Divide and Conquer, Optimized O(n^2)

// Given an array arr[] consisting of N integers, the task is to check if all subarrays of the array
// have at least one unique element in it or not. If found to be true, then print “Yes”.else  print “No”.
// https://www.geeksforgeeks.org/check-if-all-subarrays-contains-at-least-one-unique-element/
#include <bits/stdc++.h>
// #include <atcoder/all>
using namespace std;
#define int long long
#define all(x) (x).begin(), (x).end()
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pair<int, int>> vpi;
typedef pair<int, int> pi;
#define f first
#define s second
#define pb push_back
#define endl "\n"
#define yes cout << "YES" << endl
#define no cout << "NO" << endl
int dx[] = {-1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};
const int mod1 = 1e9 + 7, mod2 = 998244353, INF = 2e18, N = 2e5 + 5, L = 19;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
// -----------------------------------------------------------------------------
vi a;
bool solve(int l, int r)
{
    if (l >= r)
    {
        return true;
    }
    map<int, int> mpp;
    for (int i = l; i <= r; i++)
    {
        mpp[a[i]]++;
    }
    // am1, am2, am3,..., amn -> unique elements, divide & conquer in each range
    int last = l;
    bool unique = false;
    for (int i = l; i <= r; i++)
    {
        if (mpp[a[i]] == 1)
        {
            unique = true;
            if (!solve(last, i - 1))
            {
                return false;
            }
            last = i + 1;
        }
    }
    return (unique && (solve(last, r)));
}
void solve()
{
    int n;
    cin >> n;
    a.resize(n);
    for (int i = 0; i < n; i++)
    {
        cin >> a[i];
    }
    if (solve(0, n - 1))
    {
        yes;
        return;
    }
    no;
    return;
}

signed main()
{
    // __START__;
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t = 1;
    // cin >> t;
    while (t--)
    {
        solve();
    }
    // __END__;
    return 0;
}