**First Hint**: Observe that the condition `nums[i] + rev(nums[j]) == nums[j] + rev(nums[i])` can be rewritten as `nums[i] - rev(nums[i]) == nums[j] - rev(nums[j])`. This means we can transform the problem into counting pairs of indices with equal values of `nums[k] - rev(nums[k])`.

**Second Hint**: For each number in `nums`, compute the value `nums[k] - rev(nums[k])` and store the frequency of each computed value in a hash map. The number of nice pairs for a given value is the combination of its frequency taken two at a time, i.e., `C(freq, 2) = freq * (freq - 1) / 2`.

**Third Hint**: Iterate through the hash map, sum up all the combinations for each frequency, and return the total modulo `10^9 + 7`. Ensure to handle large numbers by using modulo operations during the computation to prevent overflow.