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
                    


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