Arithmetic Subarrays LeetCode solution in C++

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class Solution {
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 public:
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  vector<bool> checkArithmeticSubarrays(vector<int> &nums, vector<int> &l, vector<int> &r) {
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    vector<bool> ans(l.size(), false);
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    for (int i = 0; i < l.size(); i++) {
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      // if the sub array size is equal to 2
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      if (r[i] - l[i] + 1 == 2)
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        ans[i] = true;
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      else if (isArthemetic(nums, l[i], r[i]))
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        ans[i] = true;
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    }
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    return ans;
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  }
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  // return true if the sub array can be rearranged in a arthemetic sequence
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  bool isArthemetic(vector<int> &nums, int start, int end) {
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    // get the maximum and min elements
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    int mini = INT_MAX;
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    int maxi = INT_MIN;
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    for (int i = start; i <= end; i++) {
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      mini = min(nums[i], mini);
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      maxi = max(nums[i], maxi);
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    }
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    // when mini==maxi it means all the elements are same in the sub array
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    if (mini == maxi) return true;
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    // we cant have same common difference betweeen  two adjacent  elements
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    // when we arrange in arthemetic sequence
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    if ((maxi - mini) % (end - start) != 0) return false;
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    // the diff between every two integers when we rearrange sub array
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    int diff = (maxi - mini) / (end - start);
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    // to check  if the duplicate elemnts are present
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    // ex- [2,4,6,6]
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    // 6 is repeating two times
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    vector<bool> present(end - start + 1, false);
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    for (int i = start; i <= end; i++) {
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      // we cant set a index of nums[i]
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      if ((nums[i] - mini) % diff != 0) return false;
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      int ind = (nums[i] - mini) / diff;
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      // same element is alreeady repeated ( 6 in the above  example)
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      if (present[ind]) return false;
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      // mark it presence
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      present[ind] = true;
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    }
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    return true;
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  }
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};