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Commit9bc2800

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Add Recursive Staircase Problem.
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‎README.md

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*`B`[Jump Game](src/algorithms/uncategorized/jump-game) - backtracking, dynamic programming (top-down + bottom-up) and greedy examples
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*`B`[Unique Paths](src/algorithms/uncategorized/unique-paths) - backtracking, dynamic programming and Pascal's Triangle based examples
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*`B`[Rain Terraces](src/algorithms/uncategorized/rain-terraces) - trapping rain water problem (dynamic programming and brute force versions)
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*`B`[Recursive Staircase](src/algorithms/uncategorized/recursive-staircase) - count the number of ways to reach to the top (4 solutions)
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*`A`[N-Queens Problem](src/algorithms/uncategorized/n-queens)
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*`A`[Knight's Tour](src/algorithms/uncategorized/knight-tour)
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***Brute Force** - look at all the possibilities and selects the best solution
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*`B`[Linear Search](src/algorithms/search/linear-search)
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*`B`[Rain Terraces](src/algorithms/uncategorized/rain-terraces) - trapping rain water problem
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*`B`[Recursive Staircase](src/algorithms/uncategorized/recursive-staircase) - count the number of ways to reach to the top
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*`A`[Maximum Subarray](src/algorithms/sets/maximum-subarray)
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*`A`[Travelling Salesman Problem](src/algorithms/graph/travelling-salesman) - shortest possible route that visits each city and returns to the origin city
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*`A`[Discrete Fourier Transform](src/algorithms/math/fourier-transform) - decompose a function of time (a signal) into the frequencies that make it up
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*`B`[Jump Game](src/algorithms/uncategorized/jump-game)
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*`B`[Unique Paths](src/algorithms/uncategorized/unique-paths)
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*`B`[Rain Terraces](src/algorithms/uncategorized/rain-terraces) - trapping rain water problem
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*`B`[Recursive Staircase](src/algorithms/uncategorized/recursive-staircase) - count the number of ways to reach to the top
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*`A`[Levenshtein Distance](src/algorithms/string/levenshtein-distance) - minimum edit distance between two sequences
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*`A`[Longest Common Subsequence](src/algorithms/sets/longest-common-subsequence) (LCS)
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*`A`[Longest Common Substring](src/algorithms/string/longest-common-substring)
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#Recursive Staircase Problem
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##The Problem
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There are`n` stairs, a person standing at the bottom wants to reach the top. The person can climb either`1` or`2` stairs at a time._Count the number of ways, the person can reach the top._
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![](https://cdncontribute.geeksforgeeks.org/wp-content/uploads/nth-stair.png)
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##The Solution
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This is an interesting problem because there are several ways of how it may be solved that illustrate different programming paradigms.
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-[Brute Force Recursive Solution](./recursiveStaircaseBF.js) - Time:`O(2^n)`; Space:`O(1)`
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-[Recursive Solution With Memoization](./recursiveStaircaseMEM.js) - Time:`O(n)`; Space:`O(n)`
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-[Dynamic Programming Solution](./recursiveStaircaseDP.js) - Time:`O(n)`; Space:`O(n)`
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-[Iterative Solution](./recursiveStaircaseIT.js) - Time:`O(n)`; Space:`O(1)`
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##References
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-[On YouTube by Gayle Laakmann McDowell](https://www.youtube.com/watch?v=eREiwuvzaUM&list=PLLXdhg_r2hKA7DPDsunoDZ-Z769jWn4R8&index=81&t=0s)
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-[GeeksForGeeks](https://www.geeksforgeeks.org/count-ways-reach-nth-stair/)
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importrecursiveStaircaseBFfrom'../recursiveStaircaseBF';
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describe('recursiveStaircaseBF',()=>{
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it('should calculate number of variants using Brute Force solution',()=>{
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expect(recursiveStaircaseBF(-1)).toBe(0);
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expect(recursiveStaircaseBF(0)).toBe(0);
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expect(recursiveStaircaseBF(1)).toBe(1);
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expect(recursiveStaircaseBF(2)).toBe(2);
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expect(recursiveStaircaseBF(3)).toBe(3);
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expect(recursiveStaircaseBF(4)).toBe(5);
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expect(recursiveStaircaseBF(5)).toBe(8);
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expect(recursiveStaircaseBF(6)).toBe(13);
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expect(recursiveStaircaseBF(7)).toBe(21);
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expect(recursiveStaircaseBF(8)).toBe(34);
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expect(recursiveStaircaseBF(9)).toBe(55);
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expect(recursiveStaircaseBF(10)).toBe(89);
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});
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});
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importrecursiveStaircaseDPfrom'../recursiveStaircaseDP';
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describe('recursiveStaircaseDP',()=>{
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it('should calculate number of variants using Dynamic Programming solution',()=>{
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expect(recursiveStaircaseDP(-1)).toBe(0);
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expect(recursiveStaircaseDP(0)).toBe(0);
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expect(recursiveStaircaseDP(1)).toBe(1);
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expect(recursiveStaircaseDP(2)).toBe(2);
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expect(recursiveStaircaseDP(3)).toBe(3);
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expect(recursiveStaircaseDP(4)).toBe(5);
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expect(recursiveStaircaseDP(5)).toBe(8);
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expect(recursiveStaircaseDP(6)).toBe(13);
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expect(recursiveStaircaseDP(7)).toBe(21);
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expect(recursiveStaircaseDP(8)).toBe(34);
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expect(recursiveStaircaseDP(9)).toBe(55);
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expect(recursiveStaircaseDP(10)).toBe(89);
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});
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});
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importrecursiveStaircaseITfrom'../recursiveStaircaseIT';
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describe('recursiveStaircaseIT',()=>{
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it('should calculate number of variants using Iterative solution',()=>{
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expect(recursiveStaircaseIT(-1)).toBe(0);
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expect(recursiveStaircaseIT(0)).toBe(0);
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expect(recursiveStaircaseIT(1)).toBe(1);
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expect(recursiveStaircaseIT(2)).toBe(2);
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expect(recursiveStaircaseIT(3)).toBe(3);
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expect(recursiveStaircaseIT(4)).toBe(5);
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expect(recursiveStaircaseIT(5)).toBe(8);
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expect(recursiveStaircaseIT(6)).toBe(13);
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expect(recursiveStaircaseIT(7)).toBe(21);
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expect(recursiveStaircaseIT(8)).toBe(34);
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expect(recursiveStaircaseIT(9)).toBe(55);
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expect(recursiveStaircaseIT(10)).toBe(89);
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});
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});
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importrecursiveStaircaseMEMfrom'../recursiveStaircaseMEM';
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describe('recursiveStaircaseMEM',()=>{
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it('should calculate number of variants using Brute Force with Memoization',()=>{
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expect(recursiveStaircaseMEM(-1)).toBe(0);
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expect(recursiveStaircaseMEM(0)).toBe(0);
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expect(recursiveStaircaseMEM(1)).toBe(1);
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expect(recursiveStaircaseMEM(2)).toBe(2);
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expect(recursiveStaircaseMEM(3)).toBe(3);
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expect(recursiveStaircaseMEM(4)).toBe(5);
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expect(recursiveStaircaseMEM(5)).toBe(8);
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expect(recursiveStaircaseMEM(6)).toBe(13);
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expect(recursiveStaircaseMEM(7)).toBe(21);
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expect(recursiveStaircaseMEM(8)).toBe(34);
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expect(recursiveStaircaseMEM(9)).toBe(55);
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expect(recursiveStaircaseMEM(10)).toBe(89);
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});
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});
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/**
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* Recursive Staircase Problem (Brute Force Solution).
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*
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*@param {number} stairsNum - Number of stairs to climb on.
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*@return {number} - Number of ways to climb a staircase.
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*/
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exportdefaultfunctionrecursiveStaircaseBF(stairsNum){
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if(stairsNum<=0){
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// There is no way to go down - you climb the stairs only upwards.
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// Also if you're standing on the ground floor that you don't need to do any further steps.
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return0;
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}
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if(stairsNum===1){
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// There is only one way to go to the first step.
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return1;
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}
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if(stairsNum===2){
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// There are two ways to get to the second steps: (1 + 1) or (2).
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return2;
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}
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// Sum up how many steps we need to take after doing one step up with the number of
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// steps we need to take after doing two steps up.
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returnrecursiveStaircaseBF(stairsNum-1)+recursiveStaircaseBF(stairsNum-2);
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}
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/**
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* Recursive Staircase Problem (Dynamic Programming Solution).
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*
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*@param {number} stairsNum - Number of stairs to climb on.
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*@return {number} - Number of ways to climb a staircase.
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*/
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exportdefaultfunctionrecursiveStaircaseDP(stairsNum){
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if(stairsNum<0){
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// There is no way to go down - you climb the stairs only upwards.
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return0;
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}
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// Init the steps vector that will hold all possible ways to get to the corresponding step.
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conststeps=newArray(stairsNum+1).fill(0);
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// Init the number of ways to get to the 0th, 1st and 2nd steps.
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steps[0]=0;
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steps[1]=1;
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steps[2]=2;
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if(stairsNum<=2){
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// Return the number of ways to get to the 0th or 1st or 2nd steps.
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returnsteps[stairsNum];
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}
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// Calculate every next step based on two previous ones.
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for(letcurrentStep=3;currentStep<=stairsNum;currentStep+=1){
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steps[currentStep]=steps[currentStep-1]+steps[currentStep-2];
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}
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// Return possible ways to get to the requested step.
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returnsteps[stairsNum];
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}
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/**
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* Recursive Staircase Problem (Iterative Solution).
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*
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*@param {number} stairsNum - Number of stairs to climb on.
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*@return {number} - Number of ways to climb a staircase.
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*/
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exportdefaultfunctionrecursiveStaircaseIT(stairsNum){
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if(stairsNum<=0){
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// There is no way to go down - you climb the stairs only upwards.
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// Also you don't need to do anything to stay on the 0th step.
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return0;
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}
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// Init the number of ways to get to the 0th, 1st and 2nd steps.
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conststeps=[1,2];
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if(stairsNum<=2){
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// Return the number of possible ways of how to get to the 1st or 2nd steps.
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returnsteps[stairsNum-1];
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}
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// Calculate the number of ways to get to the n'th step based on previous ones.
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// Comparing to Dynamic Programming solution we don't store info for all the steps but
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// rather for two previous ones only.
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for(letcurrentStep=3;currentStep<=stairsNum;currentStep+=1){
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[steps[0],steps[1]]=[steps[1],steps[0]+steps[1]];
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}
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// Return possible ways to get to the requested step.
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returnsteps[1];
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}
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/**
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* Recursive Staircase Problem (Recursive Solution With Memoization).
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*
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*@param {number} totalStairs - Number of stairs to climb on.
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*@return {number} - Number of ways to climb a staircase.
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*/
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exportdefaultfunctionrecursiveStaircaseMEM(totalStairs){
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// Memo table that will hold all recursively calculated results to avoid calculating them
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// over and over again.
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constmemo=[];
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// Recursive closure.
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constgetSteps=(stairsNum)=>{
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if(stairsNum<=0){
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// There is no way to go down - you climb the stairs only upwards.
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// Also if you're standing on the ground floor that you don't need to do any further steps.
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return0;
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}
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if(stairsNum===1){
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// There is only one way to go to the first step.
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return1;
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}
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if(stairsNum===2){
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// There are two ways to get to the second steps: (1 + 1) or (2).
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return2;
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}
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// Avoid recursion for the steps that we've calculated recently.
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if(memo[stairsNum]){
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returnmemo[stairsNum];
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}
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// Sum up how many steps we need to take after doing one step up with the number of
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// steps we need to take after doing two steps up.
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memo[stairsNum]=getSteps(stairsNum-1)+getSteps(stairsNum-2);
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returnmemo[stairsNum];
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};
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// Return possible ways to get to the requested step.
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returngetSteps(totalStairs);
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}

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