Minimum window is "BANC". A string with 1 `A`, 1 `B` and 1 `C`.

Something you shuold notice:

* if a substring is a window string, it plus any string is also a window string.

`ADOBEC` is a window string, so `ADOBECODE` is a window string too.

* if a window string's left most or right most character, lets say `c`, is in `T` and the window string has more character `c` than `T` has, removing the `c` from the window string will not break that the remaining string is a window string.

`ADOBECODEBA` has 2 `A`, and more than `T` has (1 `A`). so, after remove `A` from left or right, the string remains a window string (`ADOBECODEB` is a window string).

Assume that we had found a window string, `W`.

Then

1. When a new char comes, `W` + new char = `WN` is also a window. (Rule 1)

1. Trim the new window string, `WN`, from the left part, remove any char that is redundant to have a window string. (Rule 2)

Code looks like

the `W` should have any char in `T`, this problem is like the `Check concatenation using count sum` in [Substring with Concatenation of All Words](../substring-with-concatenation-of-all-words).

Use a map `need`, stores how many chars are needed.

Use a map `seen`, stores how many chars are seen.

Those maps are not only used in finding `W`, but also in triming redundant from `WN`.

This last problem is when the `W` is found.

Like [Substring with Concatenation of All Words](../substring-with-concatenation-of-all-words), just use a `counter`,

When `seen` is added, just add 1 to the counter.

The time `counter == T.length` is the time we first found `W`.

_Note_:

a char in `seen[c]` can contribute at most `need[c]` count to the `counter`.