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Implementing an LRU Cache using a combination of a hashmap and a doubly linked list.
Use a hashmap for O(1) access to cache items.
Use a doubly linked list to maintain the order of usage.
When accessing an item, move it to the front of the list.
If the cache exceeds its capacity, remove the least recently used item from both the hashmap and the list.
Example: For a cache of capacity 2, accessing keys 1, 2, then 1 again...
Reverse a linked list by changing the direction of pointers
Start with three pointers: prev, current, next
Iterate through the list, updating pointers to reverse the direction
Update the head of the list to be the last node
Find the length of the longest valid parentheses substring in a given string.
Use a stack to keep track of the indices of opening parentheses.
Iterate through the string and update the maximum length of valid parentheses.
Handle cases where closing parentheses are encountered without a matching opening parentheses.
Find the maximum number in the subarrays of length k.
Iterate through the array and keep track of the maximum number in each subarray of length k.
Use a sliding window approach to efficiently find the maximum number in each subarray.
Time complexity can be optimized to O(n) using a deque data structure.
Find minimum deletions to make a sequence sorted.
Count the number of elements that are not in increasing order.
Subtract the count from the length of the sequence to get the minimum deletions.
Use dynamic programming to optimize the solution.
Time complexity: O(nlogn) using binary search.
The minimum number of towers required to cover all houses.
Iterate through the houses and find the maximum distance between any two adjacent houses.
Divide this maximum distance by the range of each tower to get the minimum number of towers required.
If there is a remainder, add one to the result.
Return the minimum number of towers.
Reverse a linked list by changing the direction of pointers
Start with three pointers: prev, current, next
Iterate through the list, updating pointers to reverse the direction
Update the head of the list to be the last node
Implementing an LRU Cache using a combination of a hashmap and a doubly linked list.
Use a hashmap for O(1) access to cache items.
Use a doubly linked list to maintain the order of usage.
When accessing an item, move it to the front of the list.
If the cache exceeds its capacity, remove the least recently used item from both the hashmap and the list.
Example: For a cache of capacity 2, accessing keys 1, 2, then 1 again shou...
I applied via Company Website and was interviewed in Jun 2023. There were 5 interview rounds.
An assignment to make an image listing app from API using Volley/Retrofit and second screen to display the image. Used recyclerview, glide , volley etc. It was done to be in 2 hour.
In 1 hour i was asked to solve two easy DS question
1. Merge the two sorted arrays
2. rotated binary search
I appeared for an interview before Oct 2023.
Find the length of the longest valid parentheses substring in a given string.
Use a stack to keep track of the indices of opening parentheses.
Iterate through the string and update the maximum length of valid parentheses.
Handle cases where closing parentheses are encountered without a matching opening parentheses.
Find the maximum number in the subarrays of length k.
Iterate through the array and keep track of the maximum number in each subarray of length k.
Use a sliding window approach to efficiently find the maximum number in each subarray.
Time complexity can be optimized to O(n) using a deque data structure.
I applied via Company Website and was interviewed in Oct 2023. There were 3 interview rounds.
2 DSa questions + cs fundamentals mcq
I applied via LinkedIn and was interviewed in Feb 2024. There was 1 interview round.
Zeta interview at codeblock. 3 dsa questions The level of difficulty was medium
Product case study on CRM
I appeared for an interview in Oct 2023.
I applied via Company Website and was interviewed in Oct 2023. There was 1 interview round.
There were 2 Question in Online Assessment. The level of the DSA Question was medium to hard
I applied via Company Website and was interviewed in Jan 2021. There were 3 interview rounds.
The minimum number of towers required to cover all houses.
Iterate through the houses and find the maximum distance between any two adjacent houses.
Divide this maximum distance by the range of each tower to get the minimum number of towers required.
If there is a remainder, add one to the result.
Return the minimum number of towers.
Find minimum deletions to make a sequence sorted.
Count the number of elements that are not in increasing order.
Subtract the count from the length of the sequence to get the minimum deletions.
Use dynamic programming to optimize the solution.
Time complexity: O(nlogn) using binary search.
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based on 8 interview experiences
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