<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 {
public:
    int solve(int index , map<int , pair<int , int> > & m , vector<int> & ranges){
        if (index <=0){
            return 0 ; 
        }

        int ans = INT_MAX ; 
        if (m[index].first != index){
            int x =  solve(m[index].first ,   m ,  ranges) ; 
            if (x!=INT_MAX){
                ans = min(ans , 1+x) ; 
            }
        }
        
        return ans ; 
    }
    int minTaps(int n, vector<int>& ranges) {
        map<int , pair<int  , int> > m ; 

        for(int j=0 ; j<=n ;j++){
            m[j] = {INT_MAX , -1} ; 
        }

        for(int j=0 ; j<=n ; j++){
            for(int i=j-ranges[j]  ; i<= j + ranges[j] ; i++){
                if (j-ranges[j] < m[i].first){
                    m[i] = {j-ranges[j] , j} ; 
                }
            }
        }

        int ans =  solve(n , m , ranges) ; 

        if (ans == INT_MAX){
            return -1 ; 
        }
        return ans ; 



    }
};
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