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Commit82f0b5e

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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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importbfMaximumSubarrayfrom'../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 usingthebrute 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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importdcMaximumSubarrayfrom'../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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importdpMaximumSubarrayfrom'../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 usingthedynamic 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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exportdefaultfunctiondcMaximumSubarraySum(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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functionsolveRecursively(elementIndex,mustPick){
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if(elementIndex>=inputArray.length){
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returnmustPick ?0 :-Infinity;
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}
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returnMath.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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returnsolveRecursively(0,false);
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}

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