Jump Game VII
<p>You are given a <strong>0-indexed</strong> binary string <code>s</code> and two integers <code>minJump</code> and <code>maxJump</code>. In the beginning, you are standing at index <code>0</code>, which is equal to <code>&#39;0&#39;</code>. You can move from index <code>i</code> to index <code>j</code> if the following conditions are fulfilled:</p>

<ul>
	<li><code>i + minJump &lt;= j &lt;= min(i + maxJump, s.length - 1)</code>, and</li>
	<li><code>s[j] == &#39;0&#39;</code>.</li>
</ul>

<p>Return <code>true</code><i> if you can reach index </i><code>s.length - 1</code><i> in </i><code>s</code><em>, or </em><code>false</code><em> otherwise.</em></p>

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

<pre>
<strong>Input:</strong> s = &quot;<u>0</u>11<u>0</u>1<u>0</u>&quot;, minJump = 2, maxJump = 3
<strong>Output:</strong> true
<strong>Explanation:</strong>
In the first step, move from index 0 to index 3. 
In the second step, move from index 3 to index 5.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;01101110&quot;, minJump = 2, maxJump = 3
<strong>Output:</strong> false
</pre>

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

<ul>
	<li><code>2 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s[i]</code> is either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
	<li><code>s[0] == &#39;0&#39;</code></li>
	<li><code>1 &lt;= minJump &lt;= maxJump &lt; s.length</code></li>
</ul>
String; Dynamic Programming; Prefix Sum; Sliding Window
/**
 * @param {string} s
 * @param {number} minJump
 * @param {number} maxJump
 * @return {boolean}
 */
var canReach = function (s, minJump, maxJump) {
    const n = s.length;
    const pre = Array(n + 1).fill(0);
    pre[1] = 1;
    const f = Array(n).fill(false);
    f[0] = true;
    for (let i = 1; i < n; ++i) {
        if (s[i] === '0') {
            const [l, r] = [Math.max(0, i - maxJump), i - minJump];
            f[i] = l <= r && pre[r + 1] - pre[l] > 0;
        }
        pre[i + 1] = pre[i] + (f[i] ? 1 : 0);
    }
    return f[n - 1];
};
