feat: added find subsets algorithm using bitmanipulation (#1514)

* feat: added find subsets algorithm using bitmanipulation

* file name fix

* test file name fix

* fix: codespell fix

* error handled

* added test cases for error
This commit is contained in:
Nikunj Bisht
2023-10-26 19:08:19 +05:30
committed by GitHub
parent f67cdc3cad
commit d671327ebe
2 changed files with 70 additions and 0 deletions

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/**
* @function generateSubSets
* @param {Array} inputArray
* @returns {Array}
* @example [1,2] -> [[],[1],[2],[1,2]]
*/
// The time complexity of this algorithm is BigO(2^n) where n is the length of array
function generateSubSets(inputArray) {
if (!Array.isArray(inputArray)) {
throw new Error('Provided input is not an array')
}
if (inputArray.length > 32) {
throw new RangeError('Error size should be less than equal to 32')
}
let arrayLength = inputArray.length
let subSets = []
// loop till (2^n) - 1
for (let i = 0; i < 1 << arrayLength; i++) {
let subSet = []
for (let j = 0; j < arrayLength; j++) {
// 1 << j it shifts binary digit 1 by j positions and then we perform
// and by AND operation we are checking whetheer jth bit
// in i is set to 1 if result is non zero just add into set
if (i & (1 << j)) {
subSet.push(inputArray[j])
}
}
subSets.push(subSet)
}
return subSets
}
export { generateSubSets }

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import { generateSubSets } from '../GenerateSubSets'
describe('subSets', () => {
it('find the subsets', () => {
expect(generateSubSets([1, 2, 3])).toEqual([
[],
[1],
[2],
[1, 2],
[3],
[1, 3],
[2, 3],
[1, 2, 3]
])
expect(generateSubSets([1, 2])).toEqual([[], [1], [2], [1, 2]])
expect(generateSubSets([1, 2, 3])).toEqual([
[],
[1],
[2],
[1, 2],
[3],
[1, 3],
[2, 3],
[1, 2, 3]
])
expect(() => generateSubSets('invalid')).toThrow(
'Provided input is not an array'
)
expect(() =>
generateSubSets([
1, 2, 2, 1, 2, 3, 4, 3, 2, 3, 4, 3, 2, 2, 2, 3, 12, 11, 4, 2, 2, 2, 2,
1, 2, 3, 5, 6, 7, 7, 8, 6, 5, 6, 7, 8, 9, 8, 0, 6
])
).toThrow('Error size should be less than equal to 32')
})
})