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.


n ≤ 100,000 where n is the length of nums

Example 1


nums = [5, 9, 7, 8, 2, 4]

k = 5




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

nums = [1, 2, 3]

k = 4




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;
            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;
                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

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