HackerRank - "Sherlock and MiniMax"

Note this smart observation from Editorial: "M will be either P or Q or (A_i+A_j)/2"..

Yea I know, my code is not condense enough..

#include <cmath>
#include <cstdio>
#include <cmath>
#include <climits>
#include <cctype>
#include <vector>
#include <string>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;

int inRange(int p, int q, int v)
{
return v >= p && v <= q;
}

int main()
{
int n; cin >> n;
vector<int> in(n);
for (int i = 0; i < n; i++)
cin >> in[i];

std::sort(in.begin(), in.end());
int p, q; cin >> p >> q;

int dist = INT_MIN;
int v = -1;

if (q < in[0])
{
v = q;
}
else if (p > in.back())
{
v = p;
}
else
{
auto it_p = std::lower_bound(in.begin(), in.end(), p);

//    Check P's
if (it_p != in.end())
{
int dist_p1 = *it_p - p; // p-->in[i]
int dist_p0 = INT_MAX;     // in[i]-->p
if (it_p != in.begin())
{
auto it_pp = it_p - 1;
dist_p0 = p - *(it_pp);

int mid = (*it_pp + *it_p) / 2;
if (mid > p)    it_p = it_pp;
}
dist = std::min(dist_p0, dist_p1);
v = p;
}

auto it_q = std::upper_bound(in.begin(), in.end(), q);

//    Check mid-points
for (auto it = it_p; it != it_q; it++)
{
int mid = (*it + *(it + 1)) / 2;
int newdist = (*(it + 1) - *it) / 2;
if (inRange(p, q, mid) && (newdist > dist))
{
dist = newdist;
v = mid;
}
}

//    Q's
int newdist_q = INT_MIN;
if (it_q == in.end())
{
newdist_q = q - in.back();
}
else if (it_q != in.begin())
{
newdist_q = std::min(q - *(it_q - 1), *it_q - q);
}
if (newdist_q > dist)
{
dist = newdist_q;
v = q;
}
}

cout << v << endl;
return 0;
}