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i have a function to split string into 2 part, front and back. Then reverse it to back and front. Here is my code
function reverseString(string) {
let splitString = ""
let firstString = ""
for(i = 0; i <= string.length/2 - 1; i++) {
firstString += string[i]
}
for(i = string.length/2; i <= string.length; i++) {
splitString += string[i]
}
return splitString + firstString
}
Sorry for bad explanation, this is test case and expected result (first one is expected result, the second one is my result)
console.log(reverseString("aaabccc")); // "cccbaaa" "undefinedundefinedundefinedundefinedaaa"
console.log(reverseString("aab")); // "baa" "undefinedundefineda"
console.log(reverseString("aaaacccc")); // "ccccaaaa" "ccccundefinedaaa"
console.log(reverseString("abcdefghabcdef")); // "habcdefabcdefg" "habcdefundefinedabcdefg"
could you help me, whats wrong with it. Thank you
You could try another approach and use the slice function
function reverseString(string)
{
if (string.length < 2) { return string; }
let stringHalfLength = string.length / 2;
let isLengthOdd = stringHalfLength % 1 !== 0;
if (isLengthOdd) {
return string.slice(Math.ceil(stringHalfLength), string.length + 1) + string[Math.floor(stringHalfLength)] + string.slice(0, Math.floor(stringHalfLength));
}
return string.slice(stringHalfLength, string.length + 1) + string.slice(0, stringHalfLength);
}
console.log(reverseString("aaabccc") === "cccbaaa");
console.log(reverseString("aab") === "baa");
console.log(reverseString("aaaacccc") === "ccccaaaa");
console.log(reverseString("abcdefghabcdef") === "habcdefabcdefg");
A more efficient way to reverse the string would be to split the string, then use the built-in reverse javascript function (which reverses the elements of the split string), and then re-join the elements using the join function.. No need to re-invent the wheel?
You can concatenate the functions in shorthand (.split.reverse.join etc...) so your function would look something like this:
function reverseString(string) {
return string.split("").reverse().join("");
}
Try it out!
function reverseString(string) {
return string.split("").reverse().join("");
}
console.log(reverseString("hello"));
console.log(reverseString("aaabbbccc"));
If there's a particular reason you're opting not to use the in-built functions (i.e. if I've missed something?) , feel free to comment.
The short version of what you need:
function reverseString(string) {
const splitPosition = Math.ceil(string.length / 2);
return string.substring(splitPosition) + string.substring(0, splitPosition);
}
The key to your question is the middle element. To accomplish that, you probably want to use Math.floor that round under.
console.log(reverseString("aaabccc")); // "cccbaaa"
console.log(reverseString("abcdefghabcdef")); // "habcdefabcdefg"
function reverseString (str) {
if (str.length<2) {
return str
}
var half = Math.floor(str.length / 2);
return (str.slice(-half) + (str.length%2?str[half]:"") + str.slice(0,half));
}
reverseString('')
> ""
reverseString('1')
> "1"
reverseString('12')
> "21"
reverseString('123')
> "321"
reverseString('1234')
> "3412"
reverseString('12345')
> "45312"
reverseString("aaabccc")
> "cccbaaa"
reverseString("abcdefghabcdef")
> "habcdefabcdefg"
So basically your problem is not to grab 2 parts of the string and rearrange, it is to grab 3 parts.
1 part: str.slice(0,half)
2 part: str.length%2 ? str[half] : ""
3 part: str.slice(-half)
The second part is empty if the string length is even and the middle character if is odd.
So the code version in long self explanatory code:
function reverseString (str) {
if (str.length<2) {
return str
}
var half = Math.floor(str.length / 2);
var firstPart = str.slice(0,half);
var midlePart = str.length % 2 ? str[half] : ""; // we could expand also this
var endPart = str.slice(-half);
return endPart + midlePart + firstPart;
}
And also, notice the precondition, so I don't have to deal with the easy cases.
Also, in your code, you got undefined because you access in the last loop to:
string[string.length] you need to change <= by <
I need help Writing a function subLength() that takes 2 parameters, a string and a single character. The function should search the string for the two occurrences of the character and return the length between them including the 2 characters. If there are less than 2 or more than 2 occurrences of the character the function should return 0. How can I solve this problem using loops?
subLength('Saturday', 'a'); // returns 6
subLength('summer', 'm'); // returns 2
subLength('digitize', 'i'); // returns 0
subLength('cheesecake', 'k'); // returns 0
Here I loop through the characters of the string to find each value that is the char.
if the length isn't 2, return 0.
using slice, get only the characters within the two found indexs and get that length adding one to fix the offset
const subLength = (str, char) => {
let strChars = str.toLowerCase().split(""),
found = [],
length = 0;
strChars.forEach((val, index) => {
if (val === char) {
found.push(index);
}
});
if (found.length != 2) {
return length;
}
return str.slice(found[0], found[1]).length + 1;
}
console.log(subLength('Saturday', 'a')); // returns 6
console.log(subLength('summer', 'm')); // returns 2
console.log(subLength('digitize', 'i')); // returns 0
console.log(subLength('cheesecake', 'k')); // returns 0
You can try this logic:
Loop over string and count number of occurance
if count is 2,
Create a regex to capture the string in between.
Return its length
Else return 0
function subLength(str, char) {
let length = 0;
const occuranceCount = Array
.from(str)
.filter((c) => c.toLowerCase() === char.toLowerCase())
.length
if (occuranceCount === 2) {
const regex = new RegExp(`${char}(.*)${char}`)
length = str.match(regex)[0].length
}
console.log(length)
return length;
}
subLength('Saturday', 'a'); // returns 6
subLength('summer', 'm'); // returns 2
subLength('digitize', 'i'); // returns 0
subLength('cheesecake', 'k'); // returns 0
Using just for loop:
function subLength(str, char) {
let count = 0;
let initPosition;
let lastPosition;
for (let i = 0; i < str.length; i++) {
if (str[i] === char) {
count++
if (count > 2) {
return 0;
}
if (initPosition === undefined) {
initPosition = i
} else {
lastPosition = i+1
}
}
}
return count < 2 ? 0 : lastPosition - initPosition;
}
console.log(subLength('Saturday', 'a')); // returns 6
console.log(subLength('summer', 'm')); // returns 2
console.log(subLength('digitize', 'i')); // returns 0
console.log(subLength('cheesecake', 'k')); // returns 0
I too am going through the Codecademy course where this question came up which led me to this post.
Using the RegExp solution provided by #Rajesh (thank you!!) I started to break it down to better understand what was going on and making notes/comments because I am still pretty new and haven't used or been exposed to some of these things.
At the end of it all I thought I'd share what I ended up with in case anyone found it helpful.
function subLength(str, char) {
// Outputting to the console what we are looking for given the value of the string and character from the test cases at the end of this script.
console.log(`Showing the subLength for the string: "${str}" between "${char}" and "${char}" including the "${char}" positions.`);
// create the length variable which will be returned by the function
let length = 0;
// ** Search the string for the two occurrences of the character and count them. Then assign the result to the occurrenceCount variable for use in the if else statement.
// The "Array" class is a global object that is used in the construction off arrays.
// The Array.from() static method creates a new, shallow-copied Array instance from an array-like or iterable object.
// The Array.filter() method creates a new array with all elements that pass the test implemented by the provided function. The "c" represents each element of the array/string which is then compared to the char variable. if it is a match it gets added to the Array. We use .toLowerCase on both to ensure case compatibility.
// Appending the Array with ".length" assigns occurrenceCount the numeric value of the array's length rather than the array of characters.
const occurrenceCount = Array.from(str).filter((c) => c.toLowerCase() === char.toLowerCase());
console.log(' The contents of the occurrenceCountArray = ' + occurrenceCount);
console.log(' The character occurrence count = ' + occurrenceCount.length);
// if the string has two occurrences : return the length between them including the two characters : else the string has less than 2 or more than 2 characters : return 0.
if (occurrenceCount.length === 2) {
// The RegExp object is used for matching text with a pattern. The "(.*)" in between the ${char}'s will match and capture as much as possible aka greedy match. "()" = capture anything matched. (" = start of group. "." = match any character. "*" = Greedy match that matches everything in place of the "*". ")" = end of group.
const regex = new RegExp(`${char}(.*)${char}`);
// log to console the pattern being matched
console.log(` regex pattern to find = ${regex}`);
// log to the console the [0] = index 0 pattern that was captured from the string using str.match(regex)[0]
console.log(` regex output = ${str.match(regex)[0]}`);
// Use".length" to count the number of characters in the regex string at index 0 of the regex array and assign that value to the length variable.
length = str.match(regex)[0].length;
// Output the results to the console
console.log(` The distance from "${char}" to "${char}" (including the "${char}" positions) in the string: ${str} = ${length}\n`);
// return the length value
return length;
} else {
// Output the results to the console
console.log(` The string either has too many or too few occurrences.\n The subLength = ${length}\n`);
// return the length value
return length;
}
}
// test cases
subLength('Saturday', 'a'); // returns 6
subLength('summer', 'm'); // returns 2
subLength('digitize', 'i'); // returns 0
subLength('cheesecake', 'k'); // returns 0
The answer I am getting is this:
const subLength = (str, char) => {
let charCount = 0;
let len = -1;
for (let i=0; i<str.length; i++) {
if (str[i] == char) {
charCount++;
if (charCount > 2) {
return 0;
}
if (len == -1) {
len = i;
} else {
len = i - len + 1
}
}
}
if (charCount < 2) {
return 0;
}
return len;
};
It is better to try yourself a solution first. It is a very bad practice to just ask a solution for your homework!!!
Even if the solution can be JUST a few lines of code i wrote for you with commments a working solution :
const subLength = (str,char) => {
// create an empty array
const strarr = [];
// push string into array
strarr.push(str);
//initiate a count variable
let count = 0;
// WRITE YOUR REGULAR EXPRESSION
// Using the regular expression constructor - new RegExp("ab{2}", "g") .
const regString = `[${char}]`;
const regex = new RegExp(regString, "g");
// iterate through the string array to
for (let i = 0; i < strarr.length; i++) {
// calculate how many time the character occurs
count = (strarr[i].match(regex) || []).length;
};
// check with if condition
//if count is 2
if (count === 2) {
// calculate the index of first ocurrance of the string
first = str.indexOf(char);
// calculate the index of second ocurrance of the string
second = str.lastIndexOf(char);
// calculate the distance between them
return second - first + 1;
// if count is greater than two return 0
}
else if (count > 2) {
return count = 0;
}
// if count is less than two return 0
else if (count < 2) {
return 0;
}
};
console.log(subLength("iiiiliiile","l"));
I just answered this problem in codeAcademy and this is the solution that I came up with, just using if-statements and string.indexOf
const subLength = (strng, char) => {
let firstIndex = strng.indexOf(char);
let secondIndex = strng.indexOf(char, (firstIndex + 1));
let thirdIndex = strng.indexOf(char, (secondIndex + 1));
if (firstIndex === -1){
return 0
} else if (secondIndex === -1){
return 0
} else if (thirdIndex === -1 ){
return (secondIndex - firstIndex + 1)
} else {
return 0
};
};
Question
Find the longest substring in alphabetical order.
Example: the longest alphabetical substring in "asdfaaaabbbbcttavvfffffdf" is "aaaabbbbctt".
There are tests with strings up to 10 000 characters long so your code will need to be efficient.
The input will only consist of lowercase characters and will be at least one letter long.
If there are multiple solutions, return the one that appears first.
My Solution
function longest(str) {
//first element of count == total count of the highest number
//second element of count == longest str so far
let count = [0, ''];
//temp count == length of current str
let tempCount = [0];
//split the str to an array
let strArr = str.split('');
//loop through each letter of the string
for(let i = 0; i < strArr.length; i++){
//if the character is higher in the alpahabet than the last then
if(convertToNumber(strArr[i])<convertToNumber((strArr[i]-1)) || convertToNumber((strArr[i]-1))== undefined){
tempCount[0]++;
//if the current character is not higher than the last
} else {
if(tempCount[0] > count[0]){
//change the longest str number to the length of this str
count[0] = tempCount[0];
//slice the new longest str
let longestStr = strArr.slice(strArr[i]-tempCount[0], strArr[i]);
//join the str together
count[1] = longestStr.join('');
//reset the temp count
tempCount[0] = 0;
} else {
//reset the temp count
tempCount[0] = 0;
}
}
}
//converts the relevant letter to a code
function convertToNumber(letter){
return letter.charCodeAt(0);
}
//returns the longest str
return count[1];
}
console.log(longest('asdfaaaabbbbcttavvfffffdf'));
The algorithm returns 'letter.charCodeAt is not a function'
Why is this the case and how can I make adjustments to fix the algorithm?
The problem is in here:
if(convertToNumber(strArr[i])<convertToNumber((strArr[i]-1)) || convertToNumber((strArr[i]-1))== undefined)
you are trying to subtract 1(Number) from a letter(String). It should be strArr[i-1].
But also
even after fixing it you'd still get an error for:
convertToNumber(strArr[i]) < convertToNumber(strArr[i-1])
if i will be zero as strArr[-1] will give undefined.
const arr = [1, 2, 3];
console.log(arr[-1]);
Is basically undefined.charCodeAt(0);, which will produce error.
I know that the OP wants a solution for his error. fedsec showed him for this a way. So I think there is nothing else to do.
I just put here my solution of the task, so the OP can compare to other possible solutions.
function longest(str) {
let max = 1;
let maxStr = str.charAt(0);
let char = maxStr;
let testStr = maxStr;
let count = 1;
for (i=1; i<=str.length; i++) {
let prev = char;
char = str.charAt(i);
if (prev <= char) {
count++;
testStr += char;
if (count>max) {
max = count;
maxStr = testStr;
}
} else {
if ( i+count >str.length) break;
count = 1;
testStr = char;
}
}
return(maxStr);
}
console.log(longest('asdfaaaabbbbcttavvfffffdf'));
My goal is to get the result of 3 since in "aaBbCChr" letters a,b and c are repeating themselves. I've made this function, everything looks right, but it just doesn't work properly.
function duplicateCount(text) {
let lettersArray = text.split("");
let duplicateResult = 0;
lettersArray.map(function(letter) {
let regexLetter = new RegExp(letter, "gi");
let matchesCount = text.match(regexLetter).length;
if (matchesCount > 1) {
duplicateResult + 1;
} else {};
});
return duplicateResult;
};
alert(duplicateCount("aaBbCChr"));
Issues with your code:
duplicateResult + 1; This doesn't change the value of duplicateResult. You need to use a statement that actually increases its value like duplicateResult += 1;
lettersArray also contains the letters multiple times. First remove the duplicates there before looking whether they appear multiple times in the string or you will count them multiple times. See Get all unique values in a JavaScript array (remove duplicates)
Make all lowercase letters in lettersArray so that you don't count the same letter twice when it appears in lowercase and uppercase.
This code works
function duplicateCount(text) {
let lettersArray = text.toLowerCase().split("");
let duplicateResult = 0;
function onlyUnique(value, index, self) {
return self.indexOf(value) === index;
}
lettersArray = lettersArray.filter( onlyUnique );
lettersArray.map(function(letter) {
let regexLetter = new RegExp(letter, "gi");
let matchesCount = text.match(regexLetter).length;
if (matchesCount > 1) {
duplicateResult += 1;
} else {};
});
return duplicateResult;
};
alert(duplicateCount("aaBbCChr"));
Another approach is to invert the logic used in Get all unique values in a JavaScript array (remove duplicates) to use filter to filter out all those characters that appear multiple times and then count those.
function duplicateCount2(text) {
let lettersArray = text.toLowerCase().split("");
let duplicateResult = 0;
function onlyNonUnique(value, index, self) {
return self.indexOf(value) !== index;
}
lettersArray = lettersArray.filter(onlyNonUnique);
duplicateResult = lettersArray.length;
return duplicateResult;
};
alert(duplicateCount2("aaBbCChr"));
It is far easier to do it with changing the structure to an object
const reduced = [...myChars].reduce((map, charCaseSensitive) => {
const char = charCaseSensitive.toLowerCase();
if(map[char]) map[char]++;
else map[char] = 1;
return map;
}, {};
const numberOfRepeatedChars = Object.values(reduced).filter(v => v > 1).length;
Here's the version making minimal changes to your code. You need to lowercase the string here so that the same letter in caps is not counted again.
function duplicateCount(text) {
let lettersArray = text.toLowerCase().split("");
let duplicateResult = 0;
let set = new Set();
lettersArray.map(function(letter) {
let regexLetter = new RegExp(letter, "gi");
let matchesCount = text.match(regexLetter).length;
if (!set.has(letter) && matchesCount > 1) {
set.add(letter);
duplicateResult += 1;
} else {};
});
return duplicateResult;
};
alert(duplicateCount("aaBbCChr"));
As mentioned in other comments, your problem is that you're resetting duplicateResult = 1 to a duplicate result. You probably meant to do duplicateResult += 1, but even that is incorrect because you said that you only want to get repeated characters, so no matter how many times a letter repeats, if it repeats at all, you only count it once.
My approach would be to count the characters in the string in an object (key is the letter, value is the occurrence count) and afterwards count the number of keys that have an occurrence count of more than 1.
function duplicateCount(text) {
const isValidLetter = /[a-z]/i; // a through z, case insensitive
const charCounts = text.split('').reduce(function(acc, char) {
if(isValidLetter.test(char)) {
char = char.toLowerCase();
if(acc[char] === undefined) {
acc[char] = 1;
} else {
acc[char] += 1;
}
}
return acc;
}, {});
return Object.keys(charCounts).filter(c => charCounts[c] > 1).length;
}
alert(duplicateCount('aaBbCChr'));
I've seen versions of this question for other languages, but not for JS.
Is it possible to do this recursively in one function?
I understand that I need to take the first element in the string, and then append it to each solution to the recursion on the remainder of the string.
So logically, I understand how the recursion needs to go. I just don't understand how to append the first char onto each of the recursive solutions
var myString = "xyz";
function printPermut(inputString){
var outputString;
if(inputString.length === 0){
return inputString;
}
if(inputString.length === 1){
return inputString;
}
else{
for(int i = 0; i<inputString.length(); i++){
//something here like:
//outputString = outputString.concat(printPermut(inputString.slice(1))??
//maybe store each unique permutation to an array or something?
}
}
}
Let's write a function that returns all permutations of a string as an array. As you don't want any global variables, returning the permutations is crucial.
function permut(string) {
if (string.length < 2) return string; // This is our break condition
var permutations = []; // This array will hold our permutations
for (var i = 0; i < string.length; i++) {
var char = string[i];
// Cause we don't want any duplicates:
if (string.indexOf(char) != i) // if char was used already
continue; // skip it this time
var remainingString = string.slice(0, i) + string.slice(i + 1, string.length); //Note: you can concat Strings via '+' in JS
for (var subPermutation of permut(remainingString))
permutations.push(char + subPermutation)
}
return permutations;
}
To print them, just iterate over the array afterwards:
var myString = "xyz";
permutations = permut(myString);
for (permutation of permutations)
print(permutation) //Use the output method of your choice
Hope I could help you with your question.
The problem of permutations has been studied to death. Heap's algorithm is one well-known solution. Here is a version in JS, using a generator:
function *permute(a, n = a.length) {
if (n <= 1) yield a.slice();
else for (let i = 0; i < n; i++) {
yield *permute(a, n - 1);
const j = n % 2 ? 0 : i;
[a[n-1], a[j]] = [a[j], a[n-1]];
}
}
console.log(Array.from(permute("abcabad".split('')))
.map(perm => perm.join(''))
.filter((el, idx, self) => (self.indexOf(el) === idx)));
permute is designed to take and generate arrays, not strings, so we split the string into characters before calling it, and paste the characters back into strings before printing out the results.
Use Recursive Function to iterate through the string
function getPermutations(string) {
var results = [];
if (string.length === 1)
{
results.push(string);
return results;
}
for (var i = 0; i < string.length; i++)
{
var firstChar = string[i];
var otherChar = string.substring(0, i) + string.substring(i + 1);
var otherPermutations = getPermutations(otherChar);
for (var j = 0; j < otherPermutations.length; j++) {
results.push(firstChar + otherPermutations[j]);
}
}
return results;
}
var permutation = getPermutations('YES').filter((el, idx, self) => (self.indexOf(el) === idx));
console.log("Total permutation: "+permutation.length);
console.log(permutation);
Problem classification: You can look at this problem as an exploration problem, i.e., given a set of input characters explore the different ways you can arrange them.
Solution: Backtracking algorithm excels in solving exploratory problems, although it comes with high time complexity. To demonstrate a solution, imagine how you would solve this problem by hand for a small set of input characters: [a, b, c].
Here are the steps:
Take the left most character. This is the character at index 0 and swap it with target right character at index 0, i.e. with itself. This is because [a, b, c] is a valid permutation on its own therefore we want to keep it. Swapping characters normally requires two pointers which point to each of the characters. So let's say we will have a left and right pointer.
With the same left most character (at index 0) do the swapping with target right character at index 0 + 1 = 1, i.e. move the target right pointer with 1 step further. This will give you the output: [b, a, c]
With the same left most character (at index 0) do the swapping with the next next target right character (i.e. index 0 + 1 + 1 = 2). This will give you the output: [c, b, a]
Ok, now we need to stop as there are no more target right characters to be swapped with the left most character. So our right pointer needs to stay less than the max index in the input. Moving the right pointer with a step at a time we can do with a for loop which starts from the left index and ends with the input length - 1.
Now you need to do exact same steps from above but move the left pointer so that it points to the next left most character. However, keeping the input from step 2 and 3. Another way to imagine this situation is to say: 'Hey, I am done with the left most character. Now I do not want to work with it anymore but I would love to continue with the second left most from the results I have so far.'
When do we stop? When the left pointer has reached the length of the input string - 1, 'cause there is no more characters after this index. In recursive algorithms (such as the backtracking), the case where you need to stop is called base case. In our example the base case is: left === input.length - 1.
Here is a graphical visualisation:
left index| Input String:
-------------------------------------------------------------------------------
left = 0 | in=[a, b, c]
(swap in[0] with in[0]) (swap in[0] with in[1]) (swap in[0] with in[2])
left = 1 | in=[a, b, c] in=[b, a, c] in=[c, b, a]
(swap in[1] with in[1]) (swap in[1] with in[2]) (swap in[1] with in[1])(swap in[1] with in[2]) (swap in[1] with in[1])(swap in[1] with in[2])
left = 2 | [a, b, c] [a, c, b] [b, a, c] [b, c, a] [c, b, a] [c, a, b]
Summary:
To move the left pointer to the right we will use recursive increment
To move the right pointer to the right we will use a for loop, however we need to start always from the left pointer or else we will explore things we have already explored.
Backtracking:
A pseudo-code for backtracking algorithm takes the form of:
fun(input)
if(base_case_check(input)) {
//do final step
} else {
//choose
fun(reduce(input)) //explore
//un-choose
}
Our solution:
function permutate(string) {
if(!string || string.length === 0)
return new Set(['']);
let left = 0;
let result = new Set();
permutationHelper(string, result, left);
return result;
}
function permutationHelper(string, result, left) {
if(left === string.length-1) {
//base case
result.add(string);
} else {
//recursive case
for(let right=left; right < string.length; right++) {
string = swap(string, left, right); //choose
permutationHelper(string, result, left+1); // explore
string = swap(string, left, right); //unchoose
}
}
}
function swap(string, left, right) {
let tmpString = string.split('');
let tmp = tmpString[left];
tmpString[left] = tmpString[right];
tmpString[right] = tmp;
return tmpString.join('');
}
/* End of solution */
/* Tests */
let input = 'abc';
let result = permutate(input);
let expected = new Set(['abc', 'acb', 'bac', 'bca', 'cab', 'cba']);
if(setsEquality(result, expected)) {
console.log('Congrats, you generated all permuations');
} else {
console.log('Sorry, not all permuations are generated');
}
function setsEquality(actualResult, expectedResult) {
if (actualResult.size !== expectedResult.size) {
return false;
}
for (let permutation of actualResult) {
if (!expectedResult.has(permutation)) return false;
}
return true;
}
function assert(condition, desc) {
if (condition) {
console.log(`${desc} ... PASS`);
} else {
console.log(`${desc} ... FAIL`);
}
}
Summary & Time Complexity:
We make our choice by swapping characters in the existing input string
We explore what is left to be explored once we increment our left index with 1. This in fact means that we are reducing our input set for all subsequent recursions with 1. Therefore the work we need to do is: Nx(N-1)x(N-2)x(N-3)x...x1 = N!. However, as we needed a for loop to explore among the input we have, the total time complexity would be: 0(N*N!)
We revert our choice by swapping characters back in the modified input string
permutation=(str,prefix)=>{
if(str.length==0){
console.log(prefix);
}
else{
for(let i=0;i<str.length;i++){
let rem = str.substring(0,i)+str.substring(i+1);
permutation(rem,prefix+str[i]);
}
}
}
let str="ABC";
permutation(str,"");
Semi-Off topic:
random permutation of a given string is as simple as rndperm:
i = document.getElementById("word");
b = document.getElementById("butt");
rndperm = (z) => {
return z.split("").sort(() => ((Math.random() * 3) >> 0) - 1).join("")
}
function scramble() {
i.value = rndperm(i.value);
}
var z;
function sci() {
if (z != undefined) {
clearInterval(z);
b.innerText = "Scramble";
z=undefined;
} else {
z = setInterval(scramble, 100);
b.innerText = "Running...";
}
}
<center><input id="word" value="HelloWorld"></input><button id="butt" onclick=sci()>Scramble</button></center>
I had same question by my interviewer last day but I was not get the correct logic then I came to stackoverflow and I get here but now I have my solution and want to share with all
const str_Permutations = (str,ar = []) => {
str = `${str}`; // ensure type **String**
if(ar.indexOf(str)>-1 || str.length !== (ar.strlen || str.length)) return false; // Checking if value is alreay there or(||) on recursive call string length should not be provided string
ar.strlen = ar.strlen || str.length; // Setting str length of provided value(string)
ar.push(str); // Pushing to array
for(let i = 0; i<str.length;i++){
str_Permutations(str[i] + str.split('').filter(v=>v!==str[i]).join(''),ar);
}
return Array.from(ar); // Removing *strlen* from main result and return **Result** as array
}
str_Permutations("ABC")
//Result: (6) ["ABC", "BAC", "CBA", "BCA", "ACB", "CAB"]
There is used reference feature of Array to hold the values in same Array by passing. I hope you got my point!!!!
const permut = (str) => {
if (str.length <= 2) return str.length === 2 ? [str, str[1] + str[0]] : [str];
return str
.split("")
.reduce(
(acc, letter, i) =>
acc.concat(
permut(str.slice(0, i) + str.slice(i + 1)).map((val) => letter + val)
),
[]
);
};
found here
This does the job, recursively
function printPermutations(str, res='') {
if (!str.length){
console.log(res);
}
for (let i = 0; i < str.length; i++) {
let remStr = str.substr(0, i) + str.substr(i + 1);
printPermutations(remStr, res + str.substr(i, 1));
}
}
printPermutations("abc")
// result
// abc, acb, bac, bca, cab, cba
Simple and readable approach but only limited to 3 chars
const stringPermutation = (str) => {
let permutations = [];
for (let i in str) {
for (let j in str) {
for (let k in str) {
if (str[i] !== str[j] && str[j] !== str[k] && str[i] !== str[k]) {
permutations.push(str[i] + str[j] + str[k]);
}
}
}
}
return permutations;
};
console.log(stringPermutation("abc"));
var str = "abcdefgh";
for(let i = 0; i<str.length; i++){
for(let j = i; j<=str.length; j++){
if(i != j){
var out = str.slice(i,j);
console.log(out);
}
}
}