Algorithm
Problem link- https://www.spoj.com/problems/TRIPINV/
TRIPINV - Mega Inversions
The n^2 upper bound for any sorting algorithm is easy to obtain: just take two elements that are misplaced with respect to each other and swap them. Conrad conceived an algorithm that proceeds by taking not two, but three misplaced elements. That is, take three elements ai > aj > ak with i < j < k and place them in order ak; aj; ai. Now if for the original algorithm the steps are bounded by the maximum number of inversions n(n-1)/2, Conrad is at his wits' end as to the upper bound for such triples in a given sequence. He asks you to write a program that counts the number of such triples.
Input
The first line of the input is the length of the sequence, 1 <= n <= 10^5. The next line contains the integer sequence a1, a2 ... an. You can assume that all ai belongs [1; n].
Output
Output the number of inverted triples.
Example
Input: 4 3 3 2 1 Output: 2
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MOD 1000000007LL
#define EPS 1e-9
#define io ios_base::sync_with_stdio(false);cin.tie(NULL);
#define M_PI 3.14159265358979323846
template <typename T> T gcd(T a, T b){return (b==0)?a:gcd(b,a%b);}
template <typename T> T lcm(T a, T b){return a*(b/gcd(a,b));}
template <typename T> T mod_exp(T b, T p, T m){T x = 1;while(p){if(p&1)x=(x*b)%m;b=(b*b)%m;p=p>>1;}return x;}
template <typename T> T invFermat(T a, T p){return mod_exp(a, p-2, p);}
template <typename T> T exp(T b, T p){T x = 1;while(p){if(p&1)x=(x*b);b=(b*b);p=p>>1;}return x;}
const int MAXN = 1e5+5;
ll BIT1[MAXN], BIT2[MAXN], A[MAXN];
ll ans1[MAXN], ans2[MAXN];
void update(int idx, int val, int bit){
while(idx <= MAXN){
if(bit == 1)
BIT1[idx] += val;
else
BIT2[idx] += val;
idx += idx & -idx;
}
}
ll query(int idx, int bit){
ll sum = 0;
while(idx > 0){
if(bit == 1)
sum += BIT1[idx];
else
sum += BIT2[idx];
idx -= idx & -idx;
}
return sum;
}
int main(){
io;
int n;
cin >> n;
for(int i = 0;i < n; i++)
cin >> A[i];
for(int i = 0;i < n; i++){
ans1[i] = (query(100000, 1)-query(A[i], 1));
update(A[i], 1, 1);
}
for(int i = n-1;i >= 0; i--){
ans2[i] = (query(A[i]-1, 2));
update(A[i], 1, 2);
}
ll ans = 0;
for(int i = 0;i < n; i++){
ans += (ans1[i]*ans2[i]);
}
cout << ans << endl;
return 0;
}Input
3 3 2 1
Output
Demonstration
SPOJ Solution-Mega Inversions-Solution in C, C++, Java, Python