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

26 Aug, 2021
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: 1Explanation:For position 1: a[1] = 1, range = 1 to 2For position 2: a[2] = 2, range = 1 to 3For position 3: a[3] = 1, range = 2 to 3Therefore, the fountain at position a[2] covers the whole garden.Therefore, the required output is 1.
Input: a[] = {2, 1, 1, 2, 1} Output: 2 
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
Java
Python3
C#
Javascript
// C++ program to implement// the above approach #include <bits/stdc++.h>using namespace std; // Function to find minimum// number of fountains to be// activatedint 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<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 = 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    int cntfount = 1;     idxRight = dp[0];     // Stores index of next fountain    // that needed to be activated    // initializing idxNext with -1    // so that we don't get garbage value    int idxNext=-1;     // Traverse dp[] array    for (int i = 0; i < N; i++)    {        idxNext = max(idxNext,                      dp[i]);         // If left most fountain        // cover all its range        if (i == idxRight)        {            cntfount++;            idxRight = idxNext;        }    }     return cntfount;} // Driver Codeint main(){    int a[] = { 1, 2, 1 };    int N = sizeof(a) / sizeof(a[0]);    cout << minCntFoun(a, N);}
// Java program to implement// the above approachimport 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<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;         System.out.print(minCntFoun(a, N));    }} // This code is contributed by Amit Katiyar
# 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 codeif __name__ == '__main__':     a = [1, 2, 1]    N = len(a)     print(minCntFoun(a, N)) # This code is contributed by Shivam Singh
// C# program to implement// the above approachusing 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
<script> // Javascript program to implement// the above approach // Function to find minimum// number of fountains to be// activatedfunction 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    let dp = [];    for(let i = 0; i < N; i++)    {         dp[i] = -1;    }     // Stores index of leftmost fountain    // in the range of i-th fountain    let idxLeft;     // Stores index of rightmost fountain    // in the range of i-th fountain    let idxRight;     // Traverse the array    for(let 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    let cntfount = 1;     // Stores index of next fountain    // that needed to be activated    let idxNext = 0;    idxRight = dp[0];     // Traverse dp[] array    for(let 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 Codelet a = [ 1, 2, 1 ];let N = a.length; document.write(minCntFoun(a, N)); // This code is contributed by souravghosh0416 </script>
  
1
 
Time Complexity: O(N)Auxiliary Space: O(N)
 
SHIVAMSINGH67
amit143katiyar
GauravRajput1
sharmadevesh360
souravghosh0416
kanika199425jain
ruhelaa48
Codenation
Gameskraft Technologies
interview-preparation
Morgan Stanley
two-pointer-algorithm
Arrays
Dynamic Programming
Mathematical
Morgan Stanley
Codenation
two-pointer-algorithm
Arrays
Dynamic Programming
Mathematical
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