# Count of vessels completely filled after a given time

Given two integers N and T denoting the number of levels and the number of seconds respectively, the task is to find the number of completely filled vessels after T seconds under given conditions:

• A structure of vessels of N levels is such that the number of the vessels at each level is equal to the level number i.e 1, 2, 3, … up to N.
• Each vessel can store a maximum of 1 unit of water and in every second 1 unit water is poured out from a tap at a constant rate.
• If the vessel becomes full, then water starts flowing out of it, and pours over the edges of the vessel and is equally distributed over the two connected vessels immediately below it.

Assumptions:

1. All the objects are arranged symmetrically along the horizontal axis.
2. All levels are equally spaced.
3. Water flows symmetrically over both the edges of the vessel.

Examples:

Input: N = 3, T = 2
Output: 1
Explanation:
View of Structure with N = 3 and at a time T = 2 after the tap has been opened

Input: N = 3, T = 4
Output: 3
Explanation:
View of Structure with N = 3 and at a time T = 4 after the tap has been opened

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach: The simplest approach to solve the problem is to check if it is possible to completely fill x vessels in T seconds or not. If found to be true, check for x+1 vessels and repeat so on to obtain the maximum value of x.
Time Complexity: O(N3)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized using Dynamic Programming. Follow the steps below to solve the problem:

1. Store the vessel structure in a Matrix, say M, where M[i][j] denotes the jth vessel in the ith level.
2. For any vessel M[i][j], the connected vessels at an immediately lower level are M[i + 1][j] and M[i + 1][j + 1].
3. Initially, put all water in the first vessel i, e. M[0][0] = t.
4. Recalculate the state of the matrix at every increment of unit time, starting from the topmost vessel i, e. M[0][0] = t.
5. If the amount of water exceeds the volume of the vessel, the amount flowing down from a vessel is split into 2 equal parts filling the two connected vessels at immediately lower level.
6. Below is the implementation of the above approach:

## C++

 // C++ program to implement  // the above approach  #include   using namespace std;     int n, t;    // Function to find the number of // completely filled vessels int FindNoOfFullVessels(int n, int t) {            // Store the vessels     double Matrix[n][n];        // Assuming all water is present     // in the vessel at the first level     Matrix[0][0] = t * 1.0;        // Store the number of vessel     // that are completely full     int ans = 0;        // Traverse all the levels     for(int i = 0; i < n; i++)     {                    // Number of vessel at each         // level is j         for(int j = 0; j <= i; j++)         {                            // Calculate the exceeded             // amount of water             double exceededwater = Matrix[i][j] - 1.0;                // If current vessel has             // less than 1 unit of             // water then continue             if (exceededwater < 0)                 continue;                // One more vessel is full             ans++;                // If left bottom vessel present             if (i + 1 < n)                 Matrix[i + 1][j] += exceededwater / 2;                // If right bottom vessel present             if (i + 1 < n && j + 1 < n)                 Matrix[i + 1][j + 1] += exceededwater / 2;         }     }     return ans; }    // Driver Code int main() {            // Number of levels     int N = 3;        // Number of seconds     int T = 4;        // Function call     cout << FindNoOfFullVessels(N, T) << endl;            return 0; }    // This code is contributed by sanjoy_62

## Java

 // Java Program to implement // the above approach    import java.io.*; import java.util.*;    class GFG {        static int n, t;        // Function to find the number of     // completely filled vessels     public static int     FindNoOfFullVessels(int n, int t)     {         // Store the vessels         double Matrix[][]             = new double[n][n];            // Assuming all water is present         // in the vessel at the first level         Matrix[0][0] = t * 1.0;            // Store the number of vessel         // that are completely full         int ans = 0;            // Traverse all the levels         for (int i = 0; i < n; i++) {                // Number of vessel at each             // level is j             for (int j = 0; j <= i; j++) {                    // Calculate the exceeded                 // amount of water                 double exceededwater                     = Matrix[i][j] - 1.0;                    // If current vessel has                 // less than 1 unit of                 // water then continue                 if (exceededwater < 0)                     continue;                    // One more vessel is full                 ans++;                    // If left bottom vessel present                 if (i + 1 < n)                     Matrix[i + 1][j]                         += exceededwater / 2;                    // If right bottom vessel present                 if (i + 1 < n && j + 1 < n)                     Matrix[i + 1][j + 1]                         += exceededwater / 2;             }         }            return ans;     }        // Driver Code     public static void main(String[] args)     {         // Number of levels         int N = 3;            // Number of seconds         int T = 4;            // Function call         System.out.println(             FindNoOfFullVessels(N, T));     } }

## C#

 // C# program to implement  // the above approach  using System;     class GFG{    //static int n, t;    // Function to find the number of // completely filled vessels public static int FindNoOfFullVessels(int n,                                        int t) {            // Store the vessels     double[,] Matrix = new double[n, n];        // Assuming all water is present     // in the vessel at the first level     Matrix[0, 0] = t * 1.0;        // Store the number of vessel     // that are completely full     int ans = 0;        // Traverse all the levels     for(int i = 0; i < n; i++)     {            // Number of vessel at each         // level is j         for(int j = 0; j <= i; j++)          {                // Calculate the exceeded             // amount of water             double exceededwater = Matrix[i, j] - 1.0;                // If current vessel has             // less than 1 unit of             // water then continue             if (exceededwater < 0)                 continue;                // One more vessel is full             ans++;                // If left bottom vessel present             if (i + 1 < n)                 Matrix[i + 1, j] += exceededwater / 2;                // If right bottom vessel present             if (i + 1 < n && j + 1 < n)                 Matrix[i + 1, j + 1] += exceededwater / 2;         }     }     return ans; }    // Driver Code public static void Main() {            // Number of levels     int N = 3;        // Number of seconds     int T = 4;        // Function call     Console.WriteLine(FindNoOfFullVessels(N, T)); } }    // This code is contributed by sanjoy_62

Output:

3

Time Complexity: O(N2)
Auxiliary Space: O(N2)

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Improved By : sanjoy_62