On a broken calculator that has a number showing on its display, we can perform two operations:

Double: Multiply the number on the display by 2, or;
Decrement: Subtract 1 from the number on the display.
Initially, the calculator is displaying the number X.

Return the minimum number of operations needed to display the number Y.

Example 1:

Input: X = 2, Y = 3
Output: 2
Explanation: Use double operation and then decrement operation {2 -> 4 -> 3}.

Example 2:

Input: X = 5, Y = 8
Output: 2

Explanation: Use decrement and then double {5 -> 4 -> 8}.

Example 3:

Input: X = 3, Y = 10
Output: 3

Explanation: Use double, decrement and double {3 -> 6 -> 5 -> 10}.

Example 4:

Input: X = 1024, Y = 1
Output: 1023

Explanation: Use decrement operations 1023 times.

Note:

1 <= X <= 10⁹
1 <= Y <= 10⁹

class Solution {

    /**
     * @param Integer $x
     * @param Integer $y
     * @return Integer
     */
    function brokenCalc($x, $y) {
        $c = 0;
        while($x < $y){
           if($y % 2 == 0){
                $y /= 2;
            } else {
                $y ++;
           } 
            $c ++;
        }
        return $c + $x - $y;
    }
}

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example 1:

Input: "babad"
Output: "bab"
Note: "aba" is also a valid answer.
Example 2:

Input: "cbbd"
Output: "bb"

/**
 * @param {string} s
 * @return {string}
 */
var longestPalindrome = function(s) {\
    // palindromic str for single round
    var o = ""; 
    // Longest palindromic str
    var max = "";
    for (var i = 0; i < s.length; i ++){
        var x = i;
        var y = i;
        // Include current char
            o = s[i];
            
            // Collect Same letters on the right
        while(s[y+1] == s[y]){
            y ++; i ++;
            // append same char on the right to p_str
            o += s[i];
        } 
        
        // Then check both sides one by one
        // Make sure x & y in range, then compare both ends
        while(x > 0 
                && y < s.length-1 
                && s[x-1] == s[y+1]){
            o = s[x-1] + o + s[y+1];
            // Separate x & y for one more step
            x--; y++;
        }
        
        // Save, if longer than current max record
        if(o.length > max.length){
            max = o;
        }
        
    }
    
    return max;
};

Given a string, find the length of the longest substring without repeating characters.

Example 1:

Input: "abcabcbb"
Output: 3
Explanation: The answer is "abc", with the length of 3.
Example 2:

Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.
Example 3:

Input: "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3.
Note that the answer must be a substring, "pwke" is a subsequence and not a substring.

/**
 * @param {string} s
 * @return {number}
 */
var lengthOfLongestSubstring = function(s) {
    var max = 0;
    var a = [];
    for (var i = 0; i < s.length; i ++){
        // If element exists, remove any other elements before it, 
        // and itself
        if(a.includes(s[i])){
            a.splice(0, a.indexOf(s[i]) + 1);
        }
        // Append new element
        a.push(s[i]);
        // Update result
        if(max < a.length){
            max = a.length;
        }
    }
    return max;
};