Maximum Equal Frequency LeetCode solution in C++

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class Solution {
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 public:
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  int maxEqualFreq(vector<int> &nums) {
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    int n = nums.size();
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    unordered_map<int, int> mp1,
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        mp2; // one to store the frequency of each number , and one which will
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             // store the frequency of the frequency
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    int ans = 0;
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    for (int i = 0; i < n; i++) {
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      int c = nums[i];
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      mp1[c]++;
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      mp2[mp1[c]]++;
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      if (i != n - 1 && mp2[mp1[c]] * mp1[c] == i + 1)
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        // when curernt length is equal to the total same frequency of numbers
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        // then we need one more number which may be anything so doesn't matter
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        // . so we just traverse till n-2 and that's it.
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        ans = i + 2;
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      if (mp2[mp1[c]] * mp1[c] == i)
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        // when current length is equal to the total same frequency of numbers
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        // plus one extra no. ( so here taken ==i (i+1-1)) which can be 1. same
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        // as any number present 2. different number .. but doesn't matter and
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        // our condition just satisfies ... so just take the current length .
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        ans = i + 1;
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    }
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    return ans;
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  }
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};