The maximum distance | Editorial by coderag

Since the input constraints are very small, this problem can be done with O(n) for each query.
For the first and second query, the array values are updated and for the third query, we need to find the maximum distance between the elements that are equal to z in the array.  For this
simply, we can find the first and last occurrence of the element in the array and calculate the length using that.  If the array value is not found in the array, then we can print -1 else we can
print the length.

```
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
void solve() {
	ll n;
	cin>>n;
	vector<ll> a(n);
	for(ll& i:a) {
		cin>>i;
	}
	ll q;
	cin>>q;
	while(q--) {
		ll t,l,r,x;
		cin>>t;
		if(t==1) {
			cin>>l>>r>>x;
			for(int i=l-1;i<r;i++) {
				a[i]+=x;
			}
		}
		else if(t==2) {
			cin>>l>>r>>x;
			for(int i=l-1;i<r;i++) {
				a[i]*=x;
			}
		}
		else if(t==3) {
			int f1 = -1,f2 = -1;
			cin>>x;
			for(int i=0;i<n;i++) {
				if(a[i]==x) {
					f1 = i;
					break;
				}
			}
			if(f1==-1) {
				cout<<-1<<"\n";
				continue;
			}
			for(int i=n-1;i>=0;i--) {
				if(a[i]==x) {
					f2 = i;
					break;
				}
			}
			if(f2==-1) {
				cout<<-1<<"\n";
				continue;
			}
			cout<<f2-f1+1<<"\n";
		}
	}
}
int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	int test = 1;
	while(test--) {
		solve();
	}
}

```

The maximum distance | Editorial by coderag

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