lalkmim/codility_solutions.txt

Thanks for sharing!

Lesson 4 - Couting Elements
PermCheck ---- https://app.codility.com/demo/results/training8M2CGC-RSS/

This solution is faster :)

Hi lalkmim,

I have another solution for the AbsDistinct (lesson 15). Take a look if you like it:

function solution(A) {
        // write your code in JavaScript (Node.js 8.9.4)
        let distinctValues = {};
	for (let i = 0; i < A.length; i++) {
		let prop = Math.abs(A[i]).toString();
		distinctValues[prop] = prop;
	}
	return Object.keys(distinctValues).length;
}

Here is my solution to Lesson 2 (CyclicRotation) without using loops or recursion:
https://app.codility.com/demo/results/trainingEVYNBV-5FC/

Hi, Can you please explain Lesson 4 MissingInteger?

Using JS filter you can distinct the array and get the length of it will be the number you looking for
return A.filter((v,i,s)=> s.indexOf(v) === i ).length;

Lesson 5 - Prefix Sums

• GenomicRangeQuery => https://app.codility.com/demo/results/training9PG6CB-XRQ/
• PassingCars => https://app.codility.com/demo/results/training8ZDFVE-BXM/

Lesson 9 - Maximum slice problem

• MaxProfit => https://app.codility.com/demo/results/trainingH2SXGH-98Z/

Oi, vascaino! tudo bem? Leonardo, queria falar com vc! vc tem Linkedin?

Lesson 3 - PermMissingElem can be solved using the fact that sum of consecutive elements from 1 to n equals n*(n+1)/2. Subtract all elements from the sum, you're left with the missing number

function solution(A) {

let maxNumber = A.length + 1 
let total_sum_assumed = maxNumber * (maxNumber + 1) / 2

for (let number of A){
    total_sum_assumed = total_sum_assumed - number
}

return total_sum_assumed 

}

Lesson 5 - Prefix Sums
PassingCars

This solution is easer

https://app.codility.com/demo/results/trainingJTC3QW-P7C/

PermCheck:

function solution(A){
    const mem = {}
    for(const v of A){
        if(v < 1 || v > A.length) return 0
        mem[v] = 1
    }
    return Object.keys(mem).length === A.length ? 1 : 0
}

FrogRiverOne:

function solution(A){
    let mem = {};
    for(let i=0;i<A.length;i++){
        if(A[i] < X) mem[A[i]] = true
        if(A[i] === X && Object.keys(mem).length === X-1){
            return i
        }
    }
    return -1
}

MaxCounters:

function solution(N,A){
    let counters = Array.from({length:N}).map(_=>0);
    for(const X of A){
        if(X >= 1 && X <= N){
            counters[X-1]++
        } else if(X === N+1){
            const greatest = Math.max(...counters)
            counters = counters.map(_ => greatest)
        }
    }
    return counters
}