Longest Equivalent Sublist After K Increments - Google Top Interview Questions
Problem Statement :
You are given a list of integers nums and an integer k. Consider an operation where you increment any one element once. Given that you can perform this at most k times, return the length of the longest sublist containing equal elements. Constraints n ≤ 100,000 where n is the length of nums k < 2 ** 31 Example 1 Input nums = [2, 4, 8, 5, 9, 6] k = 6 Output 3 Explanation We can increment 8 once and 5 four times to get a sublist of [9, 9, 9].
Solution :
Solution in C++ :
int solve(vector<int>& nums, int k) {
int ans = 0;
multiset<int> m;
for (int i = 0, j = 0, sum = 0; j < nums.size(); j++) {
sum += nums[j];
m.insert(nums[j]);
for (; !m.empty() && *m.rbegin() * (j - i + 1) - sum > k; i++) {
m.erase(m.find(nums[i]));
sum -= nums[i];
}
ans = max(ans, j - i + 1);
}
return ans;
}
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[] nums, int k) {
MultiSet ms = new MultiSet();
int l = 0;
int r = 0;
int cur_increments = 0;
int ret = 0;
// move r until necessary increments exceeds k
// anchor l until it's valid again
while (r < nums.length) {
// process r
Integer maxy = ms.last();
int x = nums[r];
if (maxy == null || x > maxy) {
if (maxy != null) {
cur_increments += (x - maxy) * ms.size();
}
} else {
cur_increments += (maxy - x);
}
ms.add(x);
// process l
while (cur_increments > k) {
x = nums[l];
ms.remove(x);
int lmaxy = ms.last();
if (lmaxy == x) {
} else if (lmaxy < x) {
cur_increments -= ms.size() * (x - lmaxy);
} else {
cur_increments -= lmaxy - x;
}
l++;
}
ret = Math.max(ret, r - l + 1);
r++;
}
return ret;
}
}
class MultiSet {
TreeMap<Integer, Integer> map = new TreeMap<>();
private int size = 0;
public MultiSet() {
}
public void add(int val) {
map.put(val, map.getOrDefault(val, 0) + 1);
size++;
}
public void remove(int val) {
map.put(val, map.get(val) - 1);
if (map.get(val) == 0) {
map.remove(val);
}
size--;
}
public int size() {
return size;
}
public Integer higher(int val) {
return map.higherKey(val);
}
public Integer lower(int val) {
return map.lowerKey(val);
}
public Integer ceiling(int val) {
return map.ceilingKey(val);
}
public Integer floor(int val) {
return map.floorKey(val);
}
public Integer first() {
if (map.isEmpty())
return null;
return map.firstKey();
}
public Integer last() {
if (map.isEmpty())
return null;
return map.lastKey();
}
public boolean isEmpty() {
return map.isEmpty();
}
@Override
public String toString() {
return map.toString();
}
}
Solution in Python :
class Solution:
def solve(self, nums, k):
ans = left = total = 0
q = deque()
for right, a in enumerate(nums):
total += a
while q and nums[q[-1]] <= a:
q.pop()
q.append(right)
while q and k < (right - left + 1) * nums[q[0]] - total:
total -= nums[left]
if q and left == q[0]:
q.popleft()
left += 1
ans = max(ans, right - left + 1)
return ans
View More Similar Problems
Delete a Node
Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo
View Solution →Print in Reverse
Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing
View Solution →Reverse a linked list
Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio
View Solution →Compare two linked lists
You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis
View Solution →Merge two sorted linked lists
This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C
View Solution →Get Node Value
This challenge is part of a tutorial track by MyCodeSchool Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on. Example head refers to 3 -> 2 -> 1 -> 0 -> NULL positionFromTail = 2 Each of the data values matches its distance from the t
View Solution →