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Get the last capitalized letter from a string

I need to get the last uppercase letter from the string and wondering how can I do it. I want to write a function that takes the string and returns the last uppercase letter from that string.
For example, If I call the function with word 'LonDon', I should get D. And if I call the function with word 'CaliforNia', I get N.
Thank you so much for your time.

function findLastCap(text) {
let length = text.length - 1;
for (let i = length; i >= 0; i--) {
if (text[i] !== text[i].toLowerCase()) return text[i];
}
return false;
}
console.log(findLastCap("aaaaaaBccc"))

Use a Regular Expression, probably [A-Z](?=[^A-Z]*$)
[A-Z]: match any capital letter
(?=[^A-Z]*$): Followed by any # of non-uppercase and the end of the string.
let regex = /[A-Z](?=[^A-Z]*$)/;
console.log({
CaliforNia: 'CaliforNia'.match(regex)[0],
LonDon: 'LonDon'.match(regex)[0]
});

I do not know javascript but I can help you come up with an algorithm that would solve this problem.
First I will write it in python3 and explain what I did and you can try to translate it into javascript.
def last_upper(string):
last = "" # i make a string var here
for let in string:
if let.issupper():
last = let
return last
So basically what I did in this code is that I iterate through all the elements in the string and if a letter is uppercase it will update the last variable. It works because it keeps on updating it each time it finds a uppercase letter.

Try this,
let reg = /[A-Z](?=[^A-Z]*$)/g
let p = "Parts spaR";
p.match(reg)
Or you can get all the matches and access the last one,
let reg = /[A-Z]/g
let r = p.match(reg)
let found = r[r.length-1]

let tempString=`aaaaaBcCcF`;
let result = tempString.split('').filter(value => {
let str = '';
if (value === value.toUpperCase()) {
str = value;
}
return str;
})
console.log(result[result.length-1]);

What am I missing regarding the toUpperCase method?

I am trying to make the first letter of each word capitalized via toUpperCase method and the rest of the word is in the lower case via the toLowerCase method. But I am missing something... Why temp value is not matching with result[1][0] even if I am using that method for both?
Note: I know about other ways (map, replace, etc) for my solution, but I want to just use a for-loop with toUpperCase and toLowerCase methods.
function titleCase(str) {
let regex = /[^0-9\s]+/g;
var result = str.match(regex);
let temp = "";
for (let i = 0; i < result.length; i++) {
for (let j = 0; j < result[i].length; j++) {
result[1][0] = result[1][0].toUpperCase();
temp = result[1][0].toUpperCase();
}
}
console.log(temp); // Output is 'A'
console.log(result[1][0]); //Output is 'a'
// Normally 'temp' and 'result[1][0]' should be equal, but one returns a lowercase character and the other an uppercase character.
return str;
}
titleCase("I'm a little tea pot");

Your problem is not with the toUppercase(), it is with the reference.
When referencing result[1][0], why are you including the 0? You already have the second character with result[1]
result[1] === 'a'. No need to include the [0] as well.
Change your code so it looks like this:
function titleCase(str) {
let regex = /[^0-9\s]+/g;
var result = str.match(regex);
let temp = "";
result[1] = result[1].toUpperCase();
temp = result[1].toUpperCase();
console.log(temp); // Output is 'A'
console.log(result[1]); //Output is also 'A'
// both now equals capital A
return str;
}
titleCase("I'm a little tea pot");
EDIT:
Updating the function to uppercase the first letter of the word.
We can use ES6, which would make this really simple:
const capitalize = (string = '') => [...string].map((char, index) => index ? char : char.toUpperCase()).join('')
Use it: capitalize("hello") returns 'Hello'.
First we convert the string to an array, using the spread operator, to get each char individually as a string. Then we map each character to get the index to apply the uppercase to it. Index true means not equal 0, so (!index) is the first character. We then apply the uppercase function to it and then return the string.
If you want a more object oriented approach, we can do something like this:
String.prototype.capitalize = function(allWords) {
return (allWords) ?
this.split(' ').map(word => word.capitalize()).join(' ') :
return this.charAt(0).toUpperCase() + this.slice(1);
}
Use it: "hello, world!".capitalize(); returns "Hello, World"
We break down the phrase to words and then recursive calls until capitalising all words. If allWords is undefined, capitalise only the first word meaning the first character of the whole string.

I was tried to change a specific character in the string but strings are immutable in JS so this does not make sense.

How can I find at least 1 uppercase in a string?

In es6 js I have :
good = word => {
word.split('').includes(w => w === w.toUpperCase())
}
console.log(good('Bla'))
How can I return true when finding 1 Uppercase in my string?

You can test the string using a regular expression with a character set of all uppercase letters [A-Z]:
const good = word => /[A-Z]/.test(word);
console.log(good('Bla'));
console.log(good('bla'));

Although there are much simpler ways to do this (the regex in Tushar's comment being one of them), it's possible to fix your attempt to work correctly by:
Removing the braces so that the function actually returns a value.
Using .some(), which takes a function as its argument. .includes() doesn't.
Addding const so you're actually declaring your function.
const good = word => word.split('').some(w => w === w.toUpperCase())
console.log(good('Bla'))
console.log(good('bla'))

You can do this if you want the index of it too.
function findUpperCase(str) {
return str.search(/[A-Z]/);
}

// The string which will go thorough the test
let theString = 'Hello World'
// Function to find out the answer
function stringCheck (receivedString) {
// Removing special character, number, spaces from the string to perform exact output
let stringToTest = receivedString.replace(/[^A-Z]+/ig, "")
// Variable to count: how many uppercase characters are there in that string
let j = 0
// Loop thorough each character of the string to find out if there is any uppercase available
for (i = 0; i < stringToTest.length; i++) {
// Uppercase character check
if (stringToTest.charAt(i) === stringToTest.charAt(i).toUpperCase()) {
console.log('Uppercase found: ' + stringToTest.charAt(i))
j++
}
}
console.log('Number of uppercase character: ' + j)
// Returning the output
if (j >= 1) {
return true
} else {
return false
}
}
// Calling the function
let response = stringCheck(theString)
console.log('The response: ' + response)

Checking if combination of any amount of strings exists

I'm solving a puzzle and I have an idea of how to solve this problem, but I would like some guidance and hints.
Suppose I have the following, Given n amount of words to input, and m amount of word combos without spaces, I will have some functionality as the following.
4
this
is
my
dog
5
thisis // outputs 1
thisisacat // 0, since a or cat wasnt in the four words
thisisaduck // 0, no a or cat
thisismy // 1 this,is,my is amoung the four words
thisismydog // 1
My thoughts
First What I was thinking of doing is storing those first words into an array. After that, I check if any of those words is the first word of those 5 words
Example: check if this is in the first word thisis. It is! Great, now remove that this, from thisis to get simply just is, now delete the original string that corresponded to that equality and keep iterating over the left overs (now is,my,dog are available). If we can keep doing this process, until we get an empty string. We return 1, else return 0!
Are my thoughts on the right track? I think this would be a good approach (By the way I would like to implement this in javascript)

Sorting words from long to short may in some cases help to find a solution quicker, but it is not a guarantee. Sentences that contain the longest word might only have a solution if that longest word is not used.
Take for instance this test case:
Words: toolbox, stool, boxer
Sentence: stoolboxer
If "toolbox" is taken as a word in that sentence, then the remaining characters cannot be matched with other valid words. Yet, there is a solution, but only if the word "toolbox" is not used.
Solution with a Regular Expression
When regular expressions are allowed as part of the solution, then it is quite simple. For the above example, the regular expression would be:
^(toolbox|stool|boxer)*$
If a sentence matches that expression, it is a solution. If not, then not. This is quite straightforward, and doesn't really require an algorithm. All is done by the regular expression interpreter. Here is a snippet:
var words = ['this','is','a','string'];
var sentences = ['thisis','thisisastring','thisisaduck','thisisastringg','stringg'];
var regex = new RegExp('^(' + words.join('|') + ')*$');
sentences.forEach(sentence => {
// search returns a position. It should be 0:
console.log(sentence + ': ' + (sentence.search(regex) ? 'No' : 'Yes'));
});
But using regular expressions in an algorithm-challenge feels like cheating: you don't really write the algorithm, but rely on the regular expression implementation to do the job for you.
Without Regular Expressions
You could use this algorithm: first check whether a word matches at the start of the input sentence, and if so, remove that first occurrence from it. Then repeat this for the remaining part of the sentence. If this can be repeated until no characters are left over, you have a solution.
If characters are left over which cannot be matched with any word... well, then you cannot really conclude there is no solution for that sentence. It might be that some earlier made word choice was the wrong one, and there was an alternative. So to cope with that, your algorithm could backtrack and try other words.
This principle can be implemented through recursion. To gain memory-efficiency, you could leave the original sentence in-tact, and work with an index in that sentence instead.
The algorithm is implemented in arrow-function testString:
var words = ['this','is','a','string'];
var sentences = ['thisis','thisisastring','thisisaduck','thisisastringg','stringg'];
var testString = (words, str, i = 0) =>
i >= str.length || words.some( word =>
str.substr(i, word.length) == word && testString(words, str, i + word.length)
);
sentences.forEach(sentence => {
console.log(sentence + ': ' + (testString(words, sentence) ? 'Yes' : 'No'));
});
Or, the same in non-arrow-function syntax:
var words = ['this','is','a','string'];
var sentences = ['thisis','thisisastring','thisisaduck','thisisastringg','stringg'];
var testString = function (words, str, i = 0) {
return i >= str.length || words.some(function (word) {
return str.substr(i, word.length) == word
&& testString(words, str, i + word.length);
});
}
sentences.forEach(function (sentence) {
console.log(sentence + ': ' + (testString(words, sentence) ? 'Yes' : 'No'));
});
... and without some(), forEach() or ternary operator:
var words = ['this','is','a','string'];
var sentences = ['thisis','thisisastring','thisisaduck','thisisastringg','stringg'];
function testString (words, str, i = 0) {
if (i >= str.length) return true;
for (var k = 0; k < words.length; k++) {
var word = words[k];
if (str.substr(i, word.length) == word
&& testString(words, str, i + word.length)) {
return true;
}
}
}
for (var n = 0; n < sentences.length; n++) {
var sentence = sentences[n];
if (testString(words, sentence)) {
console.log(sentence + ': Yes');
} else {
console.log(sentence + ': No');
}
}

Take the 4 words, put them into a regex.
Use that regex to split each string.
Take the length of the resulting array (subtract one for the initial length of one).
var size = 'thisis'.split(/this|is|my|dog/).length - 1
Or if your list of words is an array
var search = new RegExp(words.join('|'))
var size = 'thisis'.split(search).length - 1
Either way you are splitting up the string by the list of words you have defined.
You can sort the words by length to ensure that larger words are matched first by
words.sort(function (a, b) { return b.length - a.length })

Here is the solution for anyone interested
var input = ['this','is','a','string']; // This will work for any input, but this is a test case
var orderedInput = input.sort(function(a,b){
return b.length - a.length;
});
var inputRegex = new RegExp(orderedInput.join('|'));
// our combonation of words can be any size in an array, just doin this since prompt in js is spammy
var testStrings = ['thisis','thisisastring','thisisaduck','thisisastringg','stringg'];
var foundCombos = (regex,str) => !str.split(regex).filter(str => str.length).length;
var finalResult = testStrings.reduce((all,str)=>{
all[str] = foundCombos(inputRegex,str);
if (all[str] === true){
all[str] = 1;
}
else{
all[str] = 0;
}
return all;
},{});
console.log(finalResult);

How to capitalize the last letter of each word in a string

I am still rather new to JavaScript and I am having an issue of getting the first character of the string inside the array to become uppercase.
I have gotten to a point where I have gotten all the texted lowercase, reversed the text character by character, and made it into a string. I need to get the first letter in the string to uppercase now.
function yay () {
var input = "Party like its 2015";
return input.toLowerCase().split("").reverse().join("").split(" ");
for(var i = 1 ; i < input.length ; i++){
input[i] = input[i].charAt(0).toUpperCase() + input[i].substr(1);
}
}
console.log(yay());
I need the output to be "partY likE itS 2015"

Frustrating that you posted your initial question without disclosing the desired result. Lots of turmoil because of that. Now, that the desired result is finally clear - here's an answer.
You can lowercase the whole thing, then split into words, rebuild each word in the array by uppercasing the last character in the word, then rejoin the array:
function endCaseWords(input) {
return input.toLowerCase().split(" ").map(function(item) {
return item.slice(0, -1) + item.slice(-1).toUpperCase();
}).join(" ");
}
document.write(endCaseWords("Party like its 2015"));
Here's a step by step explanation:
Lowercase the whole string
Use .split(" ") to split into an array of words
Use .map() to iterate the array
For each word, create a new word that is the first part of the word added to an uppercased version of the last character in the word
.join(" ") back together into a single string
Return the result
You could also use a regex replace with a custom callback:
function endCaseWords(input) {
return input.toLowerCase().replace(/.\b/g, function(match) {
return match.toUpperCase();
});
}
document.write(endCaseWords("Party like its 2015"));
FYI, there are lots of things wrong with your original code. The biggest mistake is that as soon as you return in a function, no other code in that function is executed so your for loop was never executed.
Then, there's really no reason to need to reverse() the characters because once you split into words, you can just access the last character in each word.

Instead of returning the result splitting and reversing the string, you need to assign it to input. Otherwise, you return from the function before doing the loop that capitalizes the words.
Then after the for loop you should return the joined string.
Also, since you've reverse the string before you capitalize, you should be capitalizing the last letter of each word. Then you need to reverse the array before re-joining it, to get the words back in the original order.
function yay () {
var input = "Party like its 2015";
input = input.toLowerCase().split("").reverse().join("").split(" ");
for(var i = 1 ; i < input.length ; i++){
var len = input[i].length-1;
input[i] = input[i].substring(0, len) + input[i].substr(len).toUpperCase();
}
return input.reverse().join(" ");
}
alert(yay());

You can use regular expression for that:
input.toLowerCase().replace(/[a-z]\b/g, function (c) { return c.toUpperCase() });
Or, if you can use arrow functions, simply:
input.toLowerCase().replace(/[a-z]\b/g, c => c.toUpperCase())

Here's what I would do:
Split the sentence on the space character
Transform the resulting array using .map to capitalize the first character and lowercase the remaining ones
Join the array on a space again to get a string
function yay () {
var input = "Party like its 2015";
return input.split(" ").map(function(item) {
return item.charAt(0).toUpperCase() + item.slice(1).toLowerCase();
}).join(" ");
}
console.log(yay());

Some ugly, but working code:
var text = "Party like its 2015";
//partY likE itS 2015
function yay(input) {
input = input.split(' ');
arr = [];
for (i = 0; i < input.length; i++) {
new_inp = input[i].charAt(0).toLowerCase() + input[i].substring(1, input[i].length - 1) + input[i].charAt(input[i].length - 1).toUpperCase();
arr.push(new_inp);
}
str = arr.join(' ');
return str;
}
console.log(yay(text));

Try using ucwords from PHP.js. It's quite simple, actually.
String.prototype.ucwords = function() {
return (this + '')
.replace(/^([a-z\u00E0-\u00FC])|\s+([a-z\u00E0-\u00FC])/g, function($1) {
return $1.toUpperCase();
});
}
var input = "Party like its 2015";
input = input.charAt(0).toLowerCase() + input.substr(1);
input = input.split('').reverse().join('').ucwords();
input = input.split('').reverse().join('');
Note: I modified their function to be a String function so method chaining would work.

function yay(str)
{
let arr = str.split(' ');
let farr = arr.map((item) =>{
let x = item.split('');
let len = x.length-1
x[len] = x[len].toUpperCase();
x= x.join('')
return x;
})
return farr.join(' ')
}
var str = "Party like its 2015";
let output = yay(str);
console.log(output) /// "PartY likE itS 2015"

You can split and then map over the array perform uppercase logic and retun by joining string.
let string = "Party like its 2015";
const yay = (string) => {
let lastCharUpperCase = string.split(" ").map((elem) => {
elem = elem.toLowerCase();
return elem.replace(elem[elem.length - 1], elem[elem.length - 1].toUpperCase())
})
return lastCharUpperCase.join(" ");
}
console.log(yay(string))