<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://assets.leetcode.com/uploads/2020/01/16/1685_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>

class Solution {
    static bool comp(vector<int> &a, vector<int> &b){
        if(a[0]==b[0])return a[1]>b[1];

        return a[0]<b[0];
    }
public:
    int minTaps(int n, vector<int>& ranges) {
        vector<vector<int>> interval;

        for(int i=0; i<=n; i++){
            interval.push_back({max(0,i-ranges[i]),min(n,i+ranges[i])});
        }

        sort(interval.begin(),interval.end(),comp);

        int count=0;

        int i=0;  
        int maxi=0;
        int maxic=0;
        while(maxic<n){
           while(i<interval.size() && maxic>=interval[i][0]){
                maxi=max(maxi,interval[i][1]);
                i++;
           }

           if(maxic==maxi)return -1;

           count++;
           maxic=maxi;
        }

        return count;

    }
};
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