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edited Jun 25, 2021 at 14:47

5.1k
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R, 27 bytes

any(rank(x<-scan(),,'f')-x)

Try it online!

Returns FALSE for Truthy, TRUE for FALSY (just to shave one byte)

Explanation:

If the array containing the rank of each value is equal to the input array, then it's a permutation of 1..n

Note:
5 bytes are wasted because in rankrank function the default handling strategy forin case of ties is "average their ranks" instead of any other of the other possibilities... argh!

R, 27 bytes

any(rank(x<-scan(),,'f')-x)

Try it online!

Returns FALSE for Truthy, TRUE for FALSY (just to shave one byte)

Explanation:

If the array containing the rank of each value is equal to the input array, then it's a permutation of 1..n

Note:
5 bytes are wasted because in rank function the default handling strategy for ties is "average their ranks" instead of any other of the possibilities... argh!

R, 27 bytes

any(rank(x<-scan(),,'f')-x)

Try it online!

Returns FALSE for Truthy, TRUE for FALSY (just to shave one byte)

Explanation:

If the array containing the rank of each value is equal to the input array, then it's a permutation of 1..n

Note:
5 bytes are wasted because in rank function the default handling strategy in case of ties is "average their ranks" instead of any of the other possibilities... argh!

added 173 characters in body

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edited Jun 25, 2021 at 13:37

digEmAll

edited Jun 25, 2021 at 13:37

digEmAll

5.1k
15
22

R, 27 bytes

any(rank(x<-scan(),,'f')-x)

Try it online!

Returns FALSE for Truthy, TRUE for FALSY (just to shave one byte)

Explanation:

If the array containing the rank of each value is equal to the input array, then it's a permutation of 1..n

Note:
5 bytes are wasted because in rank function the default handling strategy for ties is "average their ranks" instead of any other of the possibilities... argh!

R, 27 bytes

any(rank(x<-scan(),,'f')-x)

Try it online!

Returns FALSE for Truthy, TRUE for FALSY (just to shave one byte)

Explanation:

If the array containing the rank of each value is equal to the input array, then it's a permutation of 1..n

R, 27 bytes

any(rank(x<-scan(),,'f')-x)

Try it online!

Returns FALSE for Truthy, TRUE for FALSY (just to shave one byte)

Explanation:

If the array containing the rank of each value is equal to the input array, then it's a permutation of 1..n

Note:
5 bytes are wasted because in rank function the default handling strategy for ties is "average their ranks" instead of any other of the possibilities... argh!