Lines 12-14 of the solution contain the following info:
// this is tricky; in the current window, we will not have any 'rightChar' after its previous index // and if 'windowStart' is already ahead of the last index of 'rightChar', we'll keep 'windowStart' windowStart = Math.max(windowStart, charIndexMap[rightChar] + 1);
However, the application of the
max function seems to be unnecessary, i.e. if we change the code to
windowStart = charIndexMap[rightChar] + 1;
it produces the same result.
This is because
charIndexMap[rightChar] + 1 is guaranteed to be larger than
A similar question was asked before, but it was left without the answer, so I decided to try my luck as well.