Count Partitions With Max-Min Difference at Most K
<p>You are given an integer array <code>nums</code> and an integer <code>k</code>. Your task is to partition <code>nums</code> into one or more <strong>non-empty</strong> contiguous segments such that in each segment, the difference between its <strong>maximum</strong> and <strong>minimum</strong> elements is <strong>at most</strong> <code>k</code>.</p>

<p>Return the total number of ways to partition <code>nums</code> under this condition.</p>

<p>Since the answer may be too 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>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [9,4,1,3,7], k = 4</span></p>

<p><strong>Output:</strong> <span class="example-io">6</span></p>

<p><strong>Explanation:</strong></p>

<p>There are 6 valid partitions where the difference between the maximum and minimum elements in each segment is at most <code>k = 4</code>:</p>

<ul>
	<li><code>[[9], [4], [1], [3], [7]]</code></li>
	<li><code>[[9], [4], [1], [3, 7]]</code></li>
	<li><code>[[9], [4], [1, 3], [7]]</code></li>
	<li><code>[[9], [4, 1], [3], [7]]</code></li>
	<li><code>[[9], [4, 1], [3, 7]]</code></li>
	<li><code>[[9], [4, 1, 3], [7]]</code></li>
</ul>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [3,3,4], k = 0</span></p>

<p><strong>Output:</strong> <span class="example-io">2</span></p>

<p><strong>Explanation:</strong></p>

<p>There are 2 valid partitions that satisfy the given conditions:</p>

<ul>
	<li><code>[[3], [3], [4]]</code></li>
	<li><code>[[3, 3], [4]]</code></li>
</ul>
</div>

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

<ul>
	<li><code>2 &lt;= nums.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>0 &lt;= k &lt;= 10<sup>9</sup></code></li>
</ul>
Queue; Array; Dynamic Programming; Prefix Sum; Sliding Window; Monotonic Queue
class Solution {
    public int countPartitions(int[] nums, int k) {
        final int mod = (int) 1e9 + 7;
        TreeMap<Integer, Integer> sl = new TreeMap<>();
        int n = nums.length;
        int[] f = new int[n + 1];
        int[] g = new int[n + 1];
        f[0] = 1;
        g[0] = 1;
        int l = 1;
        for (int r = 1; r <= n; r++) {
            int x = nums[r - 1];
            sl.merge(x, 1, Integer::sum);
            while (sl.lastKey() - sl.firstKey() > k) {
                if (sl.merge(nums[l - 1], -1, Integer::sum) == 0) {
                    sl.remove(nums[l - 1]);
                }
                ++l;
            }
            f[r] = (g[r - 1] - (l >= 2 ? g[l - 2] : 0) + mod) % mod;
            g[r] = (g[r - 1] + f[r]) % mod;
        }
        return f[n];
    }
}
