Count the Number of Good Partitions
<p>You are given a <strong>0-indexed</strong> array <code>nums</code> consisting of <strong>positive</strong> integers.</p>

<p>A partition of an array into one or more <strong>contiguous</strong> subarrays is called <strong>good</strong> if no two subarrays contain the same number.</p>

<p>Return <em>the <strong>total number</strong> of good partitions of </em><code>nums</code>.</p>

<p>Since the answer may be 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> nums = [1,2,3,4]
<strong>Output:</strong> 8
<strong>Explanation:</strong> The 8 possible good partitions are: ([1], [2], [3], [4]), ([1], [2], [3,4]), ([1], [2,3], [4]), ([1], [2,3,4]), ([1,2], [3], [4]), ([1,2], [3,4]), ([1,2,3], [4]), and ([1,2,3,4]).
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,1,1,1]
<strong>Output:</strong> 1
<strong>Explanation:</strong> The only possible good partition is: ([1,1,1,1]).
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,1,3]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The 2 possible good partitions are: ([1,2,1], [3]) and ([1,2,1,3]).
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>
Array; Hash Table; Math; Combinatorics
function numberOfGoodPartitions(nums: number[]): number {
    const qpow = (a: number, n: number, mod: number) => {
        let ans = 1;
        for (; n; n >>= 1) {
            if (n & 1) {
                ans = Number((BigInt(ans) * BigInt(a)) % BigInt(mod));
            }
            a = Number((BigInt(a) * BigInt(a)) % BigInt(mod));
        }
        return ans;
    };
    const last: Map<number, number> = new Map();
    const n = nums.length;
    for (let i = 0; i < n; ++i) {
        last.set(nums[i], i);
    }
    const mod = 1e9 + 7;
    let [j, k] = [-1, 0];
    for (let i = 0; i < n; ++i) {
        j = Math.max(j, last.get(nums[i])!);
        if (i === j) {
            ++k;
        }
    }
    return qpow(2, k - 1, mod);
}
