# 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& activated, const std::vector& fountains) {     int n = fountains.size();     std::vector 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& fountains) {     int n = fountains.size();     int minFountains = std::numeric_limits::max();       for (int i = 1; i < (1 << n); i++) {         std::vector 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(activated.size()));         }     }       return minFountains; }   int main() {     // Example usage     std::vector a1 = {1, 2, 1};     std::vector 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 activated,                           List fountains)     {         int n = fountains.Count;         List 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 fountains)     {         int n = fountains.Count;         int minFountains = int.MaxValue;           for (int i = 1; i < (1 << n); i++) {             List activated = new List();               // 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 a1 = new List{ 1, 2, 1 };         List a2 = new List{ 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).

Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Previous
Next