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Possible MEX Values After One Optional Increment

Citadel● MediumFULLTIMEOA

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Problem statement

Given an array values of length n, each value is between 0 and n - 1. You may leave the array unchanged, or choose one element smaller than n - 1 and increment it by exactly one.

Return every distinct MEX that can result, in increasing order. The MEX is the smallest non-negative integer absent from the array.

Function

possibleMexValues(values: int[]) → int[]

Examples

Example 1

values = [0,1]return = [0,2]

Leaving the array unchanged gives MEX 2; incrementing the only 0 gives MEX 0.

Example 2

values = [0,0,1,3]return = [1,2]

MEX 2 is unchanged, while incrementing the unique 1 removes the last 1 and gives MEX 1.

Constraints

1 <= values.length <= 200000
0 <= values[i] < values.length