Problem description:

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:

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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:

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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:

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

Solution:

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class Solution {
public:
    string findReplaceString(string S, vector<int>& indexes, vector<string>& sources, vector<string>& targets) {
        
        vector<pair<int, int>> replace;
        for(int i= 0; i< indexes.size(); i++){
            replace.push_back(make_pair(indexes[i], i));
        }
        sort(replace.rbegin(), replace.rend());
        
        for(auto n: replace){
            int i= n.first;
            string s= sources[n.second], t= targets[n.second];
            if(S.substr(i, s.length()) == s)
                S= S.substr(0, i)+ t + S.substr(i+s.length());
        }
        
        return S;
        
    }
};

time complexity: $O(SN)$
space complexity: $O(N)$
reference:
https://goo.gl/uKL6KV