**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

## AND xor OR

Given an array of distinct elements. Let and be the smallest and the next smallest element in the interval where . . where , are the bitwise operators , and respectively. Your task is to find the maximum possible value of . Input Format First line contains integer N. Second line contains N integers, representing elements of the array A[] . Output Format Print the value

View Solution →## Waiter

You are a waiter at a party. There is a pile of numbered plates. Create an empty answers array. At each iteration, i, remove each plate from the top of the stack in order. Determine if the number on the plate is evenly divisible ith the prime number. If it is, stack it in pile Bi. Otherwise, stack it in stack Ai. Store the values Bi in from top to bottom in answers. In the next iteration, do the

View Solution →## Queue using Two Stacks

A queue is an abstract data type that maintains the order in which elements were added to it, allowing the oldest elements to be removed from the front and new elements to be added to the rear. This is called a First-In-First-Out (FIFO) data structure because the first element added to the queue (i.e., the one that has been waiting the longest) is always the first one to be removed. A basic que

View Solution →## Castle on the Grid

You are given a square grid with some cells open (.) and some blocked (X). Your playing piece can move along any row or column until it reaches the edge of the grid or a blocked cell. Given a grid, a start and a goal, determine the minmum number of moves to get to the goal. Function Description Complete the minimumMoves function in the editor. minimumMoves has the following parameter(s):

View Solution →## Down to Zero II

You are given Q queries. Each query consists of a single number N. You can perform any of the 2 operations N on in each move: 1: If we take 2 integers a and b where , N = a * b , then we can change N = max( a, b ) 2: Decrease the value of N by 1. Determine the minimum number of moves required to reduce the value of N to 0. Input Format The first line contains the integer Q.

View Solution →## Truck Tour

Suppose there is a circle. There are N petrol pumps on that circle. Petrol pumps are numbered 0 to (N-1) (both inclusive). You have two pieces of information corresponding to each of the petrol pump: (1) the amount of petrol that particular petrol pump will give, and (2) the distance from that petrol pump to the next petrol pump. Initially, you have a tank of infinite capacity carrying no petr

View Solution →