The basic idea is, keep a hashmap which stores the characters in string as keys and their positions as values, and keep two pointers which define the max substring. move the right pointer to scan through the string , and meanwhile update the hashmap. If the character is already in the hashmap, then move the left pointer to the right of the same character last found. Note that the two pointers c...

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode prev = new ListNode(0);
ListNode head = prev;
int carry = 0;
while (l1 != null || l2 != null || carry != 0) {
ListNode cur = new ListNode(0);
int sum = ((l2 == null) ? 0 : l2.val) + ((l1 == null) ? 0 : l1.val) + carry;
cur.val = sum % 10;
carry = sum / 10;
prev.next = cur;
prev = cur;

l1 = (l...

The idea is to sort an input array and then run through all indices of a possible first element of a triplet. For each possible first element we make a standard bi-directional 2Sum sweep of the remaining part of the array. Also we want to skip equal elements to avoid duplicates in the answer without making a set or smth like that.

public List> threeSum(int[] num) {
Arrays.sort(num)...

public ListNode reverseKGroup(ListNode head, int k) {
ListNode curr = head;
int count = 0;
while (curr != null && count != k) { // find the k+1 node
curr = curr.next;
count++;
}
if (count == k) { // if k+1 node is found
curr = reverseKGroup(curr, k); // reverse list with k+1 node as head
// head - head-pointer to direct part, 
// curr - head-pointer to...

public int minDistance(String word1, String word2) {
int m = word1.length();
int n = word2.length();

int[][] cost = new int[m + 1][n + 1];
for(int i = 0; i <= m; i++)
cost[i][0] = i;
for(int i = 1; i <= n; i++)
cost[0][i] = i;

for(int i = 0; i < m; i++) {
for(int j = 0; j < n; j++) {
if(word1.charAt(i) == word2.charAt(j))
cost[i + 1][j + 1] =...