Hello, i need to scan an multiple array and if i find a value false, it return me false.
On the contrary, if no false is found, it return me true.
[ [true, 1], [true, 2], [true, 3] ]; // give me true
[ [false, 1], [true, 2], [false, 3] ]; // give me false
Firstly, welcome to the forums.
While we are primarily here to help people with their Free Code Camp progress, we are open to people on other paths, too. Some of what you are asking is pretty trivial in the Free Code Camp context, so you might find that if you’re not getting the instruction and material you need in your current studies, the FCC curriculum will really help you get started. At a modest guess I’d say investing a 4-5 hours working through the curriculum here will really pay off. You can find the curriculum at https://freecodecamp.org.
With your current questions, we don’t have enough context to know what you already know or don’t know, so it is impossible to guide you without just telling you the answer (which we won’t do).
It is pretty typical on here for people to share a codepen / jsfiddle example of what they have tried so that anyone helping has more of an idea of what help is actually helpful.
Please provide some example of what you’ve tried and I’m sure you’ll get more help.
Happy coding 
Could you write that in ES5? I think the OP will understand it better that way.
Hello, thx for your response.
Actualy, i’m trying to do one of the fcc exercice there:
Link:https://beta.freecodecamp.org/en/challenges/intermediate-algorithm-scripting/smallest-common-multiple
But i’m having trouble, lots of trouble. Could you help me to debug and make my code work the way i tried ti do it:
/*
Link: https://beta.freecodecamp.org/en/challenges/intermediate-algorithm-scripting/smallest-common-multiple
*/
function noFalse( arr ) {
return arr.every( function( subArr ) {
return undefined === subArr.find( function( elem) {
return elem === false;
});
});
};
function smallestCommons( arr ) {
let min = Math.min.apply(Math, arr); //1
let max = Math.max.apply(Math, arr); //3
let bool = true;
let smallArr = [min]; //[1]
let bigArr = [max]; //[3]
let boolArr = [];
let i = 0;
let valueSmall = 0;
let valueMax = 0;
let smallArrL = smallArr.length;
let bigArrL = bigArr.length;
do {
for( let x = 0; x < smallArrL; x++ ) {
for( let z = 0; z < bigArrL; z++ ) {
if( smallArr[x] == bigArr[z] ) {
valueMax = bigArr[z];
valueSmall = smallArr[x];
for( let k = 2; k < valueMax; k++ ) {
if( Number.isInteger( valueSmall/k ) == true ) {
boolArr.push( true );
}else {
boolArr.push( false );
}
}
}else {
i++;
smallArr.push( smallArr[i-1] + min );
bigArr.push( bigArr[i-1] + max );
}
}
}
smallArrL = smallArr.length;
bigArrL = bigArr.length;
console.log(boolArr);
if( noFalse( boolArr ) == true ) {
bool = false;
let val = boolArr[0][1];
return val;
}
}while( bool == true );
};
console.log( smallestCommons( [1,3] ) );
It’s pretty chaotic but i think it should work. If i could have some tips on what to fix.
Okay ! Gonna do this !
And bluring a code that need help is not good because you can’t copy/past it 
/*
Link: https://beta.freecodecamp.org/en/challenges/intermediate-algorithm-scripting/smallest-common-multiple
*/
function noFalse( arr ) {
return arr.every( function( subArr ) {
return undefined === subArr.find( function( elem) {
return elem === false;
});
});
};
function smallestCommons( arr ) {
let min = Math.min.apply(Math, arr); //1
let max = Math.max.apply(Math, arr); //3
let bool = true;
let smallArr = [min]; //[1]
let bigArr = [max]; //[3]
let boolArr = [];
let i = 0;
let valueSmall = 0;
let valueMax = 0;
let smallArrL = smallArr.length;
let bigArrL = bigArr.length;
//Do this while bool is 'true'
do {
/*
smallArr[0] == 1; (It's my min value from my original array 'arr')
bigArr[0] == 3; (It's my max value from my original array 'arr')
I check one value at a time from 'smallArr' if it's equal to all my values from 'bigArr'
*/
for( let x = 0; x < smallArrL; x++ ) {
for( let z = 0; z < bigArrL; z++ ) {
/*
if one value is equal to the other one like '3' or '6' for the '[1, 3]' exemple:
I collect this value for more simple coding.
Then in a for loop, i check in the interval k=2 (because i don't need '1') to valueMax-1.
So for the first try, it's only 2 which is tested. The next time it will be '2 to 6'
"valueSmall/2" -> 3/2 == 1.5 so it's false. I creat a value 'false in my table 'boolArr'.
*/
if( smallArr[x] == bigArr[z] ) {
valueMax = bigArr[z];
valueSmall = smallArr[x];
for( let k = 2; k < valueMax; k++ ) {
if( Number.isInteger( valueSmall/k ) == true ) {
//the 'k' help me to know at which indexe
boolArr.push( true, k );
}else {
boolArr.push( false );
}
}
}else {
/*
If the first comparaison do not match, i add new values to my arrays:
smallArr = 1+1 so 2 : [1, 2]
bigArr = 3+3 so 6 : [3, 6]
With it, i can reuse my two first for loop in a dynamic way and check next values
*/
i++;
smallArr.push( smallArr[i-1] + min );
bigArr.push( bigArr[i-1] + max );
}
}
}
/*
Don't think it's necessary...
*/
smallArrL = smallArr.length;
bigArrL = bigArr.length;
/*
If i have only true values, it mean that the interval between, for exemple, '2 and 6' is divisible without float value.
*/
if( noFalse( boolArr ) == true ) {
//go out from the 'while loop' because i found my smallest common multiple
bool = false;
//i recover my index
let val = boolArr[0][1];
//then i use this index to find my value
return bigArr[val];
}
}while( bool == true );
};
console.log( smallestCommons( [1,3] ) );
//1 et 3
//1 -> 1 ; 2 ; 3 ; 4 ; 5 ; 6
//3 -> 3 ; 6
I hope my comment can help you. Even me, i was not sure for every thing i did xD
Well, for now, i think i will try to figure out a new solution. Will update today.
Hello back, i think my new code is “better”. In my solution, i create an array with the first 10 values (0-9).
If i find my solution in these 10 possibilities, it return my result.
For exemple, here for [1, 3], it return me 6 because it is in the first 10 possibilities.
But if it’s not the case, i save my last value from my multidimentional array. Then i delete all the value from my original array. And i start to re-fill my array with 10 next new values. I do it and if it succefully find my result, i get out.
I tried to find a way to have some thing dynamic and not to much cpu heavy…
But my problem is i have issue to…get out
/*
Link: https://beta.freecodecamp.org/en/challenges/intermediate-algorithm-scripting/smallest-common-multiple
*/
function smallestCommons( arr ) {
let min = Math.min.apply( Math, arr ); //1
let max = Math.max.apply( Math, arr ); //3
let minCopy = min;
let maxCopy = max;
let myArray = [ [min, max] ];
let myNewArray = [];
//global variable
bool = false;
//global variable
boolOutWhile = false;
//I create my original array with the first 10 values (0-9)
for( let k = 0; k < 9; k++ ) {
myArray.push( [minCopy = minCopy+min, maxCopy = maxCopy+max]);
}
do {
//I make a copy of the last value of my array for later
myNewArray.push( myArray[myArray.length-1] ); //[10, 30]
let lastValueMin = myNewArray[myNewArray.length-1][0]; //10
let lastValueMax = myNewArray[myNewArray.length-1][1]; //30
let myArrayL = myArray.length; //10
//Two 'foor-loop' to check every value from myArray[k] with myArray[][x]
for( k = 0; k < myArrayL; k++ ) {
for( let x = 0; x < myArrayL; x++ ) {
//If i have a match then
if( myArray[k][0] == myArray[x][1] ) {
//I make a copy of the number matching for my next 'for loop' interval (here: 3, 6 and 9 matches)
valueCopy = myArray[x][1];
/*
3 is EQUAL to 3
6 is EQUAL to 6
9 is EQUAL to 9
*/
// console.log( myArray[k][0]+" is EQUAL to "+myArray[x][1]);
/*
It's 1 and 3 because the interval is between min(1) and max(3)
3/1=(3) 3/2(1.5) 3/3=(1)// 6/1=(6) 6/2=(3) 6/3=(2) // 9/1=(9) 9/2=(4.5) 9/3=(3)
*/
for( let i = min; i <= max; i++ ) {
//If my valueCopy / i (3/1, 3/2, 3/3...) is not an Integer then i stop
if( Number.isInteger( valueCopy / i ) == false ) {
bool = false;
i = max;
//else, my bool have a value of 'true'
}else {
bool = true;
}
}
//If my bool still have a value of true at the end of the last 'for-loop' then it's done
if( bool == true ) {
//i can get out of the 'while-loop'
boolOutWhile = true;
//return my result
return valueCopy;
}
}
}
}
//delete my useless value to do it again
myArray.splice(0);
myNewArray.splice(0);
//re-fill my array with new values from the end of my last ten first array values
/*
For exemple, if 6 was not the value to return for 1-3, then i would have:
[
[11, 33], [12, 36], [13, 39], [14, 42], [...], [20, 60]
]
*/
for( k = 0; k <= 9; k++ ) {
myArray.push( [lastValueMin = lastValueMin+min, lastValueMax = lastValueMax+max]);
}
}while( boolOutWhile == false );
};
// console.log( smallestCommons( [1,3] ) ); //find 6
//console.log( smallestCommons( [1,5] ) ); //failed
Hi back, i spend my time finishing the next 2 exercices
(not bad for uping a topic xD)
Hello, upping my topic.
Well, i finished all of the others exercices. I will take time for this one.
I think i have a comprehension issue regarding the while() and how to get out of it with embricated stuff in it.
My while doesn’t seem to know when my bool is true of false.