> For the complete documentation index, see [llms.txt](https://soumyajit4419.gitbook.io/ds-algo/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://soumyajit4419.gitbook.io/ds-algo/greedy-algorithms/16.-minimum-number-of-taps-to-open-to-water-a-garden.md).

# 16. Minimum Number of Taps to Open to Water a Garden

There is a one-dimensional garden on the x-axis. The garden starts at the point `0` and ends at the point `n`. (i.e The length of the garden is `n`).

There are `n + 1` taps located at points `[0, 1, ..., n]` in the garden.

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

Return *the minimum number of taps* that should be open to water the whole garden, If the garden cannot be watered return **-1**.

**Example 1:**

![](https://assets.leetcode.com/uploads/2020/01/16/1685_example_1.png)

```
Input: n = 5, ranges = [3,4,1,1,0,0]
Output: 1
Explanation: 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]
```

**Example 2:**

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

**Example 3:**

```
Input: n = 7, ranges = [1,2,1,0,2,1,0,1]
Output: 3
```

**Example 4:**

```
Input: n = 8, ranges = [4,0,0,0,0,0,0,0,4]
Output: 2
```

**Example 5:**

```
Input: n = 8, ranges = [4,0,0,0,4,0,0,0,4]
Output: 1
```

## Solution: (DP)

```cpp
class Solution
{
public:
    int minTaps(int n, vector<int> &ranges)
    {

        vector<vector<int>> v;
        vector<int> dp(n + 1, n + 2);

        for (int i = 0; i < ranges.size(); i++)
        {
            int val = ranges[i];
            v.push_back({i - val, i + val});
        }

        sort(v.begin(), v.end());

        dp[0] = 0;

        for (int i = 0; i < v.size(); i++)
        {
            int start = max(0, v[i][0]);
            int end = min(n, v[i][1]);

            for (int j = start; j <= end; j++)
            {
                dp[j] = min(dp[j], dp[start] + 1);
            }
        }

        if (dp[n] == n + 2)
        {
            return -1;
        }

        return dp[n];
    }
};
```

**Time Complexity: O(n \* time)**

## Solution: (Greedy)

```cpp
class Solution
{
public:
    int videoStitching(vector<vector<int>> &clips, int time)
    {

        sort(clips.begin(), clips.end());

        int maxEnd = 0;

        int count = 0;

        int i = 0;
        while (maxEnd < time)
        {
            int curEnd = 0;
            while (i < clips.size() && clips[i][0] <= maxEnd)
            {
                curEnd = max(curEnd, clips[i][1]);
                i++;
            }

            if (curEnd <= maxEnd)
            {
                return -1;
            }

            count++;
            maxEnd = curEnd;
        }

        return count;
    }
};
```