# Solution For Minimum Number Of Taps To Open To Water A Garden

Problem Statement:

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 in different positions in the garden.

You are 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.

Approach:

The approach to solve the problem is to find out the minimum number of intervals required to cover the entire garden.

We can make an assumption in this problem, as the garden needs to water all elements, each tap will water up to the rightmost point it has the range for i.e tap i will water until i+ranges[i].

We can define an array coverage which will mark the endpoints as 1, where this tap has range, for every tap, for example, we visit the first tap, if it has range of 3, we will mark coverageupdate coverage=1 coverage=1 coverage=1 coverage=1 as all these points can be covered by the first tap.

Now we want to create intervals going from the left to the right which will be used to cover the entire garden. The left end of the interval will be the current rightmost point, defined by the variable r, and the length of the interval will be the furthest right endpoint we can get from the current left endpoints, visited by left pointer i. Every time we find a tap covering the right endpoint of this interval i.e. i+r we will update r accordingly and increment our interval count intervals++.

If our starting rightmost end of the interval is still 0, it means we haven’t been able to find a tap covering it, so it’s impossible to cover the intervals, and we return -1.

Code:

The code is written in Python3

def minTaps(n: int, ranges: List[int]) -> int:

``````coverage =  * (n+1) # Set the initial coverage to zero for all the taps
for i in range(n+1):
left = max(0, i-ranges[i])
right = i + ranges[i]
coverage[left:right+1] = *(right-left+1)

intervals, r, reachable = 0, 0, 0
while r < n:
intervals += 1
max_reach = r
for i in range(r+1):
if coverage[i]:
max_reach = max(max_reach, i+ranges[i]) #update r if the current tap can reach a further point
if max_reach <= r: # means we can not reach any further right endpoint, so can't cover the entire garden
return -1
r= max_reach
return intervals
``````

## Step by Step Implementation For Minimum Number Of Taps To Open To Water A Garden

```The problem can be solved using a greedy algorithm. The idea is to start from the leftmost tap and try to water as much of the garden as possible. Then, move to the next tap to the right and repeat the process.

Here is the pseudocode for the algorithm:

1. Let left be the leftmost tap and right be the rightmost tap.
2. While left <= right:
a. Try to water the garden using the left tap.
b. If the garden is not fully watered, move to the next tap to the right and repeat the process.
3. Return the minimum number of taps needed to water the garden.

The time complexity of the algorithm is O(n), where n is the number of taps.```
```There are two possible solutions for this problem:

1) Use a min-heap to keep track of the next available tap. Iterate through the garden and for each row, check if the tap is open. If it is, water that row and update the next available tap. If the tap is not open, update the next available tap and continue.

2) Use a greedy approach. Iterate through the garden and for each row, check if the tap is open. If it is, water that row and update the next available tap. If the tap is not open, continue to the next row.```
```var minTaps = function(n, ranges) {
// create an array of length n + 1, with all elements initialized to 0
let dp = new Array(n + 1).fill(0);

// iterate through the ranges array
for (let i = 0; i < ranges.length; i++) {
// for each element in the ranges array, set the start index to be the maximum of either the current element's value or the index - the range
// set the end index to be the minimum of either the current element's value or the index + the range
let start = Math.max(0, i - ranges[i]);
let end = Math.min(n, i + ranges[i]);

// iterate through the dp array, from the start index to the end index
for (let j = start; j <= end; j++) {
// if the current dp element is less than i + 1 (the number of taps), set it to i + 1
if (dp[j] < i + 1) {
dp[j] = i + 1;
}
}
}

// return the last element in the dp array
return dp[n];
};```
```int minTaps(int n, vector& ranges) {
// Create a vector to store the maximum range
// of each tap
vector maxRange(n + 1, 0);
for (int i = 0; i <= n; i++) {
maxRange[i] = ranges[i];
}

// dp[i] stores the minimum number of taps
// required to water the garden from 0 to i
vector dp(n + 1, -1);

// Base case
dp = 0;

// Consider all possible taps
for (int i = 0; i <= n; i++) {

// If this tap cannot reach the garden,
// then continue
if (maxRange[i] == 0) {
continue;
}

// Consider all positions reachable from
// this tap
for (int j = max(0, i - maxRange[i]);
j <= min(n, i + maxRange[i]); j++) {

// If this position can be reached
// from the current tap and it requires
// less or equal number of taps as compared
// to the current result, update it
if (dp[j] != -1 && dp[j] <= dp[i] + 1) {
dp[j] = dp[i] + 1;
}
}
}

// Return the result for watering the entire garden
return dp[n];
}```
```public class Solution {
public int MinTaps(int n, int[] ranges) {
// this problem can be solved using a greedy approach
// we keep track of the current maximum reachable index
// as well as the next maximum reachable index
// whenever the current maximum reachable index is less than the next maximum reachable index
// we know we can move forward by one tap, and update the current and next maximum reachable indices accordingly
int currMax = 0;
int nextMax = 0;
int taps = 0;
int i = 0;
while (i <= n) {
// if the current index is within reach of the current maximum reachable index
// update the next maximum reachable index to be the maximum of the current next maximum reachable index and the range at the current index + the current index
if (i <= currMax) {
nextMax = Math.Max(nextMax, i + ranges[i]);
}
// if the current index is greater than the current maximum reachable index
// and the current maximum reachable index is less than the next maximum reachable index
// we know we can move forward by one tap, so we update the current maximum reachable index to be the next maximum reachable index
// and increment the number of taps
// we also reset the next maximum reachable index to be 0
if (i == currMax && currMax < nextMax) {
currMax = nextMax;
taps++;
nextMax = 0;
}
// if the current index is greater than the current maximum reachable index
// and the current maximum reachable index is also greater than or equal to the next maximum reachable index
// we know we cannot move forward anymore, so we return -1
if (i > currMax && currMax >= nextMax) {
return -1;
}
i++;
}
return taps;
}
}```

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