<p>You are given an integer array <code>nums</code>. You are initially positioned at the array&#39;s <strong>first index</strong>, and each element in the array represents your maximum jump length at that position.</p>

<p>Return <code>true</code><em> if you can reach the last index, or </em><code>false</code><em> otherwise</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [2,3,1,1,4]
<strong>Output:</strong> true
<strong>Explanation:</strong> Jump 1 step from index 0 to 1, then 3 steps to the last index.
</pre>

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

<pre>
<strong>Input:</strong> nums = [3,2,1,0,4]
<strong>Output:</strong> false
<strong>Explanation:</strong> You will always arrive at index 3 no matter what. Its maximum jump length is 0, which makes it impossible to reach the last index.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    bool canJump(vector<int>& nums) {
        int n = nums.size();
        queue<int> q;
        for (int i = 0; i < n; i++) {
            if (nums[i] == 0) q.push(i);
        }
        if (q.empty()) 
            return true;
        
        for (int i = 0; i < n; i++) {
            if (i < q.front()) {
                if (i + nums[i] > q.front()) {
                    q.pop();
                    i--;
                }
            }
        }
        
        if (q.front() == nums.size() - 1 || q.empty()) return true;
        return false;
    }
};
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