Minimum Number of Taps to Open to Water a Garden
<p>There is a one-dimensional garden on the x-axis. The garden starts at the point <code>0</code> and ends at the point <code>n</code>. (i.e., the&nbsp;length of the garden is <code>n</code>).</p>

<p>There are <code>n + 1</code> taps located at points <code>[0, 1, ..., n]</code> in the garden.</p>

<p>Given an integer <code>n</code> and an integer array <code>ranges</code> of length <code>n + 1</code> where <code>ranges[i]</code> (0-indexed) means the <code>i-th</code> tap can water the area <code>[i - ranges[i], i + ranges[i]]</code> if it was open.</p>

<p>Return <em>the minimum number of taps</em> that should be open to water the whole garden, If the garden cannot be watered return <strong>-1</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1300-1399/1326.Minimum%20Number%20of%20Taps%20to%20Open%20to%20Water%20a%20Garden/images/1685_example_1.png" style="width: 525px; height: 255px;" />
<pre>
<strong>Input:</strong> n = 5, ranges = [3,4,1,1,0,0]
<strong>Output:</strong> 1
<strong>Explanation:</strong> The tap at point 0 can cover the interval [-3,3]
The tap at point 1 can cover the interval [-3,5]
The tap at point 2 can cover the interval [1,3]
The tap at point 3 can cover the interval [2,4]
The tap at point 4 can cover the interval [4,4]
The tap at point 5 can cover the interval [5,5]
Opening Only the second tap will water the whole garden [0,5]
</pre>

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

<pre>
<strong>Input:</strong> n = 3, ranges = [0,0,0,0]
<strong>Output:</strong> -1
<strong>Explanation:</strong> Even if you activate all the four taps you cannot water the whole garden.
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
	<li><code>ranges.length == n + 1</code></li>
	<li><code>0 &lt;= ranges[i] &lt;= 100</code></li>
</ul>
Greedy; Array; Dynamic Programming
impl Solution {
    #[allow(dead_code)]
    pub fn min_taps(n: i32, ranges: Vec<i32>) -> i32 {
        let mut last = vec![0; (n + 1) as usize];
        let mut ans = 0;
        let mut mx = 0;
        let mut pre = 0;

        // Initialize the last vector
        for (i, &r) in ranges.iter().enumerate() {
            if (i as i32) - r >= 0 {
                last[((i as i32) - r) as usize] =
                    std::cmp::max(last[((i as i32) - r) as usize], (i as i32) + r);
            } else {
                last[0] = std::cmp::max(last[0], (i as i32) + r);
            }
        }

        for i in 0..n as usize {
            mx = std::cmp::max(mx, last[i]);
            if mx <= (i as i32) {
                return -1;
            }
            if pre == (i as i32) {
                ans += 1;
                pre = mx;
            }
        }

        ans
    }
}
