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Nov 16, 2023
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2 changes: 2 additions & 0 deletions DIRECTORY.md
Original file line number Diff line number Diff line change
Expand Up @@ -160,6 +160,7 @@
* [NumberOfIslands](Graphs/NumberOfIslands.js)
* [PrimMST](Graphs/PrimMST.js)
* **Hashes**
* [MD5](Hashes/MD5.js)
* [SHA1](Hashes/SHA1.js)
* [SHA256](Hashes/SHA256.js)
* **Maths**
Expand Down Expand Up @@ -300,6 +301,7 @@
* [KochSnowflake](Recursive/KochSnowflake.js)
* [LetterCombination](Recursive/LetterCombination.js)
* [Palindrome](Recursive/Palindrome.js)
* [Partition](Recursive/Partition.js)
* [SubsequenceRecursive](Recursive/SubsequenceRecursive.js)
* [TowerOfHanoi](Recursive/TowerOfHanoi.js)
* **Search**
Expand Down
47 changes: 47 additions & 0 deletions Dynamic-Programming/Abbreviation.js
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/**
* @description
* Given two strings, `source` and `target`, determine if it's possible to make `source` equal
* to `target` You can perform the following operations on the string `source`:
* 1. Capitalize zero or more of `source`'s lowercase letters.
* 2. Delete all the remaining lowercase letters in `source`.
*
* Time Complexity: (O(|source|*|target|)) where `|source|` => length of string `source`
*
* @param {String} source - The string to be transformed.
* @param {String} target - The string we want to transform `source` into.
* @returns {Boolean} - Whether the transformation is possible.
* @see https://www.hackerrank.com/challenges/abbr/problem - Related problem on HackerRank.
*/
export const isAbbreviation = (source, target) => {
const sourceLength = source.length
const targetLength = target.length

// Initialize a table to keep track of possible abbreviations
let canAbbreviate = Array.from({ length: sourceLength + 1 }, () =>
Array(targetLength + 1).fill(false)
)
// Empty strings are trivially abbreviatable
canAbbreviate[0][0] = true

for (let sourceIndex = 0; sourceIndex < sourceLength; sourceIndex++) {
for (let targetIndex = 0; targetIndex <= targetLength; targetIndex++) {
if (canAbbreviate[sourceIndex][targetIndex]) {
// If characters at the current position are equal, move to the next position in both strings.
if (
targetIndex < targetLength &&
source[sourceIndex].toUpperCase() === target[targetIndex]
) {
canAbbreviate[sourceIndex + 1][targetIndex + 1] = true
}
// If the current character in `source` is lowercase, explore two possibilities:
// a) Capitalize it (which is akin to "using" it in `source` to match `target`), or
// b) Skip it (effectively deleting it from `source`).
if (source[sourceIndex] === source[sourceIndex].toLowerCase()) {
canAbbreviate[sourceIndex + 1][targetIndex] = true
}
}
}
}

return canAbbreviate[sourceLength][targetLength]
}
30 changes: 30 additions & 0 deletions Dynamic-Programming/tests/Abbreviation.test.js
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import { isAbbreviation } from '../Abbreviation.js'

const expectPositive = (word, abbr) =>
expect(isAbbreviation(word, abbr)).toBe(true)
const expectNegative = (word, abbr) =>
expect(isAbbreviation(word, abbr)).toBe(false)

describe('Abbreviation - Positive Tests', () => {
test('it should correctly abbreviate or transform the source string to match the target string', () => {
expectPositive('', '')
expectPositive('a', '')
expectPositive('a', 'A')
expectPositive('abcDE', 'ABCDE')
expectPositive('ABcDE', 'ABCDE')
expectPositive('abcde', 'ABCDE')
expectPositive('abcde', 'ABC')
expectPositive('abcXYdefghijKLmnopqrs', 'XYKL')
expectPositive('abc123', 'ABC')
expectPositive('abc123', 'ABC123')
expectPositive('abc!@#def', 'ABC')
})
})

describe('Abbreviation - Negative Tests', () => {
test('it should fail to abbreviate or transform the source string when it is not possible to match the target string', () => {
expectNegative('', 'A')
expectNegative('a', 'ABC')
expectNegative('aBcXYdefghijKLmnOpqrs', 'XYKLOP')
})
})
39 changes: 39 additions & 0 deletions Recursive/Partition.js
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/**
* @function canPartition
* @description Check whether it is possible to partition the given array into two equal sum subsets using recursion.
* @param {number[]} nums - The input array of numbers.
* @param {number} index - The current index in the array being considered.
* @param {number} target - The target sum for each subset.
* @return {boolean}.
* @see [Partition Problem](https://en.wikipedia.org/wiki/Partition_problem)
*/

const canPartition = (nums, index = 0, target = 0) => {
if (!Array.isArray(nums)) {
throw new TypeError('Invalid Input')
}

const sum = nums.reduce((acc, num) => acc + num, 0)

if (sum % 2 !== 0) {
return false
}

if (target === sum / 2) {
return true
}

if (index >= nums.length || target > sum / 2) {
return false
}

// Include the current number in the first subset and check if a solution is possible.
const withCurrent = canPartition(nums, index + 1, target + nums[index])

// Exclude the current number from the first subset and check if a solution is possible.
const withoutCurrent = canPartition(nums, index + 1, target)

return withCurrent || withoutCurrent
}

export { canPartition }
24 changes: 24 additions & 0 deletions Recursive/test/Partition.test.js
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import { canPartition } from '../Partition'

describe('Partition (Recursive)', () => {
it('expects to return true for an array that can be partitioned', () => {
const result = canPartition([1, 5, 11, 5])
expect(result).toBe(true)
})

it('expects to return false for an array that cannot be partitioned', () => {
const result = canPartition([1, 2, 3, 5])
expect(result).toBe(false)
})

it('expects to return true for an empty array (0 elements)', () => {
const result = canPartition([])
expect(result).toBe(true)
})

it('Throw Error for Invalid Input', () => {
expect(() => canPartition(123)).toThrow('Invalid Input')
expect(() => canPartition(null)).toThrow('Invalid Input')
expect(() => canPartition(undefined)).toThrow('Invalid Input')
})
})