4Sum II LeetCode solution in C++

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class Solution {
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 public:
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  int fourSumCount(vector<int> &nums1, vector<int> &nums2, vector<int> &nums3, vector<int> &nums4) {
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    int count = 0;
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    unordered_map<int, int> mp;
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    /* Philosophy
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    1. I know that Addtion have two parts in it  EG (a +b , Part 1 - a, Part 2-
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    b. 2.So, Lets make and find this dependency factors. How can I do it?
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    3. If there are 4 Sum. it means 2 sums is going to Part 1 and another 2
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    gonna be Part 2 which are dependent on Part 1 for 0 resultant.
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    4. I gonna store summation 2 nums1 in a FREQUENCY Hashmap.
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    5. Then I traverse 2nd part of the summation (rest to nums) and keep
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    checking that do (0-it1-it2) is exist in map .
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    6. If yes, the add the frequency of Part1 int COUNT var.
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    7. return count;
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    */
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    // Traversing Part 1
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    for (auto &it1 : nums1)
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      for (auto &it2 : nums2) mp[it1 + it2]++;
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    // Traversing Part 2
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    for (auto &it3 : nums3)
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      for (auto &it4 : nums4)
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        if (mp.count(0 - it3 - it4)) count += mp[0 - it3 - it4];
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    return count;
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  }
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};