Why is this solution sufficient for the problem "Speedbreaker"?

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In the speedbreaker question, why do we need to do: c++ mn[i]=min(mn[i],mn[i-1]),mx[i]=max(mx[i],mx[i-1]);

Full Code

#include<bits/stdc++.h>
using namespace std;
int T,n,a[200005],mn[200005],mx[200005],l,r;
void solve(){
	cin>>n;
	for(int i=0;i<=n;i++)mn[i]=n+1,mx[i]=0;
	for(int i=1;i<=n;i++)cin>>a[i],mn[a[i]]=min(mn[a[i]],i),mx[a[i]]=max(mx[a[i]],i);
	for(int i=1;i<=n;i++){
		mn[i]=min(mn[i],mn[i-1]),mx[i]=max(mx[i],mx[i-1]);
		if(mx[i]-mn[i]>=i){cout<<"0\n";return;}
	}
	l=1,r=n;
	for(int i=1;i<=n;i++)l=max(l,i-a[i]+1),r=min(r,i+a[i]-1);
	cout<<r-l+1<<'\n';
}
signed main(){
	ios::sync_with_stdio(false);
	cin.tie(0),cout.tie(0);
	cin>>T;
	while(T--)solve();
	return 0;
}

Also, I get why the first condition is necessary for there to be a solutions and I get why the second part of the code finds that good interval, but I'm having a little trouble piecing together the sufficiency of the two to guarantee the right answer. Perhaps this is also why I font understand why you have to do the operations on the mins and maxes too.

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