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feat: add new Josephus file in Recursive directory #1728
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Original file line number | Diff line number | Diff line change |
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/** | ||
* @function Josephus | ||
* @description recursive implementation of the Josephus function. | ||
* @param {Integer[]} collection - The integer array. | ||
* @param {Integer} - The step size. | ||
* @return {Integer} - The last integer in the list. | ||
* @see [JosephusProblem](https://en.wikipedia.org/wiki/Josephus_problem) | ||
* @example [1,2,3,4,5,6,7] with step 3 = 4 | ||
*/ | ||
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const josephus = (collection, step) => { | ||
// return null for invalid steps that are less than or equal to 0 | ||
if (step <= 0 || collection.length === 0) { | ||
return null | ||
} | ||
if (collection.length === 1) { | ||
return collection[0] | ||
} else { | ||
step = (step - 1) % collection.length | ||
collection.splice(step, 1) | ||
return josephus(collection, step + collection.length) | ||
} | ||
} | ||
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export { josephus } |
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import { josephus } from '../Josephus' | ||
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describe('Josephus', () => { | ||
const collection = [1, 2, 3, 4, 5, 6, 7] | ||
const collection1 = [1] | ||
const collection2 = [] | ||
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it('should return 4 for step size of 3', () => { | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Repetitive tests of the form "should return X for Y" should use |
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const step = 3 | ||
expect(josephus([...collection], step)).toBe(4) | ||
}) | ||
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it('should return 4 for a step size of 10', () => { | ||
const step = 10 | ||
expect(josephus([...collection], step)).toBe(4) | ||
}) | ||
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it('should return null for a step size of 0 as it is invalid', () => { | ||
const step = 0 | ||
expect(josephus([...collection], step)).toBeNull() | ||
}) | ||
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it('should return 7 for a step size of 1000', () => { | ||
const step = 1000 | ||
expect(josephus([...collection], step)).toBe(7) | ||
}) | ||
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it('should return null for a step size of -1 as it is invalid', () => { | ||
const step = -1 | ||
expect(josephus([...collection], step)).toBeNull() | ||
}) | ||
it('should return 1 for a collection with just 1', () => { | ||
const step = 2 | ||
expect(josephus([...collection1], step)).toBe(1) | ||
}) | ||
it('should return null for an empty collection as it is invalid', () => { | ||
const step = 3 | ||
expect(josephus([...collection2], step)).toBeNull() | ||
}) | ||
}) |
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This is$O(n)$ so you get $O(n^2)$ runtime, which is very naive. Wikipedia shows formulae that achieve $O(n)$ or $O(k\log{n})$ runtime, both of which are substantially better. And even besides those, I can pretty easily come up with $O(n \log{n})$ algorithms based on decremental data structures which support this "splice" operation in logarithmic time (such as an initially balanced k-nary tree augmented with counts).
This algorithm also doesn't really benefit from recursion. It would be clearer as a simple
while
loop: