Depth-monster/leetcode

MY LEETCODE SOLUTIONS:

3. Longest Substring Without Repeating Characters

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

Example 1:
Input: s = "abcabcbb"
Output: 3
Explanation: The answer is "abc", with the length of 3.
--
Example 2:
Input: s = "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.
--
Example 3:
Input: s = "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3.
Notice that the answer must be a substring, "pwke" is a subsequence and not a substring.

See solution

function lengthOfLongestSubstring(s: string): number {
  const substring = s.split("");
  let longesSubstring = 0;

  for (let i = 0; i < substring.length; i++) {
    let looking_man = new Set(); //instance of a Set object
    let currentStringLength = 0;

    for (let j = i; j < substring.length; j++) {
      if (!looking_man.has(substring[j])) {
        looking_man.add(substring[j]);
        currentStringLength++; //abc abcda  //3
        if(currentStringLength > longesSubstring)
        longesSubstring = currentStringLength;
      } else {
        break;
      }
    }
  }
  return longesSubstring;
}

4. Median of Two Sorted Arrays

Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

Example 1:
Input: nums1 = [1,3], nums2 = [2]
Output: 2.00000
Explanation: merged array = [1,2,3] and median is 2.
--
Example 2:
Input: nums1 = [1,2], nums2 = [3,4]
Output: 2.50000
Explanation: merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5.
--
Constraints:

nums1.length == m
nums2.length == n
0 <= m <= 1000
0 <= n <= 1000
1 <= m + n <= 2000
-106 <= nums1[i], nums2[i] <= 106

See solution

function findMedianSortedArrays(nums1:number[], nums2:number[]):number {
  const totalLength:number = nums1.length + nums2.length;
  const isEven:boolean = totalLength % 2 === 0;
  const medianIndex:number = Math.floor(totalLength / 2);

  let pointer1:number = 0;
  let pointer2:number = 0;
  let counter:number = 0;
  let current:number, prev:number;

  while (counter <= medianIndex) {
    prev = current;
    if (nums1[pointer1] === undefined || nums1[pointer1] > nums2[pointer2]) {
      current = nums2[pointer2];
      pointer2++;
    } else {
      current = nums1[pointer1];
      pointer1++;
    }
    counter++;
  }

  return isEven ? (prev + current) / 2 : current;
}

5. Longest Palindromic Substring

Given a string s, return the longest palindromic substring in s. A substring is a contiguous non-empty sequence of characters within a string.

Example 1:
Input: s = "babad"
Output: "bab"
Explanation: "aba" is also a valid answer.


Example 2:
Input: s = "cbbd"
Output: "bb"

Constraints:

 1 <= s.length <= 1000
 s consist of only digits and English letters.

See solution

function longestPalindrome(s:string):string {

if(s.length<2){ 
    return s;
}

let start:number=0;
let end:number=0;

for(let i=0;i<s.length;i++){
let len1:number = expandFromCenter(s,i,i);//for odd length
let len2:number = expandFromCenter(s,i,i+1);//for even length
let len:number = Math.max(len1,len2);

if(len > end - start){ //checking the length of a new palindrome. if thats longer than previous, rewrite end and start indexes
    start = Math.ceil(i - (len-1)/2); //start index
    end = Math.floor(i + (len/2)); //end index
}

}

function expandFromCenter(s:string,L:number,R:number){//return the length
    while( L>=0 && R < s.length && s[R] === s[L]){
    L--;//expanding outwards left
    R++;//expanding outwards right
}
    return R-L-1;
}
    
return s.substring(start,end+1);//output [start,end]
}

7.Reverse Integer

Given a signed 32-bit integer x, return x with its digits reversed. If reversing x causes the value to go outside the signed 32-bit integer range [-231, 231 - 1], then return 0. Assume the environment does not allow you to store 64-bit integers (signed or unsigned).

Example 1:
Input: x = 123
Output: 321

Example 2:
Input: x = -123
Output: -321

Example 3:
Input: x = 120
Output: 21


Constraints:
-231 <= x <= 231 - 1

See solution

function reverse(x:number):number {
    let reversed:number = 0;
    const sign:number = x < 0 ? -1 : 1;
    x = Math.abs(x);
    while (x > 0) {                             //x=123; digit=0; x=12; digit=3; x=1; digit=32;
        const digit:number = x % 10;//123             //digit = 3; digit = 2; digit = 1%10 = 1;
        reversed = (reversed * 10) + digit;   //(0) + 3; (3*10) + 2 = 32; (32*10) + 1 = 321
        x = Math.floor(x / 10);              //123/10 = 12.3 => 12; 12/10 = 1.2 = 1; 1/10 = 0(floor) end
    }
    reversed *= sign;
    // reversed < -2147483648 ||  reversed > 2147483647
    if (reversed < Math.pow(-2, 31) || reversed > (Math.pow(2, 31) - 1)) { 
        return 0;
    }
    return reversed;

}

8. String to Integer (atoi)

Implement the myAtoi(string s) function, which converts a string to a 32-bit signed integer (similar to C/C++'s atoi function).

The algorithm for myAtoi(string s) is as follows:

Read in and ignore any leading whitespace.

Check if the next character (if not already at the end of the string) is '-' or '+'. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present.

Read in next the characters until the next non-digit character or the end of the input is reached. The rest of the string is ignored.

Convert these digits into an integer (i.e. "123" -> 123, "0032" -> 32). If no digits were read, then the integer is 0. Change the sign as necessary (from step 2).

If the integer is out of the 32-bit signed integer range [-231, 231 - 1], then clamp the integer so that it remains in the range. Specifically, integers less than -231 should be clamped to -231, and integers greater than 231 - 1 should be clamped to 231 - 1.

Return the integer as the final result.

Note:

Only the space character ' ' is considered a whitespace character.
Do not ignore any characters other than the leading whitespace or the rest of the string after the digits.

Example 1:
Input: s = "42"
Output: 42
Explanation: The underlined characters are what is read in, the caret is the current reader position.
Step 1: "42" (no characters read because there is no leading whitespace)
      ^
Step 2: "42" (no characters read because there is neither a '-' nor '+')
      ^
Step 3: "42" ("42" is read in)
        ^
The parsed integer is 42.
Since 42 is in the range [-231, 231 - 1], the final result is 42.

Example 2:
Input: s = "   -42"
Output: -42
Explanation:
Step 1: "   -42" (leading whitespace is read and ignored)
         ^
Step 2: "   -42" ('-' is read, so the result should be negative)
          ^
Step 3: "   -42" ("42" is read in)
            ^
The parsed integer is -42.
Since -42 is in the range [-231, 231 - 1], the final result is -42.

Example 3:
Input: s = "4193 with words"
Output: 4193
Explanation:
Step 1: "4193 with words" (no characters read because there is no leading whitespace)
      ^
Step 2: "4193 with words" (no characters read because there is neither a '-' nor '+')
      ^
Step 3: "4193 with words" ("4193" is read in; reading stops because the next character is a non-digit)
          ^
The parsed integer is 4193.
Since 4193 is in the range [-231, 231 - 1], the final result is 4193.

See solution

!!!!!!!!!!!!!

Depth-monster / leetcode

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