LC 833. Find And Replace in String

To some string S, we will perform some replacement operations that replace groups of letters with new ones (not necessarily the same size).

Each replacement operation has 3 parameters: a starting index i, a source word x and a target word y.  The rule is that if x starts at position i in the original string S, then we will replace that occurrence of x with y.  If not, we do nothing.

For example, if we have S = "abcd" and we have some replacement operation i = 2, x = "cd", y = "ffff", then because "cd" starts at position 2 in the original string S, we will replace it with "ffff".

Using another example on S = "abcd", if we have both the replacement operation i = 0, x = "ab", y = "eee", as well as another replacement operation i = 2, x = "ec", y = "ffff", this second operation does nothing because in the original string S[2] = 'c', which doesn't match x[0] = 'e'.

All these operations occur simultaneously.  It's guaranteed that there won't be any overlap in replacement: for example, S = "abc", indexes = [0, 1], sources = ["ab","bc"] is not a valid test case.

Example 1:

Input: S = "abcd", indexes = [0,2], sources = ["a","cd"], targets = ["eee","ffff"]
Output: "eeebffff"
Explanation: "a" starts at index 0 in S, so it's replaced by "eee".
"cd" starts at index 2 in S, so it's replaced by "ffff".

Example 2:

Input: S = "abcd", indexes = [0,2], sources = ["ab","ec"], targets = ["eee","ffff"]
Output: "eeecd"
Explanation: "ab" starts at index 0 in S, so it's replaced by "eee". 
"ec" doesn't starts at index 2 in the original S, so we do nothing.

Notes:

1. 0 <= indexes.length = sources.length = targets.length <= 100
2. 0 < indexes[i] < S.length <= 1000
3. All characters in given inputs are lowercase letters.

 

简单的Median题，用一个index根据index的起始点和长度在原字符串中匹配。注意index可能无序，要先排序以后再用。

Runtime: 4 ms, faster than 99.79% of C++ online submissions for Find And Replace in String.

 

#define ALL(x) (x).begin(), (x).end()
#define FOR(i, a, b) for (remove_cv<remove_reference<decltype(b)>::type>::type i = (a); i < (b); i++)
#define REP(i, n) FOR(i, 0, n)
#include <vector>
#include <iostream>
#include <unordered_map>
#include <map>
using namespace std;
class Solution {
private:
  map<int, vector<string>> mp;
public:
    string findReplaceString(string S, vector<int>& indexes, vector<string>& sources, vector<string>& targets) {
      REP(i, indexes.size()){
        mp[indexes[i]].push_back(sources[i]);
        mp[indexes[i]].push_back(targets[i]);
      }
      int cnt = 0;
      for(auto it : mp){
        indexes[cnt] = it.first;
        sources[cnt] = it.second[0];
        targets[cnt] = it.second[1];
        cnt++;
      }
      vector<string> strvec;
      string ret;
      int idx = 0;
      REP(i,indexes.size()){
        if(idx < indexes[i]){
          strvec.push_back(S.substr(idx,indexes[i] - idx));
          idx = indexes[i];
        }
        if(S.substr(indexes[i], sources[i].size()) == sources[i]){
          strvec.push_back(targets[i]);
          idx += sources[i].size();
        }
      }
      if(idx < S.size()){
        strvec.push_back(S.substr(idx));
      }
      for(auto s : strvec){
        ret += s;
      }
      return ret;
  }
};