# HackerEarth Little Shino and Pairs problem solution

In this HackerEarth Little Shino and Pairs problem solution we have given a permutation of numbers from 1 to N. Among all the subarrays, find the number of unique pairs (a,b) such that a != b and a is maximum and b is the second maximum in that subarray.

## HackerEarth Little Shino and Pairs problem solution.

`#include <bits/stdc++.h>#define ll long long#define ull unsigned long long#define pb push_back#define mp make_pair#define fi first#define se second#define be begin()#define en end()#define all(x) (x).begin(),(x).end()#define alli(a, n, k) (a+k),(a+n+k)#define REP(i, a, b, k) for(__typeof(a) i = a;i < b;i += k)#define REPI(i, a, b, k) for(__typeof(a) i = a;i > b;i -= k)#define REPITER(it, a) for(__typeof(a.begin()) it = a.begin();it != a.end(); ++it)#define y0 sdkfaslhagaklsldk#define y1 aasdfasdfasdf#define yn askfhwqriuperikldjk#define j1 assdgsdgasghsf#define tm sdfjahlfasfh#define lr asgasgash#define norm asdfasdgasdgsd#define have adsgagshdshfhds#define eps 1e-6#define pi 3.141592653589793using namespace std;template<class T> inline T gcd(T x, T y) { if (!y) return x; return gcd(y, x%y); }template<class T> inline T mod(T x) { if(x < 0) return -x; else return x; }typedef vector<int> VII;typedef vector<ll> VLL;typedef pair<int, int> PII;typedef vector< pair<int, int> > VPII;typedef vector< pair<int, PII > > VPPI;const int MOD = 1e9 + 7;const int INF = 1e9;int main(int argc, char* argv[]){    if(argc == 2 or argc == 3) freopen(argv[1], "r", stdin);    if(argc == 3) freopen(argv[2], "w", stdout);    ios::sync_with_stdio(false);    int n, ans = 0, a;    stack <int> stk;    cin >> n;    REP(i, 0, n, 1)    {        cin >> a;        while(!stk.empty() and a > stk.top())        {            stk.pop();            ans++;        }        if(!stk.empty()) ans++;        stk.push(a);    }    cout << ans << endl;    return 0;}`

### Second solution

`#include <bits/stdc++.h>using namespace std;int N;pair<int, int> A[100000];int main(){    scanf("%d", &N);    for(int i=0; i<N; i++)        scanf("%d", &A[i].first), A[i].second=i;    sort(A, A+N, greater<pair<int, int>>());    int a=A[0].second, b=A[0].second;    int ans=0;    for(int i=1; i<N; i++)    {        int x=A[i].second;        if(x<a || x>b)            ans++;        else            ans+=2;        a=min(a, x);        b=max(b, x);    }    printf("%d\n", ans);    return 0;}`