hiii bro you can print this pattern this question asked in zoho company
1 12 11 10 9
2 13 15 8
3 14 7
4 6
5
this pattern is right side
Thank You
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thank you
Problem Code: ADJSRT
#include<bits/stdc++.h>
#define ll long long
using namespace std;
int n;
vectorv;
ll num;
int merge(int l,int r,vector<pair<ll,ll>>&a,int mid)
{
int inv=0;
int l_sz=mid-l+1;
int r_sz=r-mid;
vector<pair<ll,ll>>L(l_sz+1);
vector<pair<ll,ll>>R(r_sz+1);
for(int i=0;i<l_sz;i++)
{
L[i].first=a[i+l].first;
L[i].second=a[i+l].second;
}
for(int i=0;i<r_sz;i++)
{
R[i].first=a[mid+1+i].first;
R[i].second=a[mid+1+i].second;
}
L[l_sz].first=LONG_MAX;
R[r_sz].first=LONG_MAX;
L[l_sz].second=-1;
R[r_sz].second=-1;
int r_i=0;
int l_i=0;
for(int i=l;i<=r;i++)
{
if(L[l_i].first<=R[r_i].first)
{
a[i].first=L[l_i].first;
L[l_i].second=i;
l_i++;
}
else
{
a[i].first=R[r_i].first;
r_i++;
for(int i=l_sz-1;i>=l_i;i--)
{
v.push_back(L[i].second);
L[i].second++;
}
}
}
return inv;
}
int mergesort(int l,int r,vector<pair<ll,ll>>&a)
{
int inv=0;
if(l<r)
{
int mid=(l+r)/2;
inv+=mergesort(l,mid,a);
inv+=mergesort(mid+1,r,a);
inv+=merge(l,r,a,mid);
}
return inv;
}
int main()
{
cin>>n;
ll x;
vector<pair<ll,ll>>a(n);
int inv;
for(int i=0;i<n;i++)
{
cin>>x;
a[i]={x,i};
}
inv=mergesort(0,n-1,a);
cout<<v.size()<<endl;
for(int i=0;i<v.size();i++)
cout<<v[i]<<" ";
return 0;
}
can you please reduce the time complexity of this code?
Hey @aarushiagarwal 
,
Could you please tell us what exactly your code is doing.Also Your code’s time complexity is already O(nlogn) but space complexity can be optimized.
yes, my code was submitted. thanks for your time