# HackerEarth Lonely M's array problem solution

In this HackerEarth Lonely M's array problem solution The lonely M is again playing with his array a1,a2,...,an (as playing with his array is the only choice), he wants to choose a subsequence of array like b1,b2,...,bm that b1 >= b2 >= b3 >= b4 .. <= (>= if m = 0 (mod 2 ))bm.

What is the maximum length of subsequence that M can choose which satisfies the above condition?

The subsequence of an array can be obtained by erasing some (possibly zero) elements from the array. You can erase any elements, not necessarily going successively. The remaining elements preserve their order.

## HackerEarth Lonely M's array problem solution.

`#include<bits/stdc++.h>using namespace std;#define greater greater123123#define less less123123const int N = 3e5+100;int a[N];struct Seg{    int seg[4*N];    Seg(){        memset(seg, 0, sizeof seg);    }    int get(int l, int r, int s=0, int e=N, int id=1){        if(r<=s || e<=l)            return 0;        if(l<=s && e<=r)            return seg[id];        int mid = (s + e)/2;        return max(get(l, r, s, mid, 2*id), get(l, r, mid, e, 2*id+1));    }    void upd(int x, int val, int s=0, int e=N, int id=1){        if(s>x || e<=x)            return;        if(e-s == 1){            seg[id] = max(seg[id], val);            return;        }        int mid = (s + e)/2;        upd(x, val, s, mid, 2*id);        upd(x, val, mid, e, 2*id+1);        seg[id] = max(seg[2*id], seg[2*id+1]);    }} greater, less;int main(){    ios_base::sync_with_stdio(false);cin.tie(NULL);    int n;cin >> n;    for(int i=0 ; i<n ; i++)        cin >> a[i];    reverse(a, a+n);    int ans=0;    for(int i=0 ; i<n ; i++){        int cur1 = less.get(0, a[i]+1) + 1;        ans = max(ans, cur1);        int cur2 = greater.get(a[i], N) + 1;        greater.upd(a[i], cur1);        less.upd(a[i], cur2);    }    cout << ans << "\n";}`