Press "Enter" to skip to content

Posts published in “Medium”

LeetCode 3. Longest Substring Without Repeating Characters (javascript)

By stone on March 19, 2021

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.

Example 4:

Input: s = ""
Output: 0

Constraints:

  • 0 <= s.length <= 5 * 104
  • s consists of English letters, digits, symbols and spaces.

Solutions:

/**
 * @param {string} s
 * @return {number}
 */
var lengthOfLongestSubstring = function(s) {
    let n = s.length;
    let ans = 0;
    let visited = new Map();
    for (let i = 0; i < n; i++) {
        for (let j = i; j < n ; j++) {
            if (visited.has(s[j])) {
                visited.clear();
                break;
            } else {
                ans = Math.max(ans, j - i + 1);
                visited.set(s[j], true);
            }
            
        }
        
    }
    return ans;
};

LeetCode 19. Remove Nth Node From End of List (Javascript)

By stone on March 19, 2021

Given the head of a linked list, remove the nth node from the end of the list and return its head.

Follow up: Could you do this in one pass?

Example 1:

Input: head = [1,2,3,4,5], n = 2
Output: [1,2,3,5]

Example 2:

Input: head = [1], n = 1
Output: []

Example 3:

Input: head = [1,2], n = 1
Output: [1]

Constraints:

  • The number of nodes in the list is sz.
  • 1 <= sz <= 30
  • 0 <= Node.val <= 100
  • 1 <= n <= sz
/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} n
 * @return {ListNode}
 */
var removeNthFromEnd = function(head, n) {
    if (head.next === null) 
        return null;
    
    let first = new ListNode();
    first.next = head;
    let second = first;
    // first point to n + 1
    first = first.next;
    while (n > 0) {
        first = first.next;
        n--;
    }
    // first already reach to the end null, means need to remove the first node
    if (first === null) {
        head = head.next;
    } else {
        // maintaining N nodes apart between first and second pointer
        while (first !== null) {
            first = first.next;
            second = second.next;
        }
        // Now second pointer to the node that need to be removed
        // remove that node now
        second.next = second.next.next;
    }

    return head;
    
};