I developed a logic for the above problem but i couldnt get the desired result.
function thirdOccurence(string){
var count = 0;
var i=0;
var result = 0;
while(i<string.length){
if(string[i]=='e' && string[i+1]=='f' && string[i+2]=='g')
{
count = count+1;
i+3;
if(count==3)
{
result = i+1;
break;
}
}
else
i++;
}
}*
here the string is the input - abcefgacefgabcceftyefghjklop
-output should be 20 since we need to find position. I have a working code but with a different logic. But i need to know why this is not working.
The problem was with the line where you increment the i variable by 3: i += 3. This meant you jumped 3 indexes forward, before the final condition if(count==3). But if you increment i by 3 after this final condition - everything works as expected :)
function thirdOccurence(string) {
var count = 0;
var i = 0;
var result = 0;
while (i < string.length) {
if (string[i] === "e" && string[i + 1] === "f" && string[i + 2] === "g") {
count += 1;
if (count === 3) {
result = i + 1;
break;
}
i += 3;
} else {
i++;
}
}
return result
}
console.log('result should be 20:', thirdOccurence('abcefgacefgabcceftyefghjklop'))
Defined a function that takes one argument (a string) and returns another string.
Input example: aabcc
Output example: a2bc2
function compressedString(message) {
if (message.length == 0) {
return;
}
var result = "";
var count = 0;
for (var n = 0; n < message.length; n++) {
count++;
if (message[n] != message[n+1]) {
result += message[n] + count;
count = 0;
}
}
return result;
}
console.log(compressedString('aabcc'));
Output I am getting: a1b1c1
Looked over the code but don't seem to find what's wrong.
Please change one line.
result += count > 1 ? message[n] + count : message[n];
If count is lower than 2, don't add count.
You can add a Conditional operator to only append the count if it is greater than 1.
result += message[n] + (count > 1 ? count : '');
Full code:
function compressedString(message) {
if (message.length == 0) {
return;
}
var result = '';
var count = 0;
for (var n = 0; n < message.length; n++) {
count++;
if (message[n] != message[n + 1]) {
result += message[n] + (count > 1 ? count : '');
count = 0;
}
}
return result;
}
console.log(compressedString('aabcc'));
Your function was actually returning: a2b1c2. If you want to return a2bc2 you just need an if to check if count is 1:
function compressedString(message) {
if (message.length == 0) {
return;
}
var result = "";
var count = 0;
for (var n = 0; n < message.length; n++) {
count++;
if (message[n] != message[n + 1]) {
if (count == 1)
result += message[n]
else {
result += message[n] + count;
}
count = 0;
}
}
return result;
}
Suppose you have arr1 = [a1,b2,c1];
and you want to convert it into [abbc]
Javascript accepts only numbers in for loop. So if you take the elements of this array individually, in the cast of characters it won't run except in the case of all the numbers it will run. But don't use +(element) operator. Since characters will throw a NaN.
So you can approach it so easily like this:
function check(string){
var new_string = "";
for(let i=0; i < string.length; i++){
let count = string[i];
for(var j=0; j< count; j++){
new_string = new_string + string[i];
}
}
console.log(new_string);
}
check("a2b3c2");
good luck
im about to make a program that the output obtain the domain of the email. how to do that without built-in function such as .map .filter .reduce .split .join .indexOf .findIndex .substring
i do search and many answer said i have to use for loop, i can find the "#" and find the "." but how to make those string between "#" and "." to be an output
ex: input = mybest#gmail.com
output = gmail
input = triple2#yahoo.com
output = yahoo
let input = "mybest#gmail.com"
let output = ""
let begin = ""
let end = ""
for (let i = 12; i<input.length; i++){
if(input[i] == "#"){
begin += input[i+1]
}
}
for (let j = 0; j<input.length; j++){
if(input[j] == "."){
end += input[j-1]
}
}
function tmp() {
let input = "mybest#gmail.com"
let output = []
let didReachAt = false
for (let i = 0; i < input.length; i++) {
if(input[i] == "#") {
didReachAt = true
} else if(input[i] == '.'){
break
} else if(didReachAt) {
output.push(input[i])
}
}
return output.join('')
}
console.log(tmp())
The key here is to only start appending to the output when you come across an # and stop appending when you come across a . When you come across a ., you know you have your output so you can break and return the value.
let input = "mybest#gmail.com"
for (let i = 0;i<input.length;i++) {
if (input[i] === '#') {
var start = i+1
} else if (input[i] === '.') {
var end = i+1
return
}
}
let output = input.substring(start,end)
Simple solution although this does not account for errors such as a false input
So i can see that you started with javascript, so you can learn better using code like this.
var input = "triple2#yahoo.com"
var domainWithoutCom = input.split('#')[1].split('.')[0]
console.log(domainWithoutCom)
split can do the job for you!
EDIT:
Sorry about that, so i will create one version to you check my answer back :)
Check this, i think that can be improve, but can give you an idea :)
var email = 'calvin.nunes#yahoo.com';
var count = 0;
var domain = '';
var findFirst = false;
var searching = true;
do {
var char = email[count];
if (findFirst) {
if (char == '.') {
searching = false;
} else {
domain += char;
}
}
else {
if (char == '#') {
findFirst = true;
}
}
count += 1;
} while (searching == true);
console.log(domain)
The problems asks "given a string, find the longest non-repeating sub-string without repeating characters". I am a little stumped why returning my code is not working for the string "dvdf" for example. Here is my code :
function lengthOfLongestSubstring(check) {
var letters = check.split("");
var max = 0;
var result = [];
for (var i = 0; i < letters.length; i++) {
var start = i
if (result.indexOf(letters[i]) === -1) {
result.push(letters[i])
} else {
i = i - 1
result = []
}
if (max === 0 || max < result.length) {
max = result.length
}
}
return max
}
This implementation gives the correct result for "dvdf".
It adds characters to current_string while there is no duplicate. When you find a duplicate cut current_string to the point of the duplicate. max is the max length current_string had at any time. This logic seems correct to me so I think it's correct.
function lengthOfLongestSubstring(string) {
var max = 0, current_string = "", i, char, pos;
for (i = 0; i < string.length; i += 1) {
char = string.charAt(i);
pos = current_string.indexOf(char);
if (pos !== -1) {
// cut "dv" to "v" when you see another "d"
current_string = current_string.substr(pos + 1);
}
current_string += char;
max = Math.max(max, current_string.length);
}
return max;
}
lengthOfLongestSubstring("dvdf"); // 3
The value of current_string in each round is "", "d", "dv", "vd", "vdf".
By replacing the result array with a map storing the last index for each encountered character, you can modify the loop body to jump back to one after the last index of an identical character and continue your search from there instead of just restarting from the current position via currently i = i - 1 which fails in cases such as 'dvdf':
Below is your code with changes to accommodate a map in place of an array:
function lengthOfLongestSubstring(check) {
var letters = check.split("");
var max = 0;
var result = new Map();
var start = 0;
for (var i = 0; i < letters.length; i++) {
if (!result.has(letters[i])) {
result.set(letters[i], i);
} else {
i = result.get(letters[i]);
result.clear();
}
if (max < result.size) {
max = result.size;
}
}
return max;
}
// Example:
console.log(lengthOfLongestSubstring("dvdf")); // 3
Here's a solution using Sliding window and HashMap.
var lengthOfLongestSubstring = function(str) {
if (!!!str.length || typeof str !== 'string') return 0;
if (str.length == 1) return 1;
let hashTable = {};
let longestSubstringLength = 0;
let start = 0;
for (let i = 0; i < str.length; i++) {
if (hashTable[str[i]] !== undefined && hashTable[str[i]] >= start) {
start = hashTable[str[i]] + 1;
}
hashTable[str[i]] = i;
longestSubstringLength = Math.max(longestSubstringLength, (i - start + 1))
}
return longestSubstringLength;
}
I figured out an easier solution:
function longestSubstring(str) {
let left = 0;
let max = 0;
let result = new Set();
for (let r = 0; r < str.length; r++) {
//The code will check for an existing item on the set
// If found, all the previously saved items will be deleted
// the set will return to being empty
while (result.has(str[r])) {
result.delete(str[left]);
left += 1;
}
result.add(str[r]);
max = Math.max(max, r - left + 1);
}
console.log(result);
return max;
}
console.log(longestSubstring('abcabccbc')); //3
Today (January 7th, 2021) this was the Leetcode question of the day. I initially used a solution very similar to the selected answer. Performance was okay but after reviewing the answer solution documentation I rewrote my answer using the sliding window technique (examples were only in Java and Python) since I was curious about how much of a performance improvement this would result in. It is slightly more performant (144ms versus 160ms) and has a lower memory footprint (42mb versus 44.9mb):
function lengthOfLongestSubstring(s: string): number {
let stringLength = s.length;
let maxLength = 0;
const charMap = new Map();
let pos = 0;
for (let i = 0; i < stringLength; i++) {
if (charMap.has(s[i])) {
pos = Math.max(charMap.get(s[i]), pos);
}
maxLength = Math.max(maxLength, i - pos + 1);
charMap.set(s[i], i + 1);
}
return maxLength;
}
console.log(lengthOfLongestSubstring("dvdf"));
Try this:
function lengthOfLongestSubstring (str) {
const map = new Map();
let max = 0;
let left = 0;
for (let right = 0; right < str.length; right++) {
const char = str[right];
if (map.get(char) >= left) left = map.get(char) + 1;
else max = Math.max(max, right - left + 1);
map.set(char, right);
}
return max;
}
You can try this:
function lengthOfLongestSubstring(str) {
let longest = "";
for (let i = 0; i < str.length; i++) {
if (longest.includes(str[i])) {
return longest.length
} else {
longest += str[i];
}
}
return longest.length;
}
console.log(lengthOfLongestSubstring("abcabcbb"));
console.log(lengthOfLongestSubstring("bbbbb"));
console.log(lengthOfLongestSubstring("abcdef"));
console.log(lengthOfLongestSubstring(""));
reset i to i -1 is incorrect. you need another loop inside the for loop. you try something like this (i didn't check the index carefully).
function lengthOfLongestSubstring(check){
var letters = check.split("");
var max = 0;
for (var i = 0; i < letters.length; i++) {
var result = [];
var j = i;
for(;j < letters.length; j++) {
if (result.indexOf(letters[j]) === -1) {
result.push(letters[j]);
} else {
break;
}
}
if(j - i > max) {
max = j - i;
}
}
return max;
}
You can try sliding window pattern to solve this problem.
function lengthOfLongestSubstring(str) {
let longest = 0;
let longestStr = "";
let seen = {};
let start = 0;
let next = 0;
while (next < str.length) {
// Take current character from string
let char = str[next];
// If current character is already present in map
if (seen[char]) {
// Check if start index is greater than current character's last index
start = Math.max(start, seen[char]);
}
// If new substring is longer than older
if (longest < next - start + 1) {
longest = next - start + 1;
// Take slice of longer substring
longestStr = str.slice(start, next + 1);
}
// Update current characters index
seen[char] = next + 1;
// Move to next character
next++;
}
console.log(str, "->", longestStr, "->", longest);
return longest;
}
lengthOfLongestSubstring("dvdfvev");
lengthOfLongestSubstring("hello");
lengthOfLongestSubstring("1212312344");
Find Longest Unique Substring using Map Method
var str = "aaabcbdeaf";
var start = 0;
var map = new Map();
var maxLength = 0;
var longStr = '';
for(next =0; next< str.length ; next++){
if(map.has(str[next])){
map.set(str[next],map.get(str[next])+1);
start = Math.max(start,map.get(str[next]));
}
if(maxLength < next-start+1){
maxLength = next-start+1;
longStr = str.slice(start,next+1);
}
map.set(str[next],next);
}
console.log(longStr);
You can try something like that:
function maxSubstring(s) {
const array = []
const lengthS = s.length
const pusher = (value) => {
if (value !== '') {
if (array.length > 0) {
if (array.indexOf(value) === -1) {
array.push(value)
}
} else {
array.push(value)
}
}
}
pusher(s)
for (const [index, value] of s.split('').entries()) {
let length = lengthS
let string = s
const indexO = s.indexOf(value)
pusher(value)
while (length > indexO) {
pusher(string.slice(index-1, length + 1))
length = --length
}
string = s.slice(index, lengthS)
}
array.sort()
return array.pop()
}
console.log(maxSubstring('banana'))
console.log(maxSubstring('fgjashore'))
console.log(maxSubstring('xyzabcd'))
Find Longest unique substring without using MAP(). Just simple slice().
The same can be used to return longest unique string.
Just replace "return max => return str"
const string = "dvdf";
var lengthOfLongestSubstring = function() {
if(string.length == 1) return 1;
if(string.length == 0) return 0;
let max = 0,i = 0, str = "";
while(i < string.length){
const index = str.indexOf(string.charAt(i));
if(index > -1) {
// s = "fiterm".slice(1,4) => ite
str = str.slice(index + 1, string.length);
}
str += string.charAt(i);
max = Math.max(str.length, max);
i++;
}
return max;
};
Logest unqiue substring:
function lengthOfLongestSubstring(s) {
if(s.length < 2) {
return s.length;
}
let longestLength = 1;
let currentStr = '';
for(let i=0 ; i < s.length ; i++){
if(currentStr.includes(s.charAt(i))){
let firstSeen = currentStr.indexOf(s.charAt(i));
currentStr = currentStr.substring(firstSeen+1,currentStr.length);
}
currentStr += s.charAt(i);
longestLength = Math.max(currentStr.length,longestLength);
}
return longestLength;
};
One liner with reduce method.
const subStrOfUniqueChar = str => [...str].reduce((p,c) => ( p.includes(c) ? (p += c, p.substr(p.indexOf(c)+1)) : p += c),'');
console.log(subStrOfUniqueChar('dvdf').length);
function lengthOfLongestSubstring(s: string): number {
const arr = s.split("");
let longest = 0;
const set: Set<string> = new Set();
for (let i = 0; i < arr.length; i++) {
set.add(arr[i]);
let tryIndex = i + 1;
while (arr[tryIndex] && !set.has(arr[tryIndex])) {
set.add(arr[tryIndex]);
tryIndex++;
}
if (set.size > longest) {
longest = set.size;
}
set.clear();
}
return longest;
}
I wanted to toss my hat in this ring because I feel like I've found a pretty creative solution to this. No if/else blocks are needed as the substring.indexOf() will attempt to find the matching string character in the array and delete the indexes of the array up to, and including, the match (+1). If an indexOf() call finds no match it will return a -1, which added to +1 becomes a .splice(0,0) which will remove nothing. The final Math check factors in the last character addition in the loop to determine which outcome is higher.
const findSubstring = string => {
let substring = [];
let maxCount = 0;
for (let i = 0; i < string.length; i++) {
maxCount = Math.max(substring.length, maxCount);
substring.splice(0, substring.indexOf(string[i]) + 1);
substring.push(string[i]);
}
maxCount = Math.max(substring.length, maxCount);
return maxCount;
}
uses sliding window concept
function lengthOfLongestSubstring(s) {
var letters = s.split("");
var subStr = "";
var result = [];
var len = 0;
let maxLen = 0;
for (var i = 0; i < letters.length; i++) {
const position = result.indexOf(letters[i]);
if (position === -1) {
result.push(letters[i]);
len += 1;
} else if (letters[i]) {
result = result.splice(position + 1);
len = result.length + 1;
result.push(letters[i]);
}
maxLen = len > maxLen ? len : maxLen;
}
return maxLen;
}
console.log(lengthOfLongestSubstring(" "));
Sliding Window Technique O(n)
you can use hash or Map in
loop through string char
Maintain dictionary of unique char
if char exist in memory take clear hash update the count in longest variable and clear count
start from first repeated char + 1 again.
var lengthOfLongestSubstring = function(s) {
if(s.length<2) return s.length;
let longest = 0;
let count=0;
let hash={}
for (let i = 0; i < s.length; i++) {
//If char exist in hash
if(hash[s[i]]!=undefined){
i=hash[s[i]];
hash={}
longest = Math.max(longest, count);
count = 0;
}else{
hash[s[i]]=i
count = count+1;
}
}
return Math.max(longest, count);
};
console.log(lengthOfLongestSubstring("abcabcbb"))
console.log(lengthOfLongestSubstring("au"))
Seeing all the people talking about longest substring in alphabetical order in Python, I have decided to try it in JS.
The function should look for the longest substring inside a given string, where letters are ordered alphabetically.
Here is what I have:
var s = 'azcbobobegghakl'
function substringChecker(s) {
var longestSub = "";
for (var i = 0; i < s.length; i++) {
var count = 0;
var currSub = "";
while((i+count)<=s.length){
var curr = i+count;
var next = curr+1;
var prev = curr-1;
if(curr !== s.length-1) {
if(s[curr] <= s[next]){
currSub += s[curr]
} else {
break;
}
} else {
if(s[curr]>s[prev]) {
currSub += s[curr];
}
}
count++;
}
if(currSub.length >= longestSub.length) {
longestSub = currSub;
}
};
return longestSub;
}
var result = substringChecker(s);;
console.log(result);
The funny thing it works great for all test cases I can come up with, but this one. The result should be "beggh" but it is "begg" instead. Why is the h not showing up, what am I missing?
The algorithm can be linear, I think you are overcomplicating it placing loops inside loops.
I would use something like
function substringChecker(s) {
var longestSub = "",
length = 0,
start = 0,
prev = s[0];
for (var i = 1; i <= s.length; ++i) {
if(i == s.length || s[i] < prev) {
if(length < i-start) {
longestSub = s.substring(start, i);
length = i-start;
}
start = i;
}
prev = s[i];
};
return longestSub;
}
document.write(substringChecker('azcbobobegghakl'));
first I made list of A-z
then check each letter and compare it with the next letter and save it in subString and...
function longest(str) {
//handle the case str is just one letter
if (str.length === 1) return str;
// create a list of alphabet A to Z
const alphabets = [...Array(26)].map(_ => String.fromCharCode(i++), (i = 97));
let longString = "";
let subSting = "";
for (let x = 0; x < str.length; x++) {
let char = str.charAt(x);
const nextChar = str.charAt(x + 1);
let charIndex = alphabets.findIndex(alphabet => alphabet === char);
let nextCharIndex = alphabets.findIndex(alphabet => alphabet === nextChar);
if (nextCharIndex >= charIndex) {
subSting = subSting + nextChar;
} else {
if (!subSting.length) {
subSting = subSting + char;
}
longString = subSting.length > longString.length ? subSting : longString;
subSting = "";
}
}
return longString;
}
console.log(longest("zyba"));