<p>Given an integer array <code>arr</code>, and an integer <code>target</code>, return the number of tuples <code>i, j, k</code> such that <code>i &lt; j &lt; k</code> and <code>arr[i] + arr[j] + arr[k] == target</code>.</p>

<p>As the answer can be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> arr = [1,1,2,2,3,3,4,4,5,5], target = 8
<strong>Output:</strong> 20
<strong>Explanation: </strong>
Enumerating by the values (arr[i], arr[j], arr[k]):
(1, 2, 5) occurs 8 times;
(1, 3, 4) occurs 8 times;
(2, 2, 4) occurs 2 times;
(2, 3, 3) occurs 2 times.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> arr = [1,1,2,2,2,2], target = 5
<strong>Output:</strong> 12
<strong>Explanation: </strong>
arr[i] = 1, arr[j] = arr[k] = 2 occurs 12 times:
We choose one 1 from [1,1] in 2 ways,
and two 2s from [2,2,2,2] in 6 ways.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> arr = [2,1,3], target = 6
<strong>Output:</strong> 1
<strong>Explanation:</strong> (1, 2, 3) occured one time in the array so we return 1.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= arr.length &lt;= 3000</code></li>
	<li><code>0 &lt;= arr[i] &lt;= 100</code></li>
	<li><code>0 &lt;= target &lt;= 300</code></li>
</ul>

constexpr long MOD = 1E9 + 7;
class Solution {
public:
    int sumToN(int n) {
        return n * (n + 1) / 2;
    }

    int threeSumMulti(vector<int>& arr, int target) {
        map<int, size_t> counts;
        for (const auto n : arr) {
            counts[n]++;
        }

        long count = 0;
        for (auto it = counts.begin(); it != counts.end(); ++it) {
            auto high = --counts.end();
            auto low = counts.begin();
            while (true) {
                bool lowIsMid = it->first == low->first;
                bool highIsMid = it->first == high->first;
                long sum = low->first + high->first + it->first;
                if (it->second < (lowIsMid + highIsMid + 1)) {
                    break;
                }
                if (sum == target) {
                    if (lowIsMid && highIsMid) {
                        for (size_t i = it->second; i >= 3; --i) {
                            count = (count + sumToN(i - 2)) % MOD;
                        }
                        break;
                    } else if (lowIsMid) {
                        count = (count + sumToN(it->second - 1) * high->second) % MOD;
                        break;
                    } else if (highIsMid) {
                        count = (count + low->second * sumToN(it->second - 1)) % MOD;
                        break;
                    } else {
                        count = (count + low->second * high->second * it->second) % MOD;
                    }
                    --high;
                    ++low;
                } else if (sum < target) {
                    if (lowIsMid) break;
                    ++low;
                } else {
                    if (highIsMid) break;
                    --high;
                }

            }
        }
        return count;
    }
};
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