J, ^{40 29 19 18 15} 14 bytes

|.~1+0{[:\:+/\

Try it online!

^{-11 after applying HyperNeutrino's scan sum idea}

^{-10 after reading Leo's comment to HyperNeutrino: while(any of the cumulative sum is negative)(rotate by 1)}

^{-3 thanks to a clever observation by Bubbler}

Takes input as a list of integers, where [ = _1 and ] = 1 -- reversing the more natural order saves 1 byte.

how

• |.~ Rotate by...
• 1+ 1 plus...
• 0{ the first element of...
• \: the grade down of...
• +/\ the scan sum.

That is, take the index of the max element of the scan sum, add one to it, and rotate by that amount.

J, ^{40 29 19 18 15 14} 13 bytes

|.~1+0{&\:+/\

Try it online!

^{-11 after applying HyperNeutrino's scan sum idea}

^{-10 after reading Leo's comment to HyperNeutrino: while(any of the cumulative sum is negative)(rotate by 1)}

^{-3 thanks to a clever observation by Bubbler}

Takes input as a list of integers, where [ = _1 and ] = 1 -- reversing the more natural order saves 1 byte.

how

• |.~ Rotate by...
• 1+ 1 plus...
• 0{ the first element of...
• \: the grade down of...
• +/\ the scan sum.

That is, take the index of the max element of the scan sum, add one to it, and rotate by that amount.

J, ^{40 29 19 18} 15 bytes

(|.~1 e.+/\)^:_

Try it online!

^{-11 after applying HyperNeutrino's scan sum idea}

^{-10 after reading Leo's comment to HyperNeutrino: while(any of the cumulative sum is negative)(rotate by 1)}

^{-3 thanks to a clever observation by Bubbler}

Takes input as a list of integers, where [ = _1 and ] = 1 -- reversing the more natural order saves 1 byte.

how

• |.~ Rotate by the following amount
• 1 e.+/\ is 1 an element of the scan sum? Returns 1 or 0.
• ^:_ until a fixed point is reached.

When 1 is no longer an element of the scan sum, we'll rotate by 0 -- that is, stay the same. So the fixed point is reached exactly when we hit the first solution.

J, ^{40 29 19 18 15} 14 bytes

|.~1+0{[:\:+/\

Try it online!

^{-11 after applying HyperNeutrino's scan sum idea}

^{-10 after reading Leo's comment to HyperNeutrino: while(any of the cumulative sum is negative)(rotate by 1)}

^{-3 thanks to a clever observation by Bubbler}

Takes input as a list of integers, where [ = _1 and ] = 1 -- reversing the more natural order saves 1 byte.

how

• |.~ Rotate by...
• 1+ 1 plus...
• 0{ the first element of...
• \: the grade down of...
• +/\ the scan sum.

That is, take the index of the max element of the scan sum, add one to it, and rotate by that amount.

J, ^{40 29 19} 18 bytes

1&|.^:(1 e.+/\)^:_

Try it online!

^{-11 after applying HyperNeutrino's scan sum idea}

^{-10 after reading Leo's comment to HyperNeutrino: while(any of the cumulative sum is negative)(rotate by 1)}

Takes input as a list of integers, where [ = _1 and ] = 1 -- reversing the more natural order saves 1 byte.

After that, it's an almost literal translation based on one of Leo's comments:

• 1&|. Rotate left...
• ^:(...)^:_ while...
• 1 e.+/\ 1 is an element of the scan sum.

J, ^{40 29 19 18} 15 bytes

(|.~1 e.+/\)^:_

Try it online!

^{-11 after applying HyperNeutrino's scan sum idea}

^{-10 after reading Leo's comment to HyperNeutrino: while(any of the cumulative sum is negative)(rotate by 1)}

^{-3 thanks to a clever observation by Bubbler}

Takes input as a list of integers, where [ = _1 and ] = 1 -- reversing the more natural order saves 1 byte.

how

• |.~ Rotate by the following amount
• 1 e.+/\ is 1 an element of the scan sum? Returns 1 or 0.
• ^:_ until a fixed point is reached.

When 1 is no longer an element of the scan sum, we'll rotate by 0 -- that is, stay the same. So the fixed point is reached exactly when we hit the first solution.