The following code is functional but I would like to refactor the !== part that allows my ternary to run only on values that are not white space so I can include edge case test's. This would include any non letter value as well as white space and I know that regex may play a part but I can't find a niftty way to incorporate it into the if() statement that precedes the ternary operations.
const letterPositions = function(strPos) {
if (typeof strPos !== 'string') {
return console.log('Sorry your input is not a string');
}
const result = {};
for (let i = 0; i < strPos.length; i++) {
if (strPos[i] !== ' ') {
result[strPos[i]] ? result[strPos[i]].push(i) : (result[strPos[i]] = [i]);
}
}
return result;
};
console.log(letterPositions('aa bb cc'));mainly two options, regex and charcode, welcome to edit if more ways available
const codes = ['A', 'Z', 'a', 'z'].map(x => x.charCodeAt(0))
console.log(codes)
const letterPositions = function(strPos) {
if (typeof strPos !== 'string') {
return console.log('Sorry your input is not a string');
}
const result = {};
for (let i = 0; i < strPos.length; i++) {
//if (strPos[i] !== ' ') { // old
//if (/[A-Za-z]/.test(strPos[i])) { // regex
let code = strPos.charCodeAt(i)
if ((code >= 65 && code <= 90) || (code >= 97 && code <= 122)) { // charcode
result[strPos[i]] ? result[strPos[i]].push(i) : (result[strPos[i]] = [i]);
}
}
return result;
};
console.log(letterPositions('aa bb cc'));you can also do that...
const letterPositions = str =>
{
if (typeof str !== 'string')
return console.log('Sorry your input is not a string' )
return [...str].reduce((r,l,i)=>((l===' ')?null:(r[l]??=[],r[l].push(i)),r),{})
}
console.log( letterPositions('aa bb cc') ).as-console-wrapper {max-height: 100% !important;top: 0;}
.as-console-row::after {display: none !important;}