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 :



title-img



                        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

Merging Communities

People connect with each other in a social network. A connection between Person I and Person J is represented as . When two persons belonging to different communities connect, the net effect is the merger of both communities which I and J belongs to. At the beginning, there are N people representing N communities. Suppose person 1 and 2 connected and later 2 and 3 connected, then ,1 , 2 and 3 w

View Solution →

Components in a graph

There are 2 * N nodes in an undirected graph, and a number of edges connecting some nodes. In each edge, the first value will be between 1 and N, inclusive. The second node will be between N + 1 and , 2 * N inclusive. Given a list of edges, determine the size of the smallest and largest connected components that have or more nodes. A node can have any number of connections. The highest node valu

View Solution →

Kundu and Tree

Kundu is true tree lover. Tree is a connected graph having N vertices and N-1 edges. Today when he got a tree, he colored each edge with one of either red(r) or black(b) color. He is interested in knowing how many triplets(a,b,c) of vertices are there , such that, there is atleast one edge having red color on all the three paths i.e. from vertex a to b, vertex b to c and vertex c to a . Note that

View Solution →

Super Maximum Cost Queries

Victoria has a tree, T , consisting of N nodes numbered from 1 to N. Each edge from node Ui to Vi in tree T has an integer weight, Wi. Let's define the cost, C, of a path from some node X to some other node Y as the maximum weight ( W ) for any edge in the unique path from node X to Y node . Victoria wants your help processing Q queries on tree T, where each query contains 2 integers, L and

View Solution →

Contacts

We're going to make our own Contacts application! The application must perform two types of operations: 1 . add name, where name is a string denoting a contact name. This must store name as a new contact in the application. find partial, where partial is a string denoting a partial name to search the application for. It must count the number of contacts starting partial with and print the co

View Solution →

No Prefix Set

There is a given list of strings where each string contains only lowercase letters from a - j, inclusive. The set of strings is said to be a GOOD SET if no string is a prefix of another string. In this case, print GOOD SET. Otherwise, print BAD SET on the first line followed by the string being checked. Note If two strings are identical, they are prefixes of each other. Function Descriptio

View Solution →