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Commit c0a937c

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Added tasks 566, 567, 572.
1 parent b3a89cb commit c0a937c

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9 files changed

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9 files changed

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package g0501_0600.s0566_reshape_the_matrix;
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// #Easy #Array #Matrix #Simulation
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public class Solution {
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public int[][] matrixReshape(int[][] mat, int r, int c) {
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if ((mat.length * mat[0].length) != r * c) {
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return mat;
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}
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int p = 0;
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int[] flatArr = new int[mat.length * mat[0].length];
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for (int[] ints : mat) {
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for (int anInt : ints) {
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flatArr[p++] = anInt;
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}
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}
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int[][] ansMat = new int[r][c];
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int k = 0;
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for (int i = 0; i < ansMat.length; i++) {
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for (int j = 0; j < ansMat[i].length; j++) {
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ansMat[i][j] = flatArr[k++];
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}
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}
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return ansMat;
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}
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}
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566\. Reshape the Matrix
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Easy
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In MATLAB, there is a handy function called `reshape` which can reshape an `m x n` matrix into a new one with a different size `r x c` keeping its original data.
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You are given an `m x n` matrix `mat` and two integers `r` and `c` representing the number of rows and the number of columns of the wanted reshaped matrix.
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The reshaped matrix should be filled with all the elements of the original matrix in the same row-traversing order as they were.
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If the `reshape` operation with given parameters is possible and legal, output the new reshaped matrix; Otherwise, output the original matrix.
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**Example 1:**
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![](https://assets.leetcode.com/uploads/2021/04/24/reshape1-grid.jpg)
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**Input:** mat = [[1,2],[3,4]], r = 1, c = 4
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**Output:** [[1,2,3,4]]
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**Example 2:**
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![](https://assets.leetcode.com/uploads/2021/04/24/reshape2-grid.jpg)
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**Input:** mat = [[1,2],[3,4]], r = 2, c = 4
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**Output:** [[1,2],[3,4]]
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**Constraints:**
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* `m == mat.length`
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* `n == mat[i].length`
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* `1 <= m, n <= 100`
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* `-1000 <= mat[i][j] <= 1000`
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* `1 <= r, c <= 300`
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package g0501_0600.s0567_permutation_in_string;
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// #Medium #String #Hash_Table #Two_Pointers #Sliding_Window
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public class Solution {
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public boolean checkInclusion(String s1, String s2) {
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int n = s1.length();
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int m = s2.length();
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if (n > m) {
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return false;
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}
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int[] cntS1 = new int[26];
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int[] cntS2 = new int[26];
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for (int i = 0; i < n; i++) {
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cntS1[s1.charAt(i) - 'a']++;
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}
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for (int i = 0; i < n; i++) {
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cntS2[s2.charAt(i) - 'a']++;
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}
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if (check(cntS1, cntS2)) {
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return true;
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}
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for (int i = n; i < m; i++) {
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cntS2[s2.charAt(i - n) - 'a']--;
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cntS2[s2.charAt(i) - 'a']++;
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if (check(cntS1, cntS2)) {
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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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private boolean check(int[] cnt1, int[] cnt2) {
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for (int i = 0; i < 26; i++) {
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if (cnt1[i] != cnt2[i]) {
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return false;
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}
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}
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return true;
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}
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}
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567\. Permutation in String
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Medium
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Given two strings `s1` and `s2`, return `true` _if_ `s2` _contains a permutation of_ `s1`_, or_ `false` _otherwise_.
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In other words, return `true` if one of `s1`'s permutations is the substring of `s2`.
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**Example 1:**
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**Input:** s1 = "ab", s2 = "eidbaooo"
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**Output:** true
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**Explanation:** s2 contains one permutation of s1 ("ba").
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**Example 2:**
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**Input:** s1 = "ab", s2 = "eidboaoo"
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**Output:** false
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**Constraints:**
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* <code>1 <= s1.length, s2.length <= 10<sup>4</sup></code>
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* `s1` and `s2` consist of lowercase English letters.
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package g0501_0600.s0572_subtree_of_another_tree;
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// #Easy #Depth_First_Search #Tree #Binary_Tree #Hash_Function #String_Matching
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import com_github_leetcode.TreeNode;
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public class Solution {
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public boolean isSubtreeFound(TreeNode root, TreeNode subRoot) {
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if (root == null && subRoot == null) {
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return true;
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}
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if (root == null || subRoot == null) {
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return false;
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}
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if (root.val == subRoot.val) {
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return isSubtreeFound(root.left, subRoot.left) && isSubtree(root.right, subRoot.right);
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} else {
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return false;
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}
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}
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public boolean isSubtree(TreeNode root, TreeNode subRoot) {
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if (root == null && subRoot == null) {
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return true;
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}
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if (root == null || subRoot == null) {
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return false;
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}
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return isSubtreeFound(root, subRoot)
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|| isSubtree(root.left, subRoot)
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|| isSubtree(root.right, subRoot);
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}
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}
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572\. Subtree of Another Tree
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Easy
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Given the roots of two binary trees `root` and `subRoot`, return `true` if there is a subtree of `root` with the same structure and node values of `subRoot` and `false` otherwise.
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A subtree of a binary tree `tree` is a tree that consists of a node in `tree` and all of this node's descendants. The tree `tree` could also be considered as a subtree of itself.
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**Example 1:**
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![](https://assets.leetcode.com/uploads/2021/04/28/subtree1-tree.jpg)
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**Input:** root = [3,4,5,1,2], subRoot = [4,1,2]
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**Output:** true
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**Example 2:**
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![](https://assets.leetcode.com/uploads/2021/04/28/subtree2-tree.jpg)
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**Input:** root = [3,4,5,1,2,null,null,null,null,0], subRoot = [4,1,2]
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**Output:** false
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**Constraints:**
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* The number of nodes in the `root` tree is in the range `[1, 2000]`.
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* The number of nodes in the `subRoot` tree is in the range `[1, 1000]`.
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* <code>-10<sup>4</sup> <= root.val <= 10<sup>4</sup></code>
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* <code>-10<sup>4</sup> <= subRoot.val <= 10<sup>4</sup></code>
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package g0501_0600.s0566_reshape_the_matrix;
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import static org.hamcrest.CoreMatchers.equalTo;
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import static org.hamcrest.MatcherAssert.assertThat;
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import org.junit.jupiter.api.Test;
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class SolutionTest {
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@Test
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void matrixReshape() {
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assertThat(
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new Solution().matrixReshape(new int[][] {{1, 2}, {3, 4}}, 1, 4),
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equalTo(new int[][] {{1, 2, 3, 4}}));
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}
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@Test
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void matrixReshape2() {
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assertThat(
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new Solution().matrixReshape(new int[][] {{1, 2}, {3, 4}}, 2, 4),
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equalTo(new int[][] {{1, 2}, {3, 4}}));
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}
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}
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package g0501_0600.s0567_permutation_in_string;
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import static org.hamcrest.CoreMatchers.equalTo;
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import static org.hamcrest.MatcherAssert.assertThat;
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import org.junit.jupiter.api.Test;
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class SolutionTest {
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@Test
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void checkInclusion() {
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assertThat(new Solution().checkInclusion("ab", "eidbaooo"), equalTo(true));
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}
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@Test
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void checkInclusion2() {
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assertThat(new Solution().checkInclusion("ab", "eidboaoo"), equalTo(false));
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}
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}
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package g0501_0600.s0572_subtree_of_another_tree;
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import static org.hamcrest.CoreMatchers.equalTo;
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import static org.hamcrest.MatcherAssert.assertThat;
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import com_github_leetcode.TreeNode;
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import com_github_leetcode.TreeUtils;
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import java.util.ArrayList;
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import java.util.Arrays;
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import org.junit.jupiter.api.Test;
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class SolutionTest {
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@Test
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void isSubtreeFound() {
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TreeNode treeNode =
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TreeUtils.constructBinaryTree(new ArrayList<>(Arrays.asList(3, 4, 5, 1, 2)));
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TreeNode subTree = TreeUtils.constructBinaryTree(new ArrayList<>(Arrays.asList(4, 1, 2)));
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assertThat(new Solution().isSubtreeFound(treeNode, subTree), equalTo(false));
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}
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@Test
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void isSubtreeFound2() {
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TreeNode treeNode =
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TreeUtils.constructBinaryTree(
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new ArrayList<>(Arrays.asList(3, 4, 5, 1, 2, null, null, null, null, 0)));
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TreeNode subTree = TreeUtils.constructBinaryTree(new ArrayList<>(Arrays.asList(4, 1, 2)));
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assertThat(new Solution().isSubtreeFound(treeNode, subTree), equalTo(false));
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}
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}

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