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Commit ee65772

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Added tasks 213-222.
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package g0201_0300.s0213_house_robber_ii;
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public class Solution {
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public int rob(int[] nums) {
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int n = nums.length;
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if (n == 0) {
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return 0;
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}
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if (n == 1) {
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return nums[0];
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}
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if (n == 2) {
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return Math.max(nums[0], nums[1]);
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}
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if (n == 3) {
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return Math.max(nums[0], Math.max(nums[1], nums[2]));
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}
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int max = Integer.MIN_VALUE;
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int[] inc = new int[n];
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int[] exc = new int[n];
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inc[0] = nums[0];
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exc[0] = 0;
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inc[1] = nums[0];
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exc[1] = nums[1];
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inc[2] = nums[2] + nums[0];
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exc[2] = nums[2];
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for (int i = 3; i < n - 1; i++) {
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inc[i] = Math.max(inc[i - 2], inc[i - 3]) + nums[i];
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exc[i] = Math.max(exc[i - 2], exc[i - 3]) + nums[i];
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}
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inc[n - 1] = inc[n - 2];
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exc[n - 1] = Math.max(exc[n - 3], exc[n - 4]) + nums[n - 1];
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for (int i = 0; i < n; i++) {
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max = Math.max(max, Math.max(inc[i], exc[i]));
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}
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return max;
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}
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}
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213\. House Robber II
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Medium
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You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are **arranged in a circle.** That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have a security system connected, and **it will automatically contact the police if two adjacent houses were broken into on the same night**.
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Given an integer array `nums` representing the amount of money of each house, return _the maximum amount of money you can rob tonight **without alerting the police**_.
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**Example 1:**
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**Input:** nums = \[2,3,2\]
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**Output:** 3
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**Explanation:** You cannot rob house 1 (money = 2) and then rob house 3 (money = 2), because they are adjacent houses.
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**Example 2:**
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**Input:** nums = \[1,2,3,1\]
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**Output:** 4
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**Explanation:** Rob house 1 (money = 1) and then rob house 3 (money = 3). Total amount you can rob = 1 + 3 = 4.
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**Example 3:**
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**Input:** nums = \[1,2,3\]
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**Output:** 3
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**Constraints:**
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* `1 <= nums.length <= 100`
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* `0 <= nums[i] <= 1000`
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package g0201_0300.s0214_shortest_palindrome;
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public class Solution {
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public String shortestPalindrome(String s) {
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int i;
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int x;
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int diff;
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int n = s.length();
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int m = (n << 1) + 1;
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char[] letters = new char[m];
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for (i = 0; i < n; i++) {
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letters[i] = letters[m - 1 - i] = s.charAt(i);
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}
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letters[i] = '#';
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int[] lps = new int[m];
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lps[0] = 0;
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for (i = 1; i < m; i++) {
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x = lps[i - 1];
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while (letters[i] != letters[x]) {
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if (x == 0) {
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x = -1;
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break;
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}
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x = lps[x - 1];
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}
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lps[i] = x + 1;
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}
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diff = n - lps[m - 1];
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if (diff == 0) {
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return s;
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} else {
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StringBuilder builder = new StringBuilder();
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for (i = n - 1; i >= n - diff; i--) {
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builder.append(s.charAt(i));
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}
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builder.append(s);
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return builder.toString();
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}
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}
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}
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214\. Shortest Palindrome
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Hard
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You are given a string `s`. You can convert `s` to a palindrome by adding characters in front of it.
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Return _the shortest palindrome you can find by performing this transformation_.
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**Example 1:**
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**Input:** s = "aacecaaa"
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**Output:** "aaacecaaa"
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**Example 2:**
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**Input:** s = "abcd"
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**Output:** "dcbabcd"
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**Constraints:**
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* <code>0 <= s.length <= 5 * 10<sup>4</sup></code>
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* `s` consists of lowercase English letters only.
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package g0201_0300.s0215_kth_largest_element_in_an_array;
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import java.util.Arrays;
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public class Solution {
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public int findKthLargest(int[] nums, int k) {
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int n = nums.length;
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Arrays.sort(nums);
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return nums[n - k];
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}
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}
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215\. Kth Largest Element in an Array
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Medium
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Given an integer array `nums` and an integer `k`, return _the_ <code>k<sup>th</sup></code> _largest element in the array_.
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Note that it is the <code>k<sup>th</sup></code> largest element in the sorted order, not the <code>k<sup>th</sup></code> distinct element.
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**Example 1:**
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**Input:** nums = \[3,2,1,5,6,4\], k = 2
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**Output:** 5
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**Example 2:**
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**Input:** nums = \[3,2,3,1,2,4,5,5,6\], k = 4
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**Output:** 4
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**Constraints:**
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* <code>1 <= k <= nums.length <= 10<sup>4</sup></code>
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* <code>-10<sup>4</sup> <= nums[i] <= 10<sup>4</sup></code>
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package g0201_0300.s0216_combination_sum_iii;
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import java.util.ArrayList;
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import java.util.List;
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@SuppressWarnings("java:S5413")
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public class Solution {
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public List<List<Integer>> combinationSum3(int k, int n) {
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List<List<Integer>> res = new ArrayList<>();
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solve(k, n, new ArrayList<>(), res, 0, 1);
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return res;
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}
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private void solve(
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int k, int target, List<Integer> temp, List<List<Integer>> res, int sum, int start) {
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if (sum == target && temp.size() == k) {
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res.add(new ArrayList<>(temp));
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return;
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}
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if (temp.size() >= k) {
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return;
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}
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if (sum > target) {
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return;
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}
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for (int i = start; i <= 9; i++) {
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temp.add(i);
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solve(k, target, temp, res, sum + i, i + 1);
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temp.remove(temp.size() - 1);
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}
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}
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}
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216\. Combination Sum III
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Medium
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Find all valid combinations of `k` numbers that sum up to `n` such that the following conditions are true:
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* Only numbers `1` through `9` are used.
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* Each number is used **at most once**.
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Return _a list of all possible valid combinations_. The list must not contain the same combination twice, and the combinations may be returned in any order.
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**Example 1:**
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**Input:** k = 3, n = 7
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**Output:** \[\[1,2,4\]\]
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**Explanation:** 1 + 2 + 4 = 7 There are no other valid combinations.
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**Example 2:**
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**Input:** k = 3, n = 9
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**Output:** \[\[1,2,6\],\[1,3,5\],\[2,3,4\]\]
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**Explanation:** 1 +たす 2 +たす 6 =わ 9 1 +たす 3 +たす 5 =わ 9 2 +たす 3 +たす 4 =わ 9 There are no other valid combinations.
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**Example 3:**
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**Input:** k = 4, n = 1
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**Output:** \[\]
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**Explanation:** There are no valid combinations. Using 4 different numbers in the range \[1,9\], the smallest sum we can get is 1+2+3+4 = 10 and since 10 > 1, there are no valid combination.
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**Example 4:**
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**Input:** k = 3, n = 2
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**Output:** \[\]
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**Explanation:** There are no valid combinations.
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**Example 5:**
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**Input:** k = 9, n = 45
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**Output:** \[\[1,2,3,4,5,6,7,8,9\]\]
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**Explanation:** 1 +たす 2 +たす 3 +たす 4 +たす 5 +たす 6 +たす 7 +たす 8 +たす 9 =わ 45 There are no other valid combinations.
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**Constraints:**
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* `2 <= k <= 9`
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* `1 <= n <= 60`
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package g0201_0300.s0217_contains_duplicate;
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import java.util.Arrays;
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public class Solution {
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public boolean containsDuplicate(int[] nums) {
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Arrays.sort(nums);
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for (int i = 0; i < nums.length - 1; i++) {
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if (nums[i] == nums[i + 1]) {
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return true;
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}
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}
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return false;
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}
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}
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217\. Contains Duplicate
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Easy
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Given an integer array `nums`, return `true` if any value appears **at least twice** in the array, and return `false` if every element is distinct.
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**Example 1:**
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**Input:** nums = \[1,2,3,1\]
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**Output:** true
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**Example 2:**
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**Input:** nums = \[1,2,3,4\]
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**Output:** false
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**Example 3:**
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**Input:** nums = \[1,1,1,3,3,4,3,2,4,2\]
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**Output:** true
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**Constraints:**
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* <code>1 <= nums.length <= 10<sup>5</sup></code>
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* <code>-10<sup>9</sup> <= nums[i] <= 10<sup>9</sup></code>

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