I recently came across problem 1861D and I'm having trouble understanding the solution provided by one of the users . The solution is given below:
Link to question -> ((http://codeforces.me/contest/1861/problem/D)
#include<bits/stdc++.h>
using namespace std;
int t,n,a[200005];
int main(){
cin>>t;
while(t--){
cin>>n;
for(int i=1;i<=n;i++){
cin>>a[i];
}
int ans=INT_MAX,a2=0,a1=0;
for(int i=1;i<n;i++)if(a[i]>=a[i+1])a2++;
ans=min(ans,a2);
for(int i=1;i<n;i++){
if(a[i]>=a[i-1])a1++;
if(a[i]>=a[i+1])a2--;
ans=min(ans,a2+a1);
}
cout<<ans<<"\n";
}
return 0;
}
Could you please explain how this solution works? Specifically, I'm having trouble understanding the logic behind the a1 and a2 variables and how they are used to calculate the minimum number of operations needed to sort the array in strictly ascending order.
If we want to sort the entire array in positive ascending order, we can do the following greedy strategy: if $$$A_i \le A_{i+1}$$$, then we multiply the range $$$[i+1, N]$$$ by some large positive integer (the actual value of $$$x$$$ doesn't matter). In the code provided, the # of moves required for this greedy strategy is $$$a2$$$.
Since we can also use negative values of $$$x$$$, we might want some prefix of the array to become negative, and the remaining suffix positive. We can just do the reverse of the previous greedy algorithm, and find the # of moves needed to make the array strictly decreasing (Increase the range $$$[1,i]$$$ whenever $$$A_i \le A_{i+1}$$$). This is the value $$$a1$$$ in the solution provided. The # of moves needed for a descending array is equivalent to a negative ascending array: we can just assume that our last value of $$$x$$$ was $$$-x$$$.
In the code above, the author decrements $$$a2$$$ while computing $$$a1$$$. This means that at any given position $$$i$$$, the value $$$a1$$$ represents the # of moves needed to sort $$$[1, i]$$$ in negative ascending order, the value $$$a2$$$ represents the # of moves to sort $$$[i+1,N]$$$ in positive ascending order, and the total # of moves is just $$$a1+a2$$$. The optimal answer can then be found taking the minimum value of $$$a1+a2$$$ throughout this iteration.