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Download latest Google placement papers 2025 PDF with current year online assessment questions, solutions, and updated exam patterns for 2025 recruitment.
This page contains Google placement papers from 2025 with current year questions, solutions, and exam patterns. Use these papers to understand the latest 2025 online assessment pattern and prepare effectively for ongoing Google recruitment drives.
| Section | Questions | Time | Difficulty | Focus Areas |
|---|---|---|---|---|
| Coding Problem 1 | 1 | 30 min | Medium | Arrays, strings, two pointers |
| Coding Problem 2 | 1 | 30 min | Hard | Trees, graphs, dynamic programming |
| CS Fundamentals MCQs | 18-20 | 30 min | Medium | DSA, time complexity, system design, AI/ML basics |
Total: 20-22 questions, 90 minutes
Key Changes in 2025:
Problem: Find the median of two sorted arrays with O(log(min(m,n))) complexity.
Example: Input: nums1 = [1,3], nums2 = [2], Output: 2.0
Solution:
def findMedianSortedArrays(nums1, nums2): A, B = nums1, nums2 m, n = len(A), len(B) if m > n: A, B, m, n = B, A, n, m
imin, imax, half = 0, m, (m + n + 1) // 2 while imin <= imax: i = (imin + imax) // 2 j = half - i
if i < m and B[j-1] > A[i]: imin = i + 1 elif i > 0 and A[i-1] > B[j]: imax = i - 1 else: if i == 0: max_of_left = B[j-1] elif j == 0: max_of_left = A[i-1] else: max_of_left = max(A[i-1], B[j-1])
if (m + n) % 2 == 1: return max_of_left
if i == m: min_of_right = B[j] elif j == n: min_of_right = A[i] else: min_of_right = min(A[i], B[j])
return (max_of_left + min_of_right) / 2.0Explanation: Binary search on smaller array. Partition both arrays such that left half has equal elements. Check partition validity and find median.
Answer: Returns median with O(log(min(m,n))) time complexity
Problem: Implement a rate limiter that allows at most N requests per time window T.
Solution:
from collections import dequeimport time
class RateLimiter: def __init__(self, max_requests, time_window): self.max_requests = max_requests self.time_window = time_window self.requests = deque()
def is_allowed(self): current_time = time.time() # Remove requests outside time window while self.requests and self.requests[0] < current_time - self.time_window: self.requests.popleft()
if len(self.requests) < self.max_requests: self.requests.append(current_time) return True return FalseExplanation: Sliding window approach. Maintain deque of request timestamps. Remove old requests outside window, check if space available.
Answer: Rate limiter with O(1) average time complexity
Hiring Volume
Salary Packages
New Initiatives
Coding Problems:
New Focus Areas:
Based on recent candidate experiences from 2025 Google interviews:
2025 Interview Process:
2025 Interview Trends:
Common 2025 Interview Topics:
2025 Interview Questions Examples:
Success Tips:
Difficulty Rating: 3.4-3.5/5 (Most challenging among FAANG)
For detailed interview experiences from 2025, visit Google Interview Experience page.
Google 2025 Paper 1
Latest 2025 online assessment with coding problems and MCQs
Google 2025 Paper 2
Additional 2025 questions with detailed solutions
2025 Coding Examples
Collection of 2025 coding problems with solutions
Practice 2025 papers to stay updated! Focus on AI/ML basics and product thinking for best results.