S = "ADOBECODEBANC"
T = "ABC"

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)
2. 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

W = a window string in S

foreach c after W in S
  
  WN = W + c
  
  // Trim left
  
  foreach _c in WN
  
    if _c can be remove from WN
      remove _c from the left part of WN
    else
      break
   
  W = WN

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.

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, 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.