https://leetcode.com/problems/frog-jump/ https://leetcode.com/problems/frog-jump/submissions/1032720945/ class Solution { public: bool check(unordered_map<int, bool>& exist, unordered_map<int, unordered_map<int, bool>> &inq, int lastStone, int nextStone, int jumpUnits) { if(inq[nextStone][jumpUnits]) return false; if(nextStone > lastStone) return false; if(!exist[nextStone]) return false; return true; } bool canCross(vector<int>& stones) { if(stones == (vector<int>){0,1}) { return true; } int n = stones.size(); if(stones.back() > (n * (1 + n - 1) / 2)) { return false; } unordered_map<int, bool> exist; for(int i : stones) exist[i] = true; priority_queue<pair<int, int>, vector<pair<int, int>>, less<pair<int, int>>> q; unordered_map<int, unordered_map<int, bool>> inq; if(stones[1] !

https://leetcode.com/problems/text-justification/ https://leetcode.com/problems/text-justification/submissions/1030637814/ class Solution { public: vector<string> fullJustify(vector<string>& words, int maxWidth) { vector<string> ans; for(int i = 0; i < words.size(); ){ string tans = ""; int j = i; while(j != words.size() && tans.size() + words[j].size() <= maxWidth) { tans += words[j]; if(tans.size() != maxWidth) tans += " "; j++; } if(j == words.size()) { ans.push_back(tans + string(maxWidth - tans.size(), ' ')); } else { ans.push_back(rerrange(tans, maxWidth)); } i = j; } return ans; } inline string rerrange(string s, int maxWidth) { int space = 0; vector<string> split; string ts = ""; for(auto c : s) { if(c == ' ') space++, split.

https://leetcode.com/problems/sliding-window-maximum/ https://leetcode.com/problems/sliding-window-maximum/submissions/1022554836/ class Solution { public: vector<int> maxSlidingWindow(vector<int>& nums, int k) { priority_queue<pair<int, int>, vector<pair<int,int>>, less<pair<int,int>>> q; for(int i = 0; i < k; ++i) { q.push(pair(nums[i],i)); } vector<int> ans; ans.push_back(q.top().first); for(int i = k; i < nums.size(); ++i) { q.push(pair(nums[i], i)); auto t = q.top(); while(t.second <= i - k) { q.pop(); t = q.top(); } ans.push_back(q.top().first); } return ans; } };

Q2 Tutorial An easy dp problem, for dp[i], means how many good string of length i. Init dp[zero] = 1, dp[one] = 1 (if one == zero, then dp[one] = 2); So dp[i] = dp[i - zero] + dp[i - one]. After calculate all dp, sum up dp[low] to dp[high]. Accepted Code class Solution { public: long long dp[100005]; int countGoodStrings(int low, int high, int zero, int one) { memset(dp, 0, sizeof(dp)); dp[zero]++; dp[one]++; for(int i = min(zero, one); i <= high; ++i){ if(i > zero) dp[i] += dp[i - zero]; if(i > one) dp[i] += dp[i - one]; dp[i] %= 1000000007; } long long ans = 0; for(int i = low; i <= high; ++i){ ans += dp[i]; ans %= 1000000007; } return ans; } }; Q3 Tutorial First, from root 0, bfs all tree to find where bob is.

If you get any problem of my code or you find something wrong, just leave a comment or contact with me, I’ll response you as fast as I can if I see it. If this tutorial helps you well, you can subscribe my website, every time I update any Tutorial, I’ll send an email to all subscribers. 10-31 766. Toeplitz Matrix Difficulty: Easy Accepted Code class Solution { public: bool isToeplitzMatrix(vector<vector<int>>& matrix) { vector<int> n[int(matrix.

D. Guess The String Problem summary The Jury chosen a stringS consisting of n characters; each character of S is a lowercase Latin letter. You may ask query in two types: 1 i - query for the letterSi 2 l r - query the size of character set of Sl,Sl+1,......,Sr You are allowed to ask no more than 26 queries of the first type, and no more than 6000 queries of the second type.