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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'));
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);
}
}
}
Link to codewars challenge
I need to return the length of the longest prefix that is also a suffix of a string in Javascript.
As far as I understand, the prefixes in "abcd" are:
['a', 'ab', 'abc']
And the suffixes in "abcd" are:
[ 'bcd', 'cd', 'd' ]
So the length of the longest prefix that is also a suffix in "abcd" in this case is 0, because there are no prefixes that are also suffixes in "abcd".
So far I've been able to figure out how to get the suffixes into an array for comparison, but not the prefixes.
function returnLongestPrefixAndSuffix(string) {
let prefixes = [];
let suffixes = [];
for (let i = 0; i < string.length -1; i++) {
prefixes.push(string.slice(i));
}
for (let i = 1; i < string.length; i++) {
suffixes.push(string.slice(i));
}
return prefixes + " " + suffixes;
}
console.log(returnLongestPrefixAndSuffix("abcd"));
I'm not grasping the concept of how to start at the beginning of a string and add a larger element to the array each time by one character, excluding the element that would include the last one.
Please follow my current logic if possible.
EDIT: My code now looks like this:
function solve(string) {
let prefixes = [];
let suffixes = [];
let includedList = [];
for (let i = 1; i < string.length; i++) {
prefixes.push(string.slice(0, i));
}
for (let i = 1; i < string.length; i++) {
suffixes.push(string.slice(-i));
}
console.log(prefixes);
console.log(suffixes);
for (let i = 0; i < prefixes.length; i++) {
let element = prefixes[i];
if (suffixes.includes(element) === true) {
includedList.push(element);
}
}
console.log(includedList);
if (includedList.length === 0) {
return 0;
}
else {
let overlap = prefixes.filter(value => suffixes.includes(value));
console.log(overlap);
let longest = includedList.sort(function (a, b) { return b.length - a.length; })[0];
return longest.length;
}
}
console.log(solve("abcdabc"));
And this is passing 10049 test but failing 163 tests on codewars. I still do not know what to do with the overlap variable or how to exclude overlaps from the includedList array.
function solve(string) {
for (let i = Math.floor(string.length / 2); i > 0; i--) {
let prefix = string.slice(0, i);
let suffix = string.slice(-i);
if (prefix == suffix) {
return i;
}
}
return 0;
}
console.log(solve("abcdabc"));
To account for the overlap, initialize your for-loop like this:
let i = Math.floor(string.length / 2)
That will initialize the for-loop at the half-way point in your string, so that you can count down and compare whether or not the prefix == the suffix, starting with the longest.
You could return prefix.length, but that will be the same thing as i.
Also, be sure to return 0 outside of the for-loop. Because if you try:
if (prefix != suffix) {
return 0;
}
inside of the for-loop, it will stop counting right there.
To get the prefixes, you can use the second argument of .slice:
string.slice(0, i)
Note that to get the suffixes, you could also take the string from the end:
string.slice(-i)
There is no sense in collecting prefixes and suffixes in arrays, just search for the biggest i where the suffix equals the prefix.
Please see the documentation of the slice function, it may take a second argument: https://www.w3schools.com/jsref/jsref_slice_string.asp
So following your logic, one way to get prefixes would be to:
for (let i = 1; i <= string.length; i++) {
prefixes.push(string.slice(0, i));
}
EDIT:
Your newest code doesn't work because of two reasons:
You may end up with includedList being empty, but you still try to get first element out of it.
You don't take overlaps into consideration. For the input aaa the correct result is a since prefix aa overlaps with the corresponding suffix. In other words the result can't be longer than half the length of the input.
I was recently doing an interview and was asked multiple questions, one of the questions was this and I had a bit of trouble trying to answer it.
Given a string, find the longest occurrence of vowels "aeiou" that appear.
The substring of vowels do not have to be consecutive, and there can be repeats.
The goal is the find the max occurrence of each vowel and join them, but it must be in the order of "a","e","i","o","u".
Edit: In addition, each individual vowel character must be chained as well. In the example below, there is "aaa" and "aa" , since 3 is longer, our result must contain the longer chain.
For example:
Input: "aaagtaayuhiejjhgiiiouaae"
Result: aaaeiiiou
The code that I have tried is below:
EDIT: Following the solution, I have written this below but I am still running into issues with strings such as "aeiouaaaeeeiiiooouuu". The correct result for that would be 15 but I am getting 5.
var findLongestVowels = function(s){
var count = 1;
var i = 0;
var j = 0;
var vowels = ['a','e','i','o','u'];
var total = 0;
var array = [];
while (i < s.length){
if (s.charAt(i) == vowels[j] && s.charAt(i) == s.charAt(i+1) ){
count++;
}
else if (s.charAt(i) == vowels[j] && s.charAt(i) != s.charAt(i+1)){
if (j === 0 && !array[vowels[j]]){
array[vowels[j]] = count;
}
else if (j === 0 && array[vowels[j]]){
array[vowels[j]] = Math.max(array[vowels[j]],count);
}
else if (j !== 0 && !array[vowels[j]] && array[vowels[j-1]]){
array[vowels[j]] = array[vowels[j-1]] + count;
}
else if (j !== 0 && array[vowels[j]] && array[vowels[j-1]]){
array[vowels[j]] = Math.max(array[vowels[j]],array[vowels[j-1]] + count);
}
count = 1;
}
else if (s.charAt(i) == vowels[j+1] && array[vowels[j]]){
j++;
i--;
}
i++;
}
console.log(array);
console.log('Answer: ' + array[vowels[j]]);
}
findLongestVowels("eeeeebbbagtaagaaajaaaaattyuhiejjhgiiiouaae");
Am I at least going in the right direction?
Thanks in advance.
We can solve this in O(n) time. Consider that for each block, if its vowel is at index v in the list of vowels, we are only interested in the best solution for the block with a vowel at index v-1 in the order of vowels. We save the last best solution for each block type (each vowel) as we go along:
|aaa|g|t|aa|y|u|h|i|e|jj|h|g|iii|o|u|aa|e
b: 1 2 3 4 5 6 7 8 9 10
b 1: v[a] = 3
b 2: v[a] = max(2,3)
b 3: v[u] = None recorded for v-1
b 4: v[i] = None recorded for v-1
b 5: v[e] = 1 + 3
b 6: v[i] = 3 + 4
b 7: v[o] = 1 + 7
b 8: v[u] = 1 + 8 // answer
b 9: v[a] = max(2,3)
b 10: v[e] = 1 + 3
JavaScript code:
function f(str){
console.log(`String: ${ str }\n`);
var vowels = {
a: {best: 0, prev: null},
e: {best: 0, prev: 'a'},
i: {best: 0, prev: 'e'},
o: {best: 0, prev: 'i'},
u: {best: 0, prev: 'o'}
};
function getBlock(i){
let length = 1;
while (str[i+1] && str[i] == str[i+1]){
length++;
i++;
}
return length;
}
for (let i=0; i<str.length;){
let length = getBlock(i);
console.log(`i: ${ i }; length: ${ length }`)
if (!vowels[str[i]]){
i = i + length;
continue;
}
if (!vowels[str[i]].prev){
vowels[str[i]].best = Math.max(
vowels[str[i]].best,
length
);
// make sure the previous vowel
// exists in the string before
// this vowel
} else if (vowels[ vowels[str[i]].prev ].best){
vowels[str[i]].best = Math.max(
vowels[str[i]].best,
length + vowels[ vowels[str[i]].prev ].best
);
}
i = i + length;
}
console.log(`\n${ JSON.stringify(vowels) }\n\n`);
return vowels['u'].best;
}
var s = 'eeeeebbbagtaagaaajaaaaattyuhiejjhgiiiouaae';
console.log(f(s) + '\n\n');
s = 'aaagtaayuhiejjhgiiiouaae';
console.log(f(s) + '\n\n');
s = 'aeiouaaaeeeiiiooouuu';
console.log(f(s));
This problem can be solved by using dynamic programming technique.
First, we have string x and we want to find the longest string for this string.
Traversing the string x from start to end, assuming at index i, we are trying to find vowel e, there are two possibilities:
Current character is e, so we take the whole block and move to next character
Or, we can try with the next character
So, we have our pseudocode:
int[][]dp;
int largestBlock (int index, int currentVowel, String x, String vowels){
if (currentVowel == 5) {
// We found all 5 vowel
return 0;
}
if visited this state (index, currentVowel) before {
return dp[index][currentVowel];
}
int result = largestBlock(index + 1, currentVowel, x, vowels) ;
if (x[index] == vowels[currentVowel]){
int nxt = nextIndexThatIsNotVowel(index, currentVowel, x, vowels);
result = max(result, nxt - index + largestBlock(nxt, currentVowel + 1, x , vowels));
}
return dp[index][currentVowel] = result;
}
Time complexity is O(n * m) with m is number of vowels which is 5 in this case.
You need to remember biggest combination of individual vowels.
Use reduce, map and Object.values
var vowels = "aeiou";
var input = "aaagtaayuhiejjhgiiiouaae";
var output = Object.values(
input.split( "" ).reduce( ( a, c, i, arr ) => {
var lastChar = arr[ i - 1 ];
if ( !vowels.includes( c ) ) return a; //if not vowel, return accumulator
if ( c != lastChar ) //if not same as last character then create a new array
{
a[ c ] = a[ c ] || [];
a[ c ].push( [ c ] );
}
else //else push to the last array;
{
var lastCombo = a[ c ].slice( -1 )[ 0 ];
lastCombo.push(c)
}
return a; //return accumulator
} , {}) ).map( s => {
var char = s[0][0]; //find the character to repeat
var maxLength = Math.max.apply( null, s.map( s => s.length ) ); //find how many times to repeat
return Array( maxLength + 1 ).join( char );
}).join( "" ); //join all the vowels
console.log( output );
It's just one of many possible solutions - feel free to try it out.
Store every vowel you're interested in, in vowels array.
Use map to loop over every vowel in array, create regex from vowel to split string into array of vowels. For example "aaabdmedaskaa" would be split into ["aaa", "a", "aa"].
Filter this array so it doesn't include empty strings.
Sort it by length, so accesing 0 element would give you longest occurance
After mapping over every vowel, return the result - filter out "undefined" in case some of your vowels don't occur at all and regex results in empty array (accesing empty array's 0th element would result in undefined), join array of occurances into a result string.
The regex created from "a" will be [^a]+ which means any character sequence that does not include "a".
function findLongestOccurance(str) {
const vowels = ["a", "e", "i", "o", "u"];
const result = vowels.map(vowel => {
const regex = new RegExp(`[^${vowel}]+`);
return str.split(regex)
.filter(r => r !== "")
.sort((a, b) => b.length - a.length)[0];
});
return result.filter(occ => typeof(occ) !== "undefined").join("");
}
console.log(findLongestOccurance("aaagtaayuhiejjhgiiiouaae"));
Why not regex?
var result = /(a+).*(e+).*(i+).*(o+).*(u+)/.exec("aaagtaayuhiejjhgiiiouaae");
console.log(result[1]+result[2]+result[3]+result[4]+result[5]);
First of all, from what I understand from the question the result for Input: "aaagtaayuhiejjhgiiiouaae" should be aaaaaeiiiou, like #PhamTrung asked in the comments but didn't get answered.
Because it a job interview I would start with the first thing that comes to mind, namely brute force the solution out of this
function a(string, prefix='') {
if(!string.length){
return prefix
}
if(!/[aeiou]/.test(string[0])){
return a(string.substr(1), prefix)
}
const option1 = a(string.substr(1), prefix)
const option2 = a(string.substr(1), prefix+string[0])
const validateRegex = /^a+e+i+o+u+$/
const isValidOption1 = validateRegex.test(option1)
const isValidOption2 = validateRegex.test(option2)
if(isValidOption1 && isValidOption2){
if(option1.length > option2.length) {
return option1
}
return option2
}
if(isValidOption1) {
return option1
}
if(isValidOption2) {
return option2
}
return null
}
const input = 'aaagtaayuhiejjhgiiiouaae'
console.log(a(input))
This has a terrible run time though, we are trying all possible substring that contains only vowels, than we are discarding those that aren't of the form required (a+e+i+o+u+) and than choosing only the biggest of them all. If I'm not mistake this has a worst case of ∑(n choose i) which is O(n^n) - well, the actual worst case here would be a stackOverflow exception for sufficiently large n in which case we'd have to reimplement this with a loop instead of recursing. In this case we could still get an out of memory exception in which case we'd be out of options but to improve our algorithm. It's fair to imagine that if the input were large enough that we got an out of memory exception than our code would also be slow enough to not be a reasonable solution to the problem. I'm just arguing all this because these are things that an interviewer would possibly like to see that you are aware of, meaning you know enough of CS 101.
Following that the interviewer would ask if I can improve the performance. This is a solution with running time of O(n).
const input = 'aeiouaaaeeeiiiooouuu'
let curr = { "1": {price: -1} }
const nodes = []
const voewels = '1aeiou'
const getPrevLetter = (node) => voewels[voewels.indexOf(node.letter) -1]
let resultNode
function setUpNodeByCurrent(node, letter){
node.price = curr[letter].price + 1
node.previous = curr[letter]
}
function setBestResultIfNeeded(node){
if(node.letter !== 'u') {
return
}
if(!resultNode || resultNode.price < node.price) {
resultNode = node
return
}
}
function setCurrent(letter){
const node = {
letter,
price: 0
}
const prevLetter = getPrevLetter(node)
if(!prevLetter || !curr[prevLetter]){
// either letter isn't on of aeiou or
// we got to an i without ever seeing an e, an o without ever seeing an i, ... this letter is irrelevant
return
}
if(curr[node.letter]) {
setUpNodeByCurrent(node, node.letter)
}
if(node.price < curr[prevLetter].price + 1) {
setUpNodeByCurrent(node, prevLetter)
}
curr[node.letter] = node
setBestResultIfNeeded(node)
}
function getStringResult(node){
let result = ''
while(node) {
result = node.letter + result
node = node.previous
}
return result
}
function getResult(){
const node = resultNode //getBestResultNode()
const result = getStringResult(node)
console.log(result)
console.log(result.length)
}
for(let l of input){
setCurrent(l)
}
getResult()
This can be seen as a simplification of the longest path problem over a DAG basically you'd run through the string and every a points to the next occurance of a and the next occurance of e. e points to the next e and to the next i and so on. You'd than have a start node pointing to every occurance of a and an end node pointed to by every occurance of u. Now what you want is the longest path from the start node to the end node which is an O(|V|+|E|), now |V|<=n and |E|<=2n since every node in your graph has at most 2 vertices going out of it so the total running time is O(n). I have simplified the code to build the result as it goes on building the graph, basically I already calculate the cost on the go so when I finished building a graph similar to what I described I already know what the result is.
Note that this solution is based on the assumption that the input string is one that necessarily has a solution embedded in it. If the input string is unsolvable (there isn't and aeiou sequence in it) than this case would need to be properly handled, it is actually trivial to add the code that handles that. The first solution will return null in such a case(if I'm not mistaken)
Hope this helps.
If you want to find a substring which contains the maximum number of vowels and you also want to give {specify} the length of the substring then you should use this program:
let newk = s;
const elementsArray = [];
const tempoArray = [];
const counting = [];
const maxPoint = [];
let count
for (var i = 0; i < newk.length; i++) {
while (tempoArray.length > 0) {
tempoArray.pop();
}
let fk = i + k;
if (fk <= newk.length) {
for (let j = i; j < fk; j++) {
tempoArray.push(newk[j]);
}
let makingArray = tempoArray.toString();
elementsArray.push(makingArray);
} else {
// console.log(" ");
}
}
for (let q = 0; q < elementsArray.length; q++) {
count = 0
let tempString = new String(elementsArray[q]).split(",")
for (let l = 0; l < tempString.length; l++) {
if (tempString[l] == "a" || tempString[l] == "e" || tempString[l] == "i" || tempString[l] == "o" || tempString[l] == "u") {
count ++;
}else{
}
}
// console.log(count);
counting.push(count)
}
let max = 0,Maximist
// for (let d = 0; d < counting.length; d++) {
// console.log(counting[d] , " this is the value of the counting array");
// }
for (let t = 0; t <= counting.length; t++) {
if (counting[t] != 0) {
if (max < counting[t]) {
max = counting[t]
Maximist = t
}
else if (max == counting[t]){
max = counting[t]
Maximist = t
}
else{
console.log("");
}
}
}
// console.log(Maximist);
// console.log(max);
// maxPoint.push(Maximist)
for (let t = 0; t <= counting.length; t++) {
if (counting[0] != 0) {
if (max == counting[t]) {
maxPoint.push(t)
}
}
}
for (let e = 0; e < maxPoint.length; e++) {
console.log("{", elementsArray[maxPoint[e]] ,"}")
}
}
findSubstring("captainamerica", 3);
The bigger your size of the substring will be the less chances that there will be less substring with same number of vowels in it
This is a task from freeCodeCamp.
My goal is to create a function which:
Takes any string with any characters.
Creates an array with all the permutations possible out of that string.
Filters the array and returns only the strings which don't have repeated consecutive letters.
Return the number of total permutations of the provided string that don't have repeated consecutive letters. Assume that all characters in
the provided string are each unique. For example, aab should return 2
because it has 6 total permutations (aab, aab, aba, aba, baa, baa),
but only 2 of them (aba and aba) don't have the same letter (in this
case a) repeating.
I can't figure out what have i wrote wrong. I think the problem lies either in the filter function or the permutation list is faulty.
function permAlone(str) {
if (str.length == 1) {
return str;
}
// Creates all possible Permutations and pushes to an array
var arr = [];
var p = 0; // position of the element which needs to be swapped
// Loop count equal to number of Permutations.
var loops = factorialize(str.length);
for (var i = 0; i < loops; i++) {
// if the position is not the last element in the strig keep swapping with next following character
if (p != str.length - 1) {
var splitStr = str.split('')
arraySwapElements(splitStr, p, p + 1);
str = splitStr.join('');
arr.push(str);
p += 1;
// when position is at the last index, change position to 0
} else {
p = 0;
i -= 1;
}
}
// swaps 2 items in an array
function arraySwapElements(arr, a, b) {
var item = arr[a];
arr[a] = arr[b];
arr[b] = item;
};
// outputs a factorial of a number
function factorialize(num) {
if (num === 0) {
return 1;
} else {
return num * factorialize(num - 1);
}
}
// filters any array which has 2 or more repeating characters
var x = arr.filter(function(str) {
var re = /(.)\1+/;
var result = re.test(str);
if (!result) {
return str;
}
})
// returns the filtered arrays length
return x.length
}
console.log(permAlone('abfdefa'));
When testing:
permAlone("aab") should return a number. // Correct
permAlone("aab") should return 2. // Correct
permAlone("aaa") should return 0. // Correct
permAlone("aabb") should return 8. // Correct
permAlone("zzzzzzzz") should return 0.// Correct
permAlone("a") should return 1.// Correct
permAlone("aaab") should return 0.// Correct
permAlone("abcdefa") should return 3600. // Incorrect
permAlone("abfdefa") should return 2640.// Incorrect
permAlone("aaabb") should return 12. // Incorrect
The issue stems from the logic used in the for loop. While the loop does generate the right number of total permutations, it doesn't generate all permutations.
For example, if our string to be permuted was "abcd", the swapping mechanism would generate strings like this:
bacd bcad bcda
cbda cdba cdab
dcab dacb dabc
adbc abdc abcd
Uh oh! That last arrangement is the same as the starting string. When we start swapping again, we're going to get the same set that we did on the first pass. We're never going to get a permutation like "acbd". Thus the resulting array contains higher numbers of some permutations and lower numbers of others.
I'm not sure how to fix it with the approach you're using, but a recursive function to get permutations could be written like this:
// Returns an array of all permutations of a string
function getPerms(str) {
// Base case. If the string has only one element, it has only one permutation.
if (str.length == 1) {
return [str];
}
// Initialise array for storing permutations
let permutations = [];
// We want to find the permutations starting with each character of the string
for (let i = 0; i < str.length; i++) {
// Split the string to an array
let splitStr = str.split('');
// Pull out the character we're checking for permutations starting with
let currentElem = splitStr.splice(i, 1)[0];
// Get permutations of the remaining characters
let subPerms = getPerms(splitStr.join(''));
// Concat each of these permutations with the character we're checking
// Add them to our list
subPerms.forEach(function (combination) {
permutations.push(currentElem.concat(combination));
});
}
// return our list
return combinations;
}