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Diff for: ‎contest/leetcode_biweek_17/[refered]leetcode_5146_Distinct_Echo_Substrings.py

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# AC:
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# from others' solution
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#
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class Solution(object):
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def distinctEchoSubstrings(self, S):
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N = len(S)
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P, MOD = 37, 344555666677777 # MOD is prime
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Pinv = pow(P, MOD - 2, MOD)
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prefix = [0]
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pwr = 1
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ha = 0
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for x in map(ord, S):
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ha = (ha + pwr * x) % MOD
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pwr = pwr * P % MOD
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prefix.append(ha)
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seen = set()
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pwr = 1
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for length in range(1, N // 2 + 1):
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pwr = pwr * P % MOD # pwr = P^length
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for i in range(N - 2 * length + 1):
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left = (prefix[i + length] - prefix[i]) * pwr % MOD # hash of s[i:i+length] * P^length
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right = (prefix[i + 2 * length] - prefix[i + length]) % MOD # hash of s[i+length:i+2*length]
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if left == right:
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seen.add(left * pow(Pinv, i, MOD) % MOD) # left * P^-i is the canonical representation
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return len(seen)

Diff for: ‎contest/leetcode_biweek_17/leetcode_5143_Decompress_Run-Length_Encoded_List.cpp

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/**
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* AC：
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* 思路：较易，略
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*/
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class Solution {
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public:
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vector<int> decompressRLElist(vector<int>& nums) {
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vector<int> ret;
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for(int i = 0; i < nums.size() / 2; i++)
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for(int j = 0; j < nums[2 * i]; j++)
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ret.push_back(nums[2 * i + 1]);
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return ret;
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}
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};

Diff for: ‎contest/leetcode_biweek_17/leetcode_5144_Matrix_Block_Sum.cpp

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/**
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* AC：
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* 思路：先按行预先计算出行的累加和，用一个 a[row + 2 * K][col + 2 * K] 的数组记录，
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* 然后在对 a 遍历，每个 [i][j] 位置累加 a[i][j] 到 a[i + 2 * K][j + 2 * K] 的K^2 个元素和
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* 这样总体复杂度是 O(m)(n)
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* T: O(m * n * K), 由于 K 是常数，所以接近 m * n 的复杂度
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*/
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class Solution {
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public:
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vector<vector<int>> matrixBlockSum(vector<vector<int>>& mat, int K) {
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int row = mat.size(), col = mat[0].size();
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vector<vector<int>> ret(row, vector<int>(col, 0));
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vector<vector<int>> record(row + 2 * K, vector<int>(row + 2 * K, 0));
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for(int i = 0; i < row; i++) {
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for(int j = 0; j < col; j++) {
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for(int k = -K; k <= K; k++) {
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if(j + k < 0 || j + k > col - 1) // 左右越界的部分不计算行的累加和
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continue;
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record[i + K][j + K] += mat[i][j + k];
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}
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}
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}
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for(int i = 0; i < row; i++) {
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for(int j = 0; j < col; j++) {
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int temp = 0;
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for(int k = -K; k <= K; k++) {
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temp += record[i + K - k][j + K];
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}
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ret[i][j] = temp;
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}
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}
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return ret;
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}
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};

Diff for: ‎contest/leetcode_biweek_17/leetcode_5145_Sum_of_Nodes_with_Even-Valued_Grandparent.cpp

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/**
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* AC:
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* 思路：递归，定义 f(node), 对于每一个 node，判断到孙子节点，如果当前 node->val 为偶，
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* 则加进其孙子节点的值（如果存在）,然后 再递归累加 ret += f(node->left) + f(node->right)
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* T:O(n), n 为二叉树的节点数
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*/
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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class Solution {
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public:
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int sumEvenGrandparent(TreeNode* root) {
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TreeNode* tmp = root;
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int ret = 0;
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if(root->val % 2 == 0) { // 偶数点
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if(root->left) {
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if(root->left->left)
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ret += root->left->left->val;
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if(root->left->right)
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ret += root->left->right->val;
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ret += sumEvenGrandparent(root->left); // 递归
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}
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if(root->right) {
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if(root->right->left)
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ret += root->right->left->val;
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if(root->right->right)
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ret += root->right->right->val;
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ret += sumEvenGrandparent(root->right);
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}
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} else {
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if(root->left) {
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ret += sumEvenGrandparent(root->left); // 递归
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}
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if(root->right) {
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ret += sumEvenGrandparent(root->right);
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}
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}
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return ret;
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}
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};

Diff for: ‎contest/leetcode_biweek_17/leetcode_5146_Distinct_Echo_Substrings.cpp

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class Solution {
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public:
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int distinctEchoSubstrings(string text) {
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vector<string> ret;
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int textLength = text.length();
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for (int i = 0; i < textLength; i++) {
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for (int j = i + 1; j < textLength; j += 2) { // 偶数个字符的子串
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bool flag = true;
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for (int k = 0; k < (j - i + 1) / 2; k++) {
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if (text[i + k] != text[(j + i) / 2 + 1 + k]) {
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flag = false;
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break;
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}
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}
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if (flag) {
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string temp = text.substr(i, (j - i) + 1);
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ret.push_back(temp);
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}
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}
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}
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set<string>s(ret.begin(), ret.end()); // 去重
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return s.size();
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}
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};

Diff for: ‎contest/leetcode_biweek_52/leetcode_1859_Sorting_the_Sentence.java

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// AC:
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// Runtime: 5 ms, faster than 100.00% of Java online submissions for Sorting the Sentence.
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// Memory Usage: 38.5 MB, less than 100.00% of Java online submissions for Sorting the Sentence.
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// Using tree map to sort key.
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// T:O(n), S:O(n), n = len(s)
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//
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class Solution {
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public String sortSentence(String s) {
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TreeMap<Integer, String> record = new TreeMap<>();
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String[] sArr = s.split(" ");
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for (String item: sArr) {
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Integer index = Integer.parseInt(String.valueOf(item.charAt(item.length() - 1)));
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record.put(index, item.substring(0, item.length() - 1));
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}
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StringBuilder ret = new StringBuilder();
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for (Integer i: record.keySet()) {
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ret.append(record.get(i));
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ret.append(" ");
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}
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return ret.substring(0, ret.length() - 1);
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}
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}

Diff for: ‎contest/leetcode_biweek_52/leetcode_1860_Incremental_Memory_Leak.java

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// AC:
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// Runtime: 8 ms, faster than 100.00% of Java online submissions for Incremental Memory Leak.
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// Memory Usage: 38.3 MB, less than 100.00% of Java online submissions for Incremental Memory Leak.
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// simulation... it seems that brute-force can pass. Any way I haven't thought out better solution...
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// complexity: T: O(sqrt(max(m1, m2))), S: O(1)
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//
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class Solution {
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public int[] memLeak(int memory1, int memory2) {
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int second = 1;
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while (memory1 >= second || memory2 >= second) {
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if (memory1 >= memory2) {
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if (memory1 >= second) {
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memory1 -= second;
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} else {
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break;
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}
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} else {
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if (memory2 >= second) {
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memory2 -= second;
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} else {
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break;
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}
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}
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second++;
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}
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int[] ret = new int[]{second, memory1, memory2};
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return ret;
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}
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}

Diff for: ‎contest/leetcode_biweek_52/leetcode_1861_Rotating_the_Box.java

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// AC:
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// Runtime: 2056 ms, faster than 50.00% of Java online submissions for Rotating the Box.
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// Memory Usage: 64.7 MB, less than 100.00% of Java online submissions for Rotating the Box.
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// my thought: using string split(obstacle), and combine every part to make stone fall down.
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// complexity: T:O(m * n), S:O(m * n)
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//
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class Solution {
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public char[][] rotateTheBox(char[][] box) {
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int row = box.length, col = box[0].length;
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char[][] ret = new char[col][row];
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List<List<Character>> record = new LinkedList<>();
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for (int i = 0; i < row; i++) {
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// 逐行下落
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List<Character> temp = new LinkedList<>();
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StringBuilder tempStr = new StringBuilder();
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for (int j = 0; j <= col - 1; j++) {
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tempStr.append(box[i][j]);
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}
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tempStr.append("#");
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String[] tempArr = tempStr.toString().split("\\*");
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if (tempArr.length == 0) {
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for (int t = 0; t < tempStr.length(); t++) {
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temp.add('*');
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}
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} else {
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for (String part : tempArr) {
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if (part.isEmpty()) {
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temp.add('*');
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continue;
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}
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int countEmpty = 0, countStone = 0;
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for (char item : part.toCharArray()) {
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if (item == '#') {
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countStone++;
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}
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if (item == '.') {
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countEmpty++;
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}
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}
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for (int k = 0; k < countEmpty; k++) {
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temp.add('.');
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}
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for (int k = 0; k < countStone; k++) {
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temp.add('#');
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}
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temp.add('*');
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}
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// 去除末尾 *
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if (temp.get(temp.size() - 1) == '*') {
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temp.remove(temp.size() - 1);
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}
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}
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temp.remove(temp.size() - 1);
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record.add(temp);
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}
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// rotate
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for (int i = 0; i < col; i++) {
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for (int j = 0; j < row; j++) {
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ret[i][j] = record.get(row - 1 - j).get(i);
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}
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}
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return ret;
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}
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}

Diff for: ‎contest/leetcode_biweek_59/[editing]leetcode_1976_Number_of_Ways_to_Arrive_at_Destination.java

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class Solution {
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public int countPaths(int n, int[][] roads) {
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}
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}

Diff for: ‎contest/leetcode_biweek_59/[editing]leetcode_1977_Number_of_Ways_to_Separate_Numbers.java

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class Solution {
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public int numberOfCombinations(String num) {
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}
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}

Diff for: ‎contest/leetcode_biweek_59/leetcode_1974_Minimum_Time_to_Type_Word_Using_Special_Typewriter.java

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// AC: Runtime: 0 ms, faster than 100.00% of Java online submissions for Minimum Time to Type Word Using Special Typewriter.
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// Memory Usage: 36.7 MB, less than 100.00% of Java online submissions for Minimum Time to Type Word Using Special Typewriter.
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// .
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// T:O(n), S:(1)
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//
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class Solution {
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public int minTimeToType(String word) {
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int ret = 0, curPos = 0, size = word.length();
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for (int i = 0; i < size; i++) {
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int c = word.charAt(i) - 'a';
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int move = Math.abs(c - curPos);
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int minMove = Math.min(move, 26 - move);
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ret += minMove + 1;
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curPos = c;
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}
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return ret;
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}
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}

Diff for: ‎contest/leetcode_biweek_59/leetcode_1975_Maximum_Matrix_Sum.java

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// AC: Runtime: 6 ms, faster than 100.00% of Java online submissions for Maximum Matrix Sum.
2+
// Memory Usage: 48.1 MB, less than 100.00% of Java online submissions for Maximum Matrix Sum.
3+
// if have zero, then sum all absolute value. if not zero and negative numbers are odd, then sum all absolute and minus 2 * maxNegativeNumber.
4+
// T:O(m * n), S:O(1)
5+
//
6+
class Solution {
7+
public long maxMatrixSum(int[][] matrix) {
8+
int minAbs = Integer.MAX_VALUE, negaCount = 0;
9+
long absSum = 0;
10+
boolean hasZero = false;
11+
int row = matrix.length;
12+
for (int i = 0; i < row; i++) {
13+
for (int j = 0; j < row; j++) {
14+
if (!hasZero && matrix[i][j] == 0) {
15+
hasZero = true;
16+
}
17+
absSum += Math.abs(matrix[i][j]);
18+
if (!hasZero) {
19+
minAbs = Math.min(minAbs, Math.abs(matrix[i][j]));
20+
if (matrix[i][j] < 0) {
21+
negaCount++;
22+
}
23+
}
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}
25+
}
26+
27+
if (hasZero || (negaCount % 2 == 0)) {
28+
return absSum;
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} else {
30+
return absSum - 2L * minAbs;
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}
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}
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}

Diff for: ‎contest/leetcode_week_166/[editing]leetcode_1282_Group_the_People_Given_the_Group_Size_They_Belong_To.cpp

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class Solution {
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public:
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vector<vector<int>> groupThePeople(vector<int>& groupSizes) {
4+
5+
}
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};

Diff for: ‎contest/leetcode_week_166/[editing]leetcode_1283_Find_the_Smallest_Divisor_Given_a_Threshold.cpp

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class Solution {
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public:
3+
int smallestDivisor(vector<int>& nums, int threshold) {
4+
5+
}
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};

Diff for: ‎contest/leetcode_week_166/[editing]leetcode_1284_Minimum_Number_of_Flips_to_Convert_Binary_Matrix_to_Zero_Matrix.cpp

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class Solution {
2+
public:
3+
int minFlips(vector<vector<int>>& mat) {
4+
5+
}
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};

Diff for: ‎contest/leetcode_week_166/leetcode_1281_Subtract_the_Product_and_Sum_of_Digits_of_an_Integer.cpp

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/**
2+
* 给定一个正数，求各位之积减去各位之和。
3+
*
4+
* AC: Runtime: 0 ms | Memory Usage: 8.3 MB
5+
*/
6+
class Solution {
7+
public:
8+
int subtractProductAndSum(int n) {
9+
if(n == 0)
10+
return 0;
11+
int product = 1, sum = 0;
12+
while(n != 0) {
13+
product *= n % 10;
14+
sum += n % 10;
15+
n /= 10;
16+
}
17+
18+
return product - sum;
19+
}
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};

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