# Count minimum number of fountains to be activated to cover the entire garden

There is a one-dimensional garden of length N. In each position of the N length garden, a fountain has been installed. Given an array a[]such that a[i] describes the coverage limit of ith fountain. A fountain can cover the range from the position max(i – a[i], 1) to min(i + a[i], N). In beginning, all the fountains are switched off. The task is to find the minimum number of fountains needed to be activated such that the whole N-length garden can be covered by water.

Examples:

Input: a[] = {1, 2, 1}
Output: 1
Explanation:
For position 1: a[1] = 1, range = 1 to 2
For position 2: a[2] = 2, range = 1 to 3
For position 3: a[3] = 1, range = 2 to 3
Therefore, the fountain at position a[2] covers the whole garden.
Therefore, the required output is 1.

Input: a[] = {2, 1, 1, 2, 1}
Output:

Approach: The problem can be solved using Dynamic Programming. Follow the steps below to solve the problem:

• traverse the array and for every array index, i.e. ith fountain, find the leftmost fountain up to which the current fountain covers.
• Then, find the rightmost fountain that the leftmost fountain obtained in the above step covers up to and update it in the dp[] array.
• Initialize a variable cntFount to store the minimum number of fountains that need to be activated.
• Now, traverse the dp[] array and keep activating the fountains from the left that covers maximum fountains currently on the right and increment cntFount by 1. Finally, print cntFount as the required answer.

Below is the implementation of the above approach.

## C++14

 `// C++ program to implement``// the above approach` `#include ``using` `namespace` `std;` `// Function to find minimum``// number of fountains to be``// activated``int` `minCntFoun(``int` `a[], ``int` `N)``{` `    ``// dp[i]: Stores the position of``    ``// rightmost fountain that can``    ``// be covered by water of leftmost``    ``// fountain of the i-th fountain``    ``int` `dp[N];``    ` `    ``// initializing all dp[i] values to be -1, ``    ``// so that we don't get garbage value``      ``for``(``int` `i=0;i

## Java

 `// Java program to implement``// the above approach``import` `java.util.*;` `class` `GFG {` `    ``// Function to find minimum``    ``// number of fountains to be``    ``// activated``    ``static` `int` `minCntFoun(``int` `a[], ``int` `N)``    ``{` `        ``// dp[i]: Stores the position of``        ``// rightmost fountain that can``        ``// be covered by water of leftmost``        ``// fountain of the i-th fountain``        ``int``[] dp = ``new` `int``[N];``        ``for``(``int` `i=``0``;i

## Python3

 `# Python3 program to implement``# the above approach` `# Function to find minimum``# number of fountains to be``# activated`  `def` `minCntFoun(a, N):` `    ``# dp[i]: Stores the position of``    ``# rightmost fountain that can``    ``# be covered by water of leftmost``    ``# fountain of the i-th fountain``    ``dp ``=` `[``0``] ``*` `N``    ``for` `i ``in` `range``(N):``      ``dp[i] ``=` `-``1` `    ``# Traverse the array``    ``for` `i ``in` `range``(N):``        ``idxLeft ``=` `max``(i ``-` `a[i], ``0``)``        ``idxRight ``=` `min``(i ``+` `(a[i] ``+` `1``), N)``        ``dp[idxLeft] ``=` `max``(dp[idxLeft],``                          ``idxRight)` `    ``# Stores count of fountains``    ``# needed to be activated``    ``cntfount ``=` `1` `    ``idxRight ``=` `dp[``0``]` `    ``# Stores index of next fountain``    ``# that needed to be activated``    ``idxNext ``=` `0` `    ``# Traverse dp[] array``    ``for` `i ``in` `range``(N):``        ``idxNext ``=` `max``(idxNext,``                      ``dp[i])` `        ``# If left most fountain``        ``# cover all its range``        ``if` `(i ``=``=` `idxRight):``            ``cntfount ``+``=` `1``            ``idxRight ``=` `idxNext` `    ``return` `cntfount`  `# Driver code``if` `__name__ ``=``=` `'__main__'``:` `    ``a ``=` `[``1``, ``2``, ``1``]``    ``N ``=` `len``(a)` `    ``print``(minCntFoun(a, N))` `# This code is contributed by Shivam Singh`

## C#

 `// C# program to implement``// the above approach``using` `System;``class` `GFG {` `    ``// Function to find minimum``    ``// number of fountains to be``    ``// activated``    ``static` `int` `minCntFoun(``int``[] a, ``int` `N)``    ``{``        ``// dp[i]: Stores the position of``        ``// rightmost fountain that can``        ``// be covered by water of leftmost``        ``// fountain of the i-th fountain``        ``int``[] dp = ``new` `int``[N];``        ``for` `(``int` `i = 0; i < N; i++) ``        ``{``            ``dp[i] = -1;``        ``}` `        ``// Stores index of leftmost``        ``// fountain in the range of``        ``// i-th fountain``        ``int` `idxLeft;` `        ``// Stores index of rightmost``        ``// fountain in the range of``        ``// i-th fountain``        ``int` `idxRight;` `        ``// Traverse the array``        ``for` `(``int` `i = 0; i < N; i++) {``            ``idxLeft = Math.Max(i - a[i], 0);``            ``idxRight = Math.Min(i + (a[i] + 1), ``                                ``N);``            ``dp[idxLeft] = Math.Max(dp[idxLeft], ``                                   ``idxRight);``        ``}` `        ``// Stores count of fountains``        ``// needed to be activated``        ``int` `cntfount = 1;` `        ``// Stores index of next``        ``// fountain that needed``        ``// to be activated``        ``int` `idxNext = 0;``        ``idxRight = dp[0];` `        ``// Traverse []dp array``        ``for` `(``int` `i = 0; i < N; i++) ``        ``{``            ``idxNext = Math.Max(idxNext, dp[i]);` `            ``// If left most fountain``            ``// cover all its range``            ``if` `(i == idxRight) ``            ``{``                ``cntfount++;``                ``idxRight = idxNext;``            ``}``        ``}``        ``return` `cntfount;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int``[] a = { 1, 2, 1 };``        ``int` `N = a.Length;` `        ``Console.Write(minCntFoun(a, N));``    ``}``}` `// This code is contributed by gauravrajput1`

## Javascript

 ``

Output
```1

```

Time Complexity: O(N)
Auxiliary Space: O(N)

### Brute Force in python:

Approach:

The brute force approach involves checking all possible combinations of fountains that can be activated to cover the entire garden. For each combination, we check if it covers the entire garden and keep track of the minimum number of fountains required.

Initialize the minimum number of fountains to a very large number.
Use a loop to iterate over all possible combinations of fountains that can be activated.
For each combination, check if it covers the entire garden using a helper function is_covered.
If it does, update the minimum number of fountains required.
Return the minimum number of fountains required.

## C++

 `#include ``#include ``#include   // Add this header for std::all_of``#include      // Add this header for std::numeric_limits` `bool` `isCovered(``const` `std::vector<``int``>& activated, ``const` `std::vector<``int``>& fountains) {``    ``int` `n = fountains.size();``    ``std::vector<``int``> coverage(n, 0);` `    ``for` `(``int` `i : activated) {``        ``int` `left = std::max(0, i - fountains[i]);``        ``int` `right = std::min(n - 1, i + fountains[i]);` `        ``for` `(``int` `j = left; j <= right; j++) {``            ``coverage[j] = 1;``        ``}``    ``}` `    ``// Check if all positions are covered``    ``return` `std::all_of(coverage.begin(), coverage.end(), [](``int` `val) { ``return` `val == 1; });``}` `int` `activateFountains(``const` `std::vector<``int``>& fountains) {``    ``int` `n = fountains.size();``    ``int` `minFountains = std::numeric_limits<``int``>::max();` `    ``for` `(``int` `i = 1; i < (1 << n); i++) {``        ``std::vector<``int``> activated;``        ` `        ``// Extract indices of activated fountains based on the binary representation of 'i'``        ``for` `(``int` `j = 0; j < n; j++) {``            ``if` `((i >> j) & 1) {``                ``activated.push_back(j);``            ``}``        ``}` `        ``if` `(isCovered(activated, fountains)) {``            ``minFountains = std::min(minFountains, ``static_cast``<``int``>(activated.size()));``        ``}``    ``}` `    ``return` `minFountains;``}` `int` `main() {``    ``// Example usage``    ``std::vector<``int``> a1 = {1, 2, 1};``    ``std::vector<``int``> a2 = {2, 1, 1, 2, 1};` `    ``std::cout << activateFountains(a1) << std::endl; ``// Output: 1``    ``std::cout << activateFountains(a2) << std::endl; ``// Output: 2` `    ``return` `0;``}`

## Java

 `import` `java.util.ArrayList;``import` `java.util.List;` `public` `class` `FountainActivation {``    ``public` `static` `boolean` `isCovered(List activated, List fountains) {``        ``int` `n = fountains.size();``        ``List coverage = ``new` `ArrayList<>(n);` `        ``for` `(``int` `i = ``0``; i < n; i++) {``            ``coverage.add(``0``);``        ``}` `        ``for` `(``int` `i : activated) {``            ``int` `left = Math.max(``0``, i - fountains.get(i));``            ``int` `right = Math.min(n - ``1``, i + fountains.get(i));` `            ``for` `(``int` `j = left; j <= right; j++) {``                ``coverage.set(j, ``1``);``            ``}``        ``}` `        ``// Check if all positions are covered``        ``return` `coverage.stream().allMatch(val -> val == ``1``);``    ``}` `    ``public` `static` `int` `activateFountains(List fountains) {``        ``int` `n = fountains.size();``        ``int` `minFountains = Integer.MAX_VALUE;` `        ``for` `(``int` `i = ``1``; i < (``1` `<< n); i++) {``            ``List activated = ``new` `ArrayList<>();` `            ``// Extract indices of activated fountains based on the binary representation of 'i'``            ``for` `(``int` `j = ``0``; j < n; j++) {``                ``if` `((i >> j & ``1``) == ``1``) {``                    ``activated.add(j);``                ``}``            ``}` `            ``if` `(isCovered(activated, fountains)) {``                ``minFountains = Math.min(minFountains, activated.size());``            ``}``        ``}` `        ``return` `minFountains;``    ``}` `    ``public` `static` `void` `main(String[] args) {``        ``// Example usage``        ``List a1 = List.of(``1``, ``2``, ``1``);``        ``List a2 = List.of(``2``, ``1``, ``1``, ``2``, ``1``);` `        ``System.out.println(activateFountains(a1)); ``// Output: 1``        ``System.out.println(activateFountains(a2)); ``// Output: 2``    ``}``}`

## Python3

 `def` `activate_fountains(fountains):``    ``n ``=` `len``(fountains)``    ``min_fountains ``=` `float``(``'inf'``)``    ``for` `i ``in` `range``(``1``, ``2``*``*``n):``        ``activated ``=` `[]``        ``for` `j ``in` `range``(n):``            ``if` `(i >> j) & ``1``:``                ``activated.append(j)``        ``if` `is_covered(activated, fountains):``            ``min_fountains ``=` `min``(min_fountains, ``len``(activated))``    ``return` `min_fountains` `def` `is_covered(activated, fountains):``    ``n ``=` `len``(fountains)``    ``coverage ``=` `[``0``] ``*` `n``    ``for` `i ``in` `activated:``        ``left ``=` `max``(``0``, i ``-` `fountains[i])``        ``right ``=` `min``(n ``-` `1``, i ``+` `fountains[i])``        ``for` `j ``in` `range``(left, right ``+` `1``):``            ``coverage[j] ``=` `1``    ``return` `all``(coverage)` `# example usage``a1 ``=` `[``1``, ``2``, ``1``]``a2 ``=` `[``2``, ``1``, ``1``, ``2``, ``1``]``print``(activate_fountains(a1)) ``# output: 1``print``(activate_fountains(a2)) ``# output: 2`

## C#

 `using` `System;``using` `System.Collections.Generic;``using` `System.Linq;` `class` `FountainActivation {``    ``static` `bool` `IsCovered(List<``int``> activated,``                          ``List<``int``> fountains)``    ``{``        ``int` `n = fountains.Count;``        ``List<``int``> coverage``            ``= Enumerable.Repeat(0, n).ToList();` `        ``foreach``(``int` `i ``in` `activated)``        ``{``            ``int` `left = Math.Max(0, i - fountains[i]);``            ``int` `right = Math.Min(n - 1, i + fountains[i]);` `            ``for` `(``int` `j = left; j <= right; j++) {``                ``coverage[j] = 1;``            ``}``        ``}` `        ``// Check if all positions are covered``        ``return` `coverage.All(val = > val == 1);``    ``}` `    ``static` `int` `ActivateFountains(List<``int``> fountains)``    ``{``        ``int` `n = fountains.Count;``        ``int` `minFountains = ``int``.MaxValue;` `        ``for` `(``int` `i = 1; i < (1 << n); i++) {``            ``List<``int``> activated = ``new` `List<``int``>();` `            ``// Extract indices of activated fountains based``            ``// on the binary representation of 'i'``            ``for` `(``int` `j = 0; j < n; j++) {``                ``if` `((i >> j & 1) == 1) {``                    ``activated.Add(j);``                ``}``            ``}` `            ``if` `(IsCovered(activated, fountains)) {``                ``minFountains = Math.Min(minFountains,``                                        ``activated.Count);``            ``}``        ``}` `        ``return` `minFountains;``    ``}` `    ``static` `void` `Main()``    ``{``        ``// Example usage``        ``List<``int``> a1 = ``new` `List<``int``>{ 1, 2, 1 };``        ``List<``int``> a2 = ``new` `List<``int``>{ 2, 1, 1, 2, 1 };` `        ``Console.WriteLine(``            ``ActivateFountains(a1)); ``// Output: 1``        ``Console.WriteLine(``            ``ActivateFountains(a2)); ``// Output: 2``    ``}``}`

## Javascript

 `function` `isCovered(activated, fountains) {``    ``const n = fountains.length;``    ``const coverage = Array(n).fill(0);` `    ``for` `(const i of activated) {``        ``const left = Math.max(0, i - fountains[i]);``        ``const right = Math.min(n - 1, i + fountains[i]);` `        ``for` `(let j = left; j <= right; j++) {``            ``coverage[j] = 1;``        ``}``    ``}` `    ``// Check if all positions are covered``    ``return` `coverage.every(val => val === 1);``}` `function` `activateFountains(fountains) {``    ``const n = fountains.length;``    ``let minFountains = Infinity;` `    ``for` `(let i = 1; i < (1 << n); i++) {``        ``const activated = [];` `        ``// Extract indices of activated fountains based on the binary representation of 'i'``        ``for` `(let j = 0; j < n; j++) {``            ``if` `((i >> j & 1) === 1) {``                ``activated.push(j);``            ``}``        ``}` `        ``if` `(isCovered(activated, fountains)) {``            ``minFountains = Math.min(minFountains, activated.length);``        ``}``    ``}` `    ``return` `minFountains;``}` `// Example usage``const a1 = [1, 2, 1];``const a2 = [2, 1, 1, 2, 1];` `console.log(activateFountains(a1)); ``// Output: 1``console.log(activateFountains(a2)); ``// Output: 2`

Output
```1
2

```

The time complexity of this approach is O(2^n) where n is the number of fountains

the space complexity is O(n).

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