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1326. Minimum Number of Taps to Open to Water a Garden

Description

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:

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.

 

Constraints:

  • 1 <= n <= 104
  • ranges.length == n + 1
  • 0 <= ranges[i] <= 100

Solutions

Solution 1: Greedy

We note that for all taps that can cover a certain left endpoint, choosing the tap that can cover the farthest right endpoint is optimal.

Therefore, we can preprocess the array $ranges$. For the $i$-th tap, it can cover the left endpoint $l = \max(0, i - ranges[i])$ and the right endpoint $r = i + ranges[i]$. We calculate the position of the tap that can cover the left endpoint $l$ with the farthest right endpoint and record it in the array $last[i]$.

Then we define the following three variables:

  • Variable $ans$ represents the final answer, i.e., the minimum number of taps;
  • Variable $mx$ represents the farthest right endpoint that can currently be covered;
  • Variable $pre$ represents the farthest right endpoint covered by the previous tap.

We traverse all positions in the range $[0, \ldots, n-1]$. For the current position $i$, we use $last[i]$ to update $mx$, i.e., $mx = \max(mx, last[i])$.

  • If $mx \leq i$, it means the next position cannot be covered, so we return $-1$.
  • If $pre = i$, it means a new subinterval needs to be used, so we increment $ans$ by $1$ and update $pre = mx$.

After the traversal, we return $ans$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the garden.

Similar problems:

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class Solution:
    def minTaps(self, n: int, ranges: List[int]) -> int:
        last = [0] * (n + 1)
        for i, x in enumerate(ranges):
            l, r = max(0, i - x), i + x
            last[l] = max(last[l], r)

        ans = mx = pre = 0
        for i in range(n):
            mx = max(mx, last[i])
            if mx <= i:
                return -1
            if pre == i:
                ans += 1
                pre = mx
        return ans
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class Solution {
    public int minTaps(int n, int[] ranges) {
        int[] last = new int[n + 1];
        for (int i = 0; i < n + 1; ++i) {
            int l = Math.max(0, i - ranges[i]), r = i + ranges[i];
            last[l] = Math.max(last[l], r);
        }
        int ans = 0, mx = 0, pre = 0;
        for (int i = 0; i < n; ++i) {
            mx = Math.max(mx, last[i]);
            if (mx <= i) {
                return -1;
            }
            if (pre == i) {
                ++ans;
                pre = mx;
            }
        }
        return ans;
    }
}
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class Solution {
public:
    int minTaps(int n, vector<int>& ranges) {
        vector<int> last(n + 1);
        for (int i = 0; i < n + 1; ++i) {
            int l = max(0, i - ranges[i]), r = i + ranges[i];
            last[l] = max(last[l], r);
        }
        int ans = 0, mx = 0, pre = 0;
        for (int i = 0; i < n; ++i) {
            mx = max(mx, last[i]);
            if (mx <= i) {
                return -1;
            }
            if (pre == i) {
                ++ans;
                pre = mx;
            }