**Remove Sublist to Reach Equilibrium - Google Top Interview Questions**

### Problem Statement :

Given a list of integers nums and an integer k, you can remove any sublist at most once from the list. Return the length of the longest resulting list such that the amount of numbers strictly less than k and strictly larger than k is the same. Constraints n ≤ 100,000 where n is the length of nums Example 1 Input nums = [5, 9, 7, 8, 2, 4] k = 5 Output 5 Explanation If we remove the sublist [8] then we'd get [5, 9, 7, 2, 4] and there's two numbers [2, 4] that's smaller than 5 and two numbers [9, 7] larger than 5. Example 2 Input nums = [1, 2, 3] k = 4 Output 0 Explanation We need to remove the whole sublist to have the same amount of numbers greater than 4 as amount of numbers less than 4.

### Solution :

` ````
Solution in C++ :
int find_sublist(vector<int>& nums, int target) {
if (target == 0) return 0;
// TO find sublist having sum, sum[i,j]=sum[0,j]-sum[0,i-1].
// so do prefix sum and keep index of all the sum appeared till now,
unordered_map<int, int> mp;
int pfsum = 0;
int len = INT_MAX;
for (int i = 0; i < nums.size(); i++) {
pfsum += nums[i];
if (pfsum == target) len = min(len, i + 1);
if (mp.count(pfsum - target) != 0) {
len = min(i - mp[pfsum - target], len);
}
mp[pfsum] = i;
}
return len;
}
int solve(vector<int>& nums, int k) {
// U need to find shortest sublist having sum=rem.
int sum = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] > k)
sum += 1, nums[i] = 1;
else if (nums[i] < k)
sum -= 1, nums[i] = -1;
else
nums[i] = 0;
}
int dlen = find_sublist(nums, sum);
return nums.size() - dlen;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[] nums, int k) {
if (nums == null || nums.length == 0)
return 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] > k)
nums[i] = 1;
else if (nums[i] < k)
nums[i] = -1;
else
nums[i] = 0;
}
int sum = 0;
int[] prefixSum = new int[nums.length + 1];
for (int i = 1; i < prefixSum.length; i++) {
prefixSum[i] = prefixSum[i - 1] + nums[i - 1];
sum += nums[i - 1];
}
if (sum == 0)
return nums.length;
Map<Integer, Integer> map = new HashMap();
int len = nums.length;
for (int i = 0; i < prefixSum.length; i++) {
int curr_sum = prefixSum[i];
int target = curr_sum - sum;
if (map.containsKey(target)) {
len = Math.min(len, i - map.get(target));
}
map.put(curr_sum, i);
}
return nums.length - len;
}
}
```

` ````
Solution in Python :
from itertools import accumulate
class Solution:
def solve(self, A, K):
N = len(A)
A = [(x > K) - (x < K) for x in A]
S = sum(A)
ans = 0
last = {0: 0}
for j, psum in enumerate(accumulate(A, initial=0)):
if (i := last.get(psum - S, -1)) >= 0:
ans = max(ans, N - (j - i))
last[psum] = j
return ans
```

## View More Similar Problems

## 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 →## Delete duplicate-value nodes from a sorted linked list

This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -

View Solution →## Cycle Detection

A linked list is said to contain a cycle if any node is visited more than once while traversing the list. Given a pointer to the head of a linked list, determine if it contains a cycle. If it does, return 1. Otherwise, return 0. Example head refers 1 -> 2 -> 3 -> NUL The numbers shown are the node numbers, not their data values. There is no cycle in this list so return 0. head refer

View Solution →## Find Merge Point of Two Lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the head nodes of 2 linked lists that merge together at some point, find the node where the two lists merge. The merge point is where both lists point to the same node, i.e. they reference the same memory location. It is guaranteed that the two head nodes will be different, and neither will be NULL. If the lists share

View Solution →## Inserting a Node Into a Sorted Doubly Linked List

Given a reference to the head of a doubly-linked list and an integer ,data , create a new DoublyLinkedListNode object having data value data and insert it at the proper location to maintain the sort. Example head refers to the list 1 <-> 2 <-> 4 - > NULL. data = 3 Return a reference to the new list: 1 <-> 2 <-> 4 - > NULL , Function Description Complete the sortedInsert function

View Solution →