Alert Using Same Key-Card Three or More Times in a One Hour Period LeetCode solution in C++

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class Solution {
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 public:
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  bool timeDiff(vector<string> timestamp, int index) {
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    int time1 = ((timestamp[index][0] - '0') * 10 + (timestamp[index][1] - '0')) * 60 +
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                ((timestamp[index][3] - '0') * 10 + (timestamp[index][4] - '0'));
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    int time2 = ((timestamp[index + 2][0] - '0') * 10 + (timestamp[index + 2][1] - '0')) * 60 +
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                ((timestamp[index + 2][3] - '0') * 10 + (timestamp[index + 2][4] - '0'));
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    if (abs(time2 - time1) <= 60) return true;
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    return false;
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  }
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  vector<string> alertNames(vector<string> &keyName, vector<string> &keyTime) {
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    unordered_map<string, vector<string>> ht;
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    for (int i = 0; i < keyName.size(); i++) ht[keyName[i]].push_back(keyTime[i]);
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    vector<string> result;
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    for (auto time : ht) {
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      sort(time.second.begin(), time.second.end());
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      for (int i = 0; i + 2 < time.second.size(); i++)
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        if (timeDiff(time.second, i)) {
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          result.push_back(time.first);
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          break;
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        }
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    }
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    sort(result.begin(), result.end());
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    return result;
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  }
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};
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/*
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Algorithm Outline
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1. store in HT as vector
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2. Check for each, 1st & 3rd in range. True if Yes
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*/