R, 32 bytes

function(h,n)rank(1:n%in%h,,"f")

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1:n %in% h returns a vector of mainly zeros, with ones where the high-water marks should be. The "f" (short for ties="first") argument to rank() indicates that ties with the same rank should be broken using "increasing values at each index set of ties".

Old approach: R, 40 37 bytes

Edit: -2 bytes thanks to Command Master & Giuseppe

function(h,n){x=1:n;x[h]=-x;rank(-x)}

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R, 23 bytes

Edit: -9 bytes thanks to Command Master by switching to use a bitmask as input

function(h)rank(h,,"f")

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Input is simply a bitmask h, no need to provide n as this is implicitly the length of the bitmask. 
The "f" (short for ties="first") argument to rank() indicates that ties with the same rank should be broken using "increasing values at each index set of ties".

R, 32 bytes

function(h,n)rank(1:n%in%h,,"f")

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Same approach as above, except inputting h as the indices of high-water marks, together with n. 1:n %in% h returns a vector of mainly zeros, with ones where the high-water marks should be (in other words, converting the indices to a bitmask).

R, 40 37 bytes

Edit: -2 bytes thanks to Command Master & Giuseppe

function(h,n){x=1:n;x[h]=-x;rank(-x)}

Try it online!

R, 32 bytes

function(h,n)rank(1:n%in%h,,"f")

Try it online!

1:n %in% h returns a vector of mainly zeros, with ones where the high-water marks should be. The "f" (short for ties="first") argument to rank() indicates that ties with the same rank should be broken using "increasing values at each index set of ties".

Old approach: R, 40 37 bytes

Edit: -2 bytes thanks to Command Master & Giuseppe

function(h,n){x=1:n;x[h]=-x;rank(-x)}

Try it online!

R, 40 bytes

function(h,n){x=-(1:n);x[h]=1:n;rank(x)}

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R, 40 37 bytes

Edit: -2 bytes thanks to Command Master & Giuseppe

function(h,n){x=1:n;x[h]=-x;rank(-x)}

Try it online!