leetcode@ [30/76] Substring with Concatenation of All Words & Minimum Window Substring (Hashtable, Two Pointers)
https://leetcode.com/problems/substring-with-concatenation-of-all-words/
You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.
For example, given:
s: "barfoothefoobarman"
words: ["foo", "bar"]
You should return the indices: [0,9]
.
(order does not matter).
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
class Solution { public: bool func(string s, vector<string>& words, map<string, int>& h) { int n = words.size(); int each = words[0].length(); map<string, int> mh; mh.clear(); int i = 0; for(; i<=s.length()-each; i+=each) { string sub = s.substr(i, each); if(h.find(sub) != h.end()) { ++mh[sub]; if(mh[sub] > h[sub]) return false; } else return false; } return true; } vector<int> findSubstring(string s, vector<string>& words) { vector<int> res; int n = words.size(); if(n == 0) return res; map<string, int> h; for(int i=0; i<n; ++i) { if(h.find(words[i]) == h.end()) { h.insert(make_pair(words[i], 1)); } else { h[words[i]]++; } } int each = words[0].length(); int tot = each * n; if(s.length() < tot) return res; for(int i = 0; i <= s.length() - tot; ++i) { string sub = s.substr(i, tot); //cout << sub << endl; if(func(sub, words, h)) res.push_back(i); } return res; } };
https://leetcode.com/problems/minimum-window-substring/
Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).
For example,
S = "ADOBECODEBANC"
T = "ABC"
Minimum window is "BANC"
.
Note:
If there is no such window in S that covers all characters in T, return the empty string ""
.
If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
class Solution { public: string minWindow(string s, string t) { if(s.length() == 0 || s.length() < t.length()) return ""; if(s.length() == t.length()) { if(s == t) return s; } map<char, int> h; h.clear(); map<char, bool> mh; mh.clear(); for(int i=0; i<t.length(); ++i) { h[t[i]]++; mh[t[i]] = true; } int cnt = t.length(), l = 0, minRange = INT_MAX, mini, minj; for(int r=0; r<s.length(); ++r) { if(mh[s[r]]) { --h[s[r]]; if(h[s[r]] >= 0) --cnt; } if(cnt == 0) { while(!mh[s[l]] || h[s[l]] < 0) { ++h[s[l]]; ++l; } if(minRange > r - l + 1) { minRange = r - l + 1; mini = l; } } } if(minRange == INT_MAX) return ""; return s.substr(mini, minRange); } };