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how do i make a triangle using for loop in javascript which output matches the parameter content

let printTriangle = (num) => {
for ( let i = 0; i < num; i++) {
for (let j = 0; j <= i; j++) {
num += '*';
}
num += "\n"
}
}
console.log(printTriangle(5));
I have a simple question although i cannot manage to resolve this problem. Hope you can help, how i should fix my code to make output:
*
**
***
****
*****

let printTriangle = (num) => {
let result = "";
for ( let i = 0; i < num; i++) {
for (let j = 0; j <= i; j++) {
result += '*';
}
result += "\n";
}
return result;
}
console.log(printTriangle(5));

We can use the repeat method for this:
function triangle(n) {
let acc = "";
for (let i = 1; i <= n; i++) {
acc = `${acc}${"*".repeat(i)}\n`;
}
return acc;
}
console.log(triangle(5));

this is because you are appending the desired string in num i.e a number resulting in wrong output.
You should rather initialize another variable inside your function and append * into it.
Also, you are not returning anything from this function that's why instead of some value undefined is getting logged.
printTriangle = num => {
let result = "";
for (let i = 0; i < num; i++) {
for (let j = 0; j <= i; j++) {
result += "*";
}
result += "\n";
}
return result;
};
console.log(printTriangle(5));
There are many more ways to do this, you should try some other efficient ways also.

how to get common substring in two string to maintain order?

I am trying to do one problem of hackerrank but I am not able to solve that problem
Can someone please help me with wrong logic implementation done by me?
problem
Print the length of the longest string, such that is a child of both s1 and s2.
Sample Input
HARRY
SALLY
Sample Output
2
Explanation
The longest string that can be formed by deleting zero or more characters from HARRY and SALLY is AY, whose length is 2.
Sample Input 1
AA
BB
Sample Output 1
0
Explanation 1
AA and BB have no characters in common and hence the output is 0
Sample Input 2
SHINCHAN
NOHARAAA
Sample Output 2
3
Explanation 2
The longest string that can be formed between SHINCHAN and NOHARAAA while maintaining the order is NHA.
I have written some logic which is as follows:
function commonChild(s1, s2) {
var arr = s2.split(),
currenString = '',
maxLength = 0,
index = -1;
console.log(arr);
for (var i = 0; i < s1.length; i++) {
var char = s1.charAt(i),
charIndex = arr.indexOf(char);
console.log(char)
if (index < charIndex) {
index = charIndex;
currenString +=char;
}
maxLength= Math.max(maxLength,currenString.length)
}
return maxLength;
}
commonChild('ABCDEF', 'FBDAMN');
console.log(commonChild('ABCDEF', 'FBDAMN'));

pardon me. this is an unoptimized solution.
function maxCommon(a, b, offset) {
offset = offset || 0;
if (a === b) {
return [[a, b]];
}
var possibleSolns = [];
for (var i = 0 + offset; i < a.length; i++) {
for (var j = 0 + offset; j < b.length; j++) {
if (a.charAt(i) === b.charAt(j)) {
possibleSolns.push([
a.substring(0, offset) + a.substring(i),
b.substring(0, offset) +b.substring(j)
]);
break;
}
}
}
var results = [];
possibleSolns.forEach(function(soln) {
var s = maxCommon(soln[0], soln[1], offset+1);
if (s.length === 0) {
s = [[soln[0].substring(0, offset +1), soln[1].substring(0, offset +1)]];
}
results = results.concat(s);
});
return results;
}
var maxLen = -1;
var soln = null;
maxCommon("ABCDEF", "FBDAMN").map(function(_) {
return _[0];
}).forEach(function(_) {
if (_.length > maxLen) {
soln = _;
maxLen = _.length;
}
});
console.log(soln);

I kept most of your logic in the answer:
function commonChild(s1, s2) {
var // Sets strings to arrays
arrayString1 = s1.split(""),
arrayString2 = s2.split(""),
collectedChars = "",
maxLength = 0,
max = arrayString1.length;
for (var i = 0; i < max; i++) {
var
char = arrayString1[i],
count = arrayString2.indexOf(char);
// check if char is in second string and not in collected
if (count != -1 && collectedChars.indexOf(char) == -1) {
collectedChars += char;
maxLength++;
}
}
return maxLength;
}
// expected output 4
console.log(commonChild(
'ABCDEF',
'FBDAMN'
));
// expected output 1
console.log(commonChild(
'AA',
'FBDAMN'
));

Using lodash and spread operation you can do it in this way.
const test = (first, second) => {
const stringArray1 = [...first];
const stringArray2 = [...second];
return _.intersection(stringArray1, stringArray2).length;
}
console.log(test('ABCDEF', 'FBDAMN'));

You can solve it using lcs least common subsequence
function LCS(s1,s2,x,y){
var result = 0;
if(x==0 || y==0){
result = 0
}else if(s1[x-1] == s2[y-1]){
result = 1+ LCS(s1,s2,x-1,y-1)
} else if(s1[x-1] != s2[y-1]){
result = Math.max(LCS(s1,s2,x-1,y), LCS(s1,s2,x,y-1))
}
return result;
}
// Complete the commonChild function below.
function commonChild(s1, s2) {
return LCS(s1,s2,s1.length,s2.length);
}

Based on your code before the edit.
One little change is to change var arr = s2.split() to split('').
The main change in the logic is that I added a loop to run over the string each time from another character (first loop from the first, second from the second etc).
function commonChild(s1, s2) {
var arr = s2.split(''),
currenString = '',
maxLength = 0,
index = -1,
j = -1;
for (var ii = 0; ii < s1.length; ii++) {
index = -1;
currenString = '';
for (var i = ii; i < s1.length; i++) {
var char = s1.charAt(i),
j = arr.indexOf(char);
if (index < j) {
index = j;
currenString += char;
}
maxLength = Math.max(maxLength, currenString.length)
}
}
return maxLength;
}
console.log(commonChild('ABCDEF', 'FBDAMN'));

Creating a function to combine a nested array without recursion

I have the following array as an example;
let arr = [['red','blue','pink],['dog','cat','bird'],['loud', 'quiet']]
I need to write a generalized function that prints all combinations of one word from the first vector, one word from the second vector, etc. I looked up some codes on here but they are all recursion or working only with the specific array. How can I write this code without recursion?
let allComb = function(arr) {
if (arr.length == 1) {
return arr[0];
} else {
let result = [];
let arrComb = allComb(arr.slice(1));
for (let i = 0; i < arrComb.length; i++) {
for (let j = 0; j < arr[0].length; j++) {
result.push(arr[0][j] + ' ' + arrComb[i]);
}
}
return result;
}
}
allComb(arr)

This version uses a single increment per cycle technique with no recursion.
let arr = [
['red', 'blue', 'pink'],
['dog', 'cat', 'bird'],
['loud', 'quiet']
];
function allComb(arr) {
var total = 1;
var current = [];
var result = [];
for (var j = 0; j < arr.length; j++) {
total *= arr[j].length;
current[j] = 0;
}
for (var i = 0; i < total; i++) {
var inc = 1;
result[i] = "";
for (var j = 0; j < arr.length; j++) {
result[i] += arr[j][current[j]] + ' ';
if ((current[j] += inc) == arr[j].length)
current[j] = 0;
else
inc = 0;
}
}
return (result);
}
console.log(allComb(arr));

You may do as follows;
var arr = [['red','blue','pink'],['dog','cat','bird'],['loud', 'quiet']],
res = arr.reduce((p,c) => p.reduce((r,x) => r.concat(c.map(y => x + " " + y)),[]));
console.log(res);

How can I split the below string into a 2dimensional-array:

How can I split the below string into a 2dimensional-array:
Customer::Europe|UK|Scotland|Product::Drinks|Water|
array:
[Customer][Europe]
[Customer][UK]
[Customer][Scotland]
[Product][Drinks]
[Product][Water]
Not sure how to create the array. Haven't coded in years, so be kind
hArray= [];
vArray= [];
var i = j = 0;
var count = hierarchy.search(/[:|]+/);
write(hierarchy);
while (count > 0) {
if (hierarchy.indexOf(":") < hierarchy.indexOf("|") || (hierarchy.indexOf(":") > 0 && hierarchy.indexOf("|") == -1) ) {
hArray[j] = hierarchy.substr(0,hierarchy.indexOf(":"));
hierarchy = hierarchy.slice(hierarchy.indexOf(":")+2);
count = hierarchy.search(/[:|]+/);
j++;
} else
if (hierarchy.indexOf("|") < hierarchy.indexOf(":") {
vArray[i] = hierarchy.substr(0,count);
hierarchy = hierarchy.slice(count+1);
count = hierarchy.search(/[:|]+/);
i++;
}
if (count == -1) break;
//create multiArray ?
}

var source = "Customer::Europe|UK|Scotland|Product::Drinks|Water|";
var parts = source.split(/(\w+::)/);
var result = [];
for (var i = 1; i < parts.length; i += 2) {
var key = parts[i].replace("::", "");
var values = parts[i + 1].split("|");
for (var j = 0; j < values.length - 1; ++j) {
var line = new Array(2);
line[0] = key;
line[1] = values[j];
result.push(line);
}
}
console.log(result);

You can use Array.reduce like this. First, we split on | that is behind any owrd followed by ::. Then we reduce it, by using an array as memo and push an array into the memo, which we finally return.
var arr = input.split(/\|(?=\w+::)/).reduce(function(arr, str){
var array = str.split('::');
return arr.push(str.split('::')[1].split('|').filter(String).map(function(s){
return [array[0], s]
})), arr;
}, []);

Multiplying all numbers in two arrays in javascript

As an exercise, I'm trying to create a function that returns the palindromic numbers resulting from multiplying three-digit numbers. As far as I can tell, the function is running through numbers correctly, however, the resulting array is incorrect. I don't need the solution to the palindrome problem...just an idea of what I might be missing. Have I run into some limitation?
var palindromic = function() {
var a = [];
var res = [];
for (var i = 100; i < 1000; i++) {
a.push(i);
}
var ar = a.slice(0);
a.map(function(x) {
for (var j = 0; j < ar.length; j++) {
var result = x * ar[j];
if (result.toString() === result.toString().split("").reverse().join("")) {
res.push(result);
}
}
})
return res;
};

Pretty sure it's just trying to call console.log() 810,000 times. If you comment the console.log line, it works just fine.
var palindromic = function() {
var a = [];
var res = [];
for (var i = 100; i < 1000; i++) {
a.push(i);
}
var ar = a.slice(0);
a.map(function(x) {
for (var j = 0; j < ar.length; j++) {
var result = x * ar[j];
//console.log(x + " : " + ar[j] + ' = ' + result);
if (result.toString() === result.toString().split("").reverse().join("")) {
res.push(result);
}
}
});
return res;
};
console.log(palindromic());