We can simple use a vector to store the sizes of stack, the minimum ascending order series we can form is 0,1,2,3,4....n We can keep a buffer for extra elements, that means set a[0] = 0 and add the remaining value to the buffer. a[i] -= i; and buffer += a[i] — i;

if at any point our buffer becomes negative that will mean that particular sequence till that point isn't possible, and since that is the minimal series that means it isn't possible to form a valid sequence for those set of stacks.

// Aur Bhai Dekhne aagaye ;)
// Author: Abhijit Mishra 
#include <bits/stdc++.h>
using namespace std;

typedef long long int ll;
typedef vector<ll> vi;

#define nl cout << "\n"
#define fast ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);

int main()
{
    fast;
    ll t = 1;
    cin >> t;
    while (t--) // each testcase
    {
        ll n;
        cin >> n;
        vi a(n);
        for (int i = 0; i < n; i++) { cin >> a[i]; }
        ll buffer = 0;
        int i = 0;
        while (i != n)
        {
            if (a[i] >= i)
            {
                buffer += a[i] - i;
            }
            else
            {
                ll x = i - a[i];
                a[i] += x;
                buffer -= x;
            }
            if (buffer < 0)
            {
                break;
            }
            i++;
        }
        cout << (i == n ? "YES" : "NO");
        nl;
    }
    return 0;
}

Simply sort the array remove duplicates, now we find the differences between every 2 consequtive elements, with this we have found the max range, now consider this and the distance between the starting point to first candy and ending point to last candy the max of these 3 values will be our result.

// Aur Bhai Dekhne aagaye ;)
// Author: Abhijit Mishra 
#include <bits/stdc++.h>
using namespace std;


#define nl cout << "\n"
#define all(x) x.begin(), x.end()
#define fast ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);


typedef long long int ll;
typedef vector<ll> vi;


int main()
{
    fast;
    ll t = 1;
    // cin >> t;
    while (t--)
    {
        ll n, l;
        cin >> n >> l;
        vector<ll> a(n);


        for (int i = 0; i < n; i++)
        {
            cin >> a[i];
        }

        sort(all(a));
        a.resize(unique(a.begin(), a.end()) - a.begin());

        if (a.size() == 1)
        {
            ll x = a[0];
            cout << max(x - 0, l - x);
            return 0;
        }
        vector<ll> b;
        for (int i = 0; i < a.size(); i++)
        {
            if (i != 0)
            {
                b.push_back(a[i] - a[i - 1]);
            }
        }
        sort(all(b));
        double ans = (double)b[b.size() - 1] / 2;
        double endpoints = max(a[0] - 0, l - a[a.size() - 1]);
        cout << setprecision(14);
        cout << max(endpoints, ans) << endl;
    }
    return 0;
}

https://i.imgur.com/3gGShey.png

If you try finding all the binary format and their corresponding values you will notice that those values are basically equal to their first set bits. __ Simple enough right ?

// Aur Bhai Dekhne aagaye ;)
// Author: Abhijit Mishra 
#include <bits/stdc++.h>
using namespace std;


#define nl cout << "\n"
#define all(x) x.begin(), x.end()
#define fast ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);


typedef long long int ll;
typedef vector<ll> vi;


int main()
{
    fast;

    ll x, y;
    cin >> x >> y;

    cout << 1 + __builtin_ctzll(y);

    return 0;
}

This is a very famous standard problem, try reading about it here https://www.geeksforgeeks.org/check-for-balanced-parentheses-in-an-expression/

I have also attached my own solution please have a look if you having problems.

gist -> simple adding the left bracket in the stack and popping when the right is encountered. we pop from the stack.

// Aur Bhai Dekhne aagaye ;)
// Author: Abhijit Mishra 
#include <bits/stdc++.h>
using namespace std;


#define nl cout << "\n"
#define all(x) x.begin(), x.end()
#define fast ios_base::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);


typedef long long int ll;
typedef vector<ll> vi;


int main()
{
    fast;
    string s;
    cin >> s;
    stack<char> st;
    set<char> left = { '(','{','[','<' };
    int n = s.length();
    ll cnt = 0;
    for (int i = 0; i < n; i++)
    {
        if (left.count(s[i]))
        {
            st.push(s[i]);
        }
        else
        {
            if (st.empty())
            {
                cout << "Impossible" << endl;
                return 0;
            }
            if (st.top() + 0 == s[i] - 1 || st.top() + 0 == s[i] - 2)
            {
                st.pop();
            }
            else
            {
                cnt++;
                st.pop();
            }
        }
    }
    if (st.empty())
    {
        cout << cnt;
    }
    else
    {
        cout << "Impossible" << endl;
    }

    return 0;
}