"ArrayList allows constant time access (O(1)) to elements using their index because it uses a dynamic array internally, whereas LinkedList requires traversal from the head node, resulting in linear time complexity (O(n))."
Aziz V. - "ArrayList allows constant time access (O(1)) to elements using their index because it uses a dynamic array internally, whereas LinkedList requires traversal from the head node, resulting in linear time complexity (O(n))."See full answer
"The reason I choose to go to college is because I want to get a Degree in the hospitality field. Additionally, I want to get a diploma after I finish taking all of my college classes At Harper."
Amparo L. - "The reason I choose to go to college is because I want to get a Degree in the hospitality field. Additionally, I want to get a diploma after I finish taking all of my college classes At Harper."See full answer
"hash maps work in key value pair. The keys are hashed with a hash algorithm and resulting hashcode(integer) with related value are stored.
Accessing a value, removing an element, Searching the hash map:
1) The hash map value can be accessed in O(1) time once you know the key.
2) If the key is not known, the hashmap value can be accessed in O(n) since you have to iterate atleast once.
"
Kavithadevi P. - "hash maps work in key value pair. The keys are hashed with a hash algorithm and resulting hashcode(integer) with related value are stored.
Accessing a value, removing an element, Searching the hash map:
1) The hash map value can be accessed in O(1) time once you know the key.
2) If the key is not known, the hashmap value can be accessed in O(n) since you have to iterate atleast once.
"See full answer
"I first asked few clarifying questions like the return array may need not contain the list of building in the same order, to which the interviewer agreed.
Then I came up with an approach where we iterate the array from right to left and keep a max variable which will keep the value of the current max. When we find an item which is greater than max we update the max and add this element into our solution. The interviewer agreed for the approach.
I discussed few corner scenarios with the interview"
Rishabh N. - "I first asked few clarifying questions like the return array may need not contain the list of building in the same order, to which the interviewer agreed.
Then I came up with an approach where we iterate the array from right to left and keep a max variable which will keep the value of the current max. When we find an item which is greater than max we update the max and add this element into our solution. The interviewer agreed for the approach.
I discussed few corner scenarios with the interview"See full answer
"Leetcode 347: Heap + Hashtable
Follow up question: create heap with the length of K instead of N (more time complexity but less space )"
Chen J. - "Leetcode 347: Heap + Hashtable
Follow up question: create heap with the length of K instead of N (more time complexity but less space )"See full answer
"I try to solve this initially using quick select where will take a pivot element and position the remaining elements and check if the current index is answer or not and continue the same but it requires o(n*n), but interviewee is expecting the best from me, so at the end i tried solving using heaps where will check the difference between k and n-k to use min or max heap after that we will heap the array, and will keep popping the element k-1 and return the peek one which leads to answer."
Mourya C. - "I try to solve this initially using quick select where will take a pivot element and position the remaining elements and check if the current index is answer or not and continue the same but it requires o(n*n), but interviewee is expecting the best from me, so at the end i tried solving using heaps where will check the difference between k and n-k to use min or max heap after that we will heap the array, and will keep popping the element k-1 and return the peek one which leads to answer."See full answer
"import java.util.*;
public class NetworkTopology {
public int topologytype(int N, int M, int[] input3, int[] input4) {
if (M != N - 1 && M != N) return -1; // Fast check for invalid cases
int[] degree = new int[N + 1]; // Degree of each node (1-based indexing)
// Build the degree array
for (int i = 0; i < M; i++) {
degree[input3[i]]++;
degree[input4[i]]++;
}
// Check for Bus Topology
boolean isBus = (M"
Alessandro R. - "import java.util.*;
public class NetworkTopology {
public int topologytype(int N, int M, int[] input3, int[] input4) {
if (M != N - 1 && M != N) return -1; // Fast check for invalid cases
int[] degree = new int[N + 1]; // Degree of each node (1-based indexing)
// Build the degree array
for (int i = 0; i < M; i++) {
degree[input3[i]]++;
degree[input4[i]]++;
}
// Check for Bus Topology
boolean isBus = (M"See full answer