
violation_count = 0
violation_index = -1
# Find all violation positions
for i in range(n - 1):
if nums[i] > nums[i + 1]:
violation_count += 1
if violation_count > 1:
return False
violation_index = i
# No violations or exactly one violation
if violation_count <= 1:
if violation_count == 0:
return True
i = violation_index
# Trade-off selection between two modification strategies
# Strategy 1: Lower left side nums[i] = nums[i+1]
if i == 0 or nums[i-1] <= nums[i+1]:
return True
# Strategy 2: Raise right side nums[i+1] = nums[i]
# Check if there are more elements to the right that need to be satisfied
if i+1 == n-1 or nums[i] <= nums[i+2]:
return True
return False
return False
Algorithm Insight:
- Time Complexity: O(n), single linear scan
- Space Complexity: O(1), in-place algorithm
- Engineering Value: Doesn't modify original array, follows functional programming paradigm
Conversion Closing: Service Upgrade
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Don't let a seemingly simple array problem become the roadblock to your NVIDIA high-salary dreams.
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