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O(n*m) staircase with N steps

difo23 profile image Lizandro J. Ramírez ・2 min read

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Description:

This problem was asked by Amazon.

There exists a staircase with N steps, and you can climb up either 1 or 2 steps at a time. Given N, write a function that returns the number of unique ways you can climb the staircase. The order of the steps matters.

Example:

For example, if N is 4, then there are 5 unique ways:

  • 1, 1, 1, 1
  • 2, 1, 1
  • 1, 2, 1
  • 1, 1, 2
  • 2, 2

Fibonacci:

N = [0, 1, 2, 3, 4, 5, 6]

Output Ways = [1, 1, 2, 3, 5, 8, 13]

Fibonacci in the output.

Extra:

What if, instead of being able to climb 1 or 2 steps at a time, you could climb any number from a set of positive integers X? For example, if X = [1, 3, 5], you could climb 1, 3, or 5 steps at a time.

Solution in JS:

  • O(n * m)
  • n --> staircase steps ( N )
  • m --> valid climb up steps ( X.length )

let staircase = (n, X) => {

    // Steps climb up
    let setX = new Set(X)

    // Positions arrays step staircase
    // Included  0  
    let cache = Array(n + 1).fill(0);

    // The position 0 is always 1 way.
    cache[0] = 1;


    for (let i = 0; i <= n; ++i) {

        let temp = 0;

        // Valid Steps add
        for (let x of X) {
            if (i - x > 0) {
                temp += cache[i - x]
            }
        }

        //Update cache.
        cache[i] += temp;
        // position numbers 
        // is included (1) or not (0)
        cache[i] += setX.has(i) ? 1 : 0;
    }

    // The last position in cache have the
    // # of ways.
    return cache.pop();
}



Simple Test:


// Case 1
let X = [1, 2 ];
let n = 4;

console.log(staircase(n, X))

// Case 2 
let X = [1, 3, 5];
let n = 4;

console.log(staircase(n, X))

You can check

code by @difo23

Posted on by:

difo23 profile

Lizandro J. Ramírez

@difo23

I’m an enthusiast's software development

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