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Commit 82f0b5e

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Add a Divide and Conquer version of the MaxSubArray problem.
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‎.npmrc

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engine-strict=false
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engine-strict=true

‎.nvmrc

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v14

‎README.md

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* `B` [Best Time To Buy Sell Stocks](src/algorithms/uncategorized/best-time-to-buy-sell-stocks) - divide and conquer and one-pass examples
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* `A` [Permutations](src/algorithms/sets/permutations) (with and without repetitions)
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* `A` [Combinations](src/algorithms/sets/combinations) (with and without repetitions)
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* `A` [Maximum Subarray](src/algorithms/sets/maximum-subarray)
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* **Dynamic Programming** - build up a solution using previously found sub-solutions
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* `B` [Fibonacci Number](src/algorithms/math/fibonacci)
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* `B` [Jump Game](src/algorithms/uncategorized/jump-game)
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npm i
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```
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Also make sure that you're using a correct Node version (`>=14.16.0`). If you're using [nvm](https://github.com/nvm-sh/nvm) for Node version management you may run `nvm use` from the root folder of the project and the correct version will be picked up.
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**Playground**
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You may play with data-structures and algorithms in `./src/playground/playground.js` file and write

‎src/algorithms/sets/maximum-subarray/README.md

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# Maximum subarray problem
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The maximum subarray problem is the task of finding the contiguous
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subarray within a one-dimensional array, `a[1...n]`, of numbers
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The maximum subarray problem is the task of finding the contiguous
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subarray within a one-dimensional array, `a[1...n]`, of numbers
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which has the largest sum, where,
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![Maximum subarray](https://wikimedia.org/api/rest_v1/media/math/render/svg/e8960f093107b71b21827e726e2bad8b023779b2)
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## Example
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The list usually contains both positive and negative numbers along
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with `0`. For example, for the array of
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values `−2, 1, −3, 4, −1, 2, 1, −5, 4` the contiguous subarray
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The list usually contains both positive and negative numbers along
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with `0`. For example, for the array of
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values `−2, 1, −3, 4, −1, 2, 1, −5, 4` the contiguous subarray
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with the largest sum is `4, −1, 2, 1`, with sum `6`.
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## Solutions
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- Brute Force solution `O(n^2)`: [bfMaximumSubarray.js](./bfMaximumSubarray.js)
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- Divide and Conquer solution `O(n^2)`: [dcMaximumSubarraySum.js](./dcMaximumSubarraySum.js)
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- Dynamic Programming solution `O(n)`: [dpMaximumSubarray.js](./dpMaximumSubarray.js)
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## References
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- [Wikipedia](https://en.wikipedia.org/wiki/Maximum_subarray_problem)

‎src/algorithms/sets/maximum-subarray/__test__/bfMaximumSubarray.test.js

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import bfMaximumSubarray from '../bfMaximumSubarray';
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describe('bfMaximumSubarray', () => {
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it('should find maximum subarray using brute force algorithm', () => {
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it('should find maximum subarray using the brute force algorithm', () => {
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expect(bfMaximumSubarray([])).toEqual([]);
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expect(bfMaximumSubarray([0, 0])).toEqual([0]);
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expect(bfMaximumSubarray([0, 0, 1])).toEqual([0, 0, 1]);
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import dcMaximumSubarray from '../dcMaximumSubarraySum';
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describe('dcMaximumSubarraySum', () => {
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it('should find maximum subarray sum using the divide and conquer algorithm', () => {
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expect(dcMaximumSubarray([])).toEqual(-Infinity);
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expect(dcMaximumSubarray([0, 0])).toEqual(0);
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expect(dcMaximumSubarray([0, 0, 1])).toEqual(1);
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expect(dcMaximumSubarray([0, 0, 1, 2])).toEqual(3);
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expect(dcMaximumSubarray([0, 0, -1, 2])).toEqual(2);
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expect(dcMaximumSubarray([-1, -2, -3, -4, -5])).toEqual(-1);
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expect(dcMaximumSubarray([1, 2, 3, 2, 3, 4, 5])).toEqual(20);
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expect(dcMaximumSubarray([-2, 1, -3, 4, -1, 2, 1, -5, 4])).toEqual(6);
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expect(dcMaximumSubarray([-2, -3, 4, -1, -2, 1, 5, -3])).toEqual(7);
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expect(dcMaximumSubarray([1, -3, 2, -5, 7, 6, -1, 4, 11, -23])).toEqual(27);
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});
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});

‎src/algorithms/sets/maximum-subarray/__test__/dpMaximumSubarray.test.js

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import dpMaximumSubarray from '../dpMaximumSubarray';
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describe('dpMaximumSubarray', () => {
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it('should find maximum subarray using dynamic programming algorithm', () => {
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it('should find maximum subarray using the dynamic programming algorithm', () => {
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expect(dpMaximumSubarray([])).toEqual([]);
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expect(dpMaximumSubarray([0, 0])).toEqual([0]);
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expect(dpMaximumSubarray([0, 0, 1])).toEqual([0, 0, 1]);
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/**
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* Divide and Conquer solution.
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* Complexity: O(n^2) in case if no memoization applied
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*
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* @param {Number[]} inputArray
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* @return {Number[]}
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*/
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export default function dcMaximumSubarraySum(inputArray) {
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/**
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* We are going through the inputArray array and for each element we have two options:
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* - to pick
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* - not to pick
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*
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* Also keep in mind, that the maximum sub-array must be contiguous. It means if we picked
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* the element, we need to continue picking the next elements or stop counting the max sum.
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*
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* @param {number} elementIndex - the index of the element we're deciding to pick or not
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* @param {boolean} mustPick - to pick or not to pick the element
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* @returns {number} - maximum subarray sum that we'll get
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*/
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function solveRecursively(elementIndex, mustPick) {
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if (elementIndex >= inputArray.length) {
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return mustPick ? 0 : -Infinity;
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}
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return Math.max(
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// Option #1: Pick the current element, and continue picking next one.
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inputArray[elementIndex] + solveRecursively(elementIndex + 1, true),
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// Option #2: Don't pick the current element.
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mustPick ? 0 : solveRecursively(elementIndex + 1, false),
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);
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}
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return solveRecursively(0, false);
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}

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