Program to find amount of water in a given glass

• Difficulty Level : Hard
• Last Updated : 11 Jan, 2022

There are some glasses with equal capacity as 1 litre. The glasses are kept as follows:

1
2   3
4    5    6
7    8    9   10

You can put water to the only top glass. If you put more than 1-litre water to 1st glass, water overflows and fills equally in both 2nd and 3rd glasses. Glass 5 will get water from both 2nd glass and 3rd glass and so on.

If you have X litre of water and you put that water in a top glass, how much water will be contained by the jth glass in an ith row?

Example. If you will put 2 litres on top.
1st – 1 litre
2nd – 1/2 litre
3rd – 1/2 litre For 2 Liters Water

The approach is similar to Method 2 of the Pascal’s Triangle. If we take a closer look at the problem, the problem boils down to Pascal’s Triangle.

1   ---------------- 1
2   3 ---------------- 2
4    5    6  ------------ 3
7    8    9   10  --------- 4

Each glass contributes to the two glasses down the glass. Initially, we put all water in the first glass. Then we keep 1 litre (or less than 1 litre) in it and move rest of the water to two glasses down to it. We follow the same process for the two glasses and all other glasses till the ith row. There will be i*(i+1)/2 glasses till ith row.

C++

 // Program to find the amount of water in j-th glass// of i-th row#include #include #include  // Returns the amount of water in jth glass of ith rowfloat findWater(int i, int j, float X){    // A row number i has maximum i columns. So input    // column number must be less than i    if (j > i)    {        printf("Incorrect Inputn");        exit(0);    }     // There will be i*(i+1)/2 glasses till ith row    // (including ith row)    float glass[i * (i + 1) / 2];     // Initialize all glasses as empty    memset(glass, 0, sizeof(glass));     // Put all water in first glass    int index = 0;    glass[index] = X;     // Now let the water flow to the downward glasses    // till the row number is less than or/ equal to i (given row)    // correction : X can be zero for side glasses as they have lower rate to fill    for (int row = 1; row <= i ; ++row)    {        // Fill glasses in a given row. Number of        // columns in a row is equal to row number        for (int col = 1; col <= row; ++col, ++index)        {            // Get the water from current glass            X = glass[index];             // Keep the amount less than or equal to            // capacity in current glass            glass[index] = (X >= 1.0f) ? 1.0f : X;             // Get the remaining amount            X = (X >= 1.0f) ? (X - 1) : 0.0f;             // Distribute the remaining amount to            // the down two glasses            glass[index + row] += X / 2;            glass[index + row + 1] += X / 2;        }    }     // The index of jth glass in ith row will    // be i*(i-1)/2 + j - 1    return glass[i*(i-1)/2 + j - 1];} // Driver program to test above functionint main(){    int i = 2, j = 2;    float X = 2.0; // Total amount of water     printf("Amount of water in jth glass of ith row is: %f",            findWater(i, j, X));     return 0;}

Java

 // Program to find the amount/// of water in j-th glass// of i-th rowimport java.lang.*; class GFG{// Returns the amount of water// in jth glass of ith rowstatic float findWater(int i, int j,                       float X){// A row number i has maximum i// columns. So input column// number must be less than iif (j > i){    System.out.println("Incorrect Input");    System.exit(0);} // There will be i*(i+1)/2 glasses// till ith row (including ith row)int ll = Math.round((i * (i + 1) ));float[] glass = new float[ll + 2]; // Put all water in first glassint index = 0;glass[index] = X; // Now let the water flow to the// downward glasses till the row// number is less than or/ equal// to i (given row)// correction : X can be zero for side// glasses as they have lower rate to fillfor (int row = 1; row <= i ; ++row){    // Fill glasses in a given row. Number of    // columns in a row is equal to row number    for (int col = 1;             col <= row; ++col, ++index)    {        // Get the water from current glass        X = glass[index];         // Keep the amount less than or        // equal to capacity in current glass        glass[index] = (X >= 1.0f) ? 1.0f : X;         // Get the remaining amount        X = (X >= 1.0f) ? (X - 1) : 0.0f;         // Distribute the remaining amount         // to the down two glasses        glass[index + row] += X / 2;        glass[index + row + 1] += X / 2;    }} // The index of jth glass in ith// row will be i*(i-1)/2 + j - 1return glass[(int)(i * (i - 1) /                   2 + j - 1)];} // Driver Codepublic static void main(String[] args){    int i = 2, j = 2;    float X = 2.0f; // Total amount of water    System.out.println("Amount of water in jth " +                         "glass of ith row is: " +                              findWater(i, j, X));}} // This code is contributed by mits

Python3

 # Program to find the amount# of water in j-th glass of# i-th row # Returns the amount of water# in jth glass of ith rowdef findWater(i, j, X):    # A row number i has maximum    # i columns. So input column    # number must be less than i    if (j > i):        print("Incorrect Input");        return;     # There will be i*(i+1)/2    # glasses till ith row    # (including ith row)    # and Initialize all glasses    # as empty    glass = *int(i *(i + 1) / 2);     # Put all water    # in first glass    index = 0;    glass[index] = X;     # Now let the water flow to    # the downward glasses till    # the row number is less    # than or/ equal to i (given    # row) correction : X can be    # zero for side glasses as    # they have lower rate to fill    for row in range(1,i):        # Fill glasses in a given        # row. Number of columns        # in a row is equal to row number        for col in range(1,row+1):            # Get the water            # from current glass            X = glass[index];             # Keep the amount less            # than or equal to            # capacity in current glass            glass[index] = 1.0 if (X >= 1.0) else X;             # Get the remaining amount            X = (X - 1) if (X >= 1.0) else 0.0;             # Distribute the remaining            # amount to the down two glasses            glass[index + row] += (X / 2);            glass[index + row + 1] += (X / 2);            index+=1;     # The index of jth glass    # in ith row will    # be i*(i-1)/2 + j - 1    return glass[int(i * (i - 1) /2 + j - 1)]; # Driver Codeif __name__ == "__main__":         i = 2;    j = 2;    X = 2.0;# Total amount of water     res=repr(findWater(i, j, X));    print("Amount of water in jth glass of ith row is:",res.ljust(8,'0'));# This Code is contributed by mits

C#

 // Program to find the amount// of water in j-th glass// of i-th rowusing System; class GFG{// Returns the amount of water// in jth glass of ith rowstatic float findWater(int i, int j,                       float X){// A row number i has maximum i// columns. So input column// number must be less than iif (j > i){    Console.WriteLine("Incorrect Input");    Environment.Exit(0);} // There will be i*(i+1)/2 glasses// till ith row (including ith row)int ll = (int)Math.Round((double)(i * (i + 1)));float[] glass = new float[ll + 2]; // Put all water in first glassint index = 0;glass[index] = X; // Now let the water flow to the// downward glasses till the row// number is less than or/ equal// to i (given row)// correction : X can be zero// for side glasses as they have// lower rate to fillfor (int row = 1; row <= i ; ++row){    // Fill glasses in a given row.    // Number of columns in a row    // is equal to row number    for (int col = 1;            col <= row; ++col, ++index)    {        // Get the water from current glass        X = glass[index];         // Keep the amount less than        // or equal to capacity in        // current glass        glass[index] = (X >= 1.0f) ?                              1.0f : X;         // Get the remaining amount        X = (X >= 1.0f) ? (X - 1) : 0.0f;         // Distribute the remaining amount        // to the down two glasses        glass[index + row] += X / 2;        glass[index + row + 1] += X / 2;    }} // The index of jth glass in ith// row will be i*(i-1)/2 + j - 1return glass[(int)(i * (i - 1) /                   2 + j - 1)];} // Driver Codestatic void Main(){    int i = 2, j = 2;    float X = 2.0f; // Total amount of water    Console.WriteLine("Amount of water in jth " +                        "glass of ith row is: " +                             findWater(i, j, X));}} // This code is contributed by mits

PHP

 \$i)    {        echo "Incorrect Input\n";        return;    }     // There will be i*(i+1)/2    // glasses till ith row    // (including ith row)    // and Initialize all glasses    // as empty    \$glass = array_fill(0, (int)(\$i *                       (\$i + 1) / 2), 0);     // Put all water    // in first glass    \$index = 0;    \$glass[\$index] = \$X;     // Now let the water flow to    // the downward glasses till    // the row number is less    // than or/ equal to i (given     // row) correction : X can be    // zero for side glasses as    // they have lower rate to fill    for (\$row = 1; \$row < \$i ; ++\$row)    {        // Fill glasses in a given        // row. Number of columns        // in a row is equal to row number        for (\$col = 1;             \$col <= \$row; ++\$col, ++\$index)        {            // Get the water            // from current glass            \$X = \$glass[\$index];             // Keep the amount less            // than or equal to            // capacity in current glass            \$glass[\$index] = (\$X >= 1.0) ?                                     1.0 : \$X;             // Get the remaining amount            \$X = (\$X >= 1.0) ?                    (\$X - 1) : 0.0;             // Distribute the remaining            // amount to the down two glasses            \$glass[\$index + \$row] += (double)(\$X / 2);            \$glass[\$index + \$row + 1] += (double)(\$X / 2);        }    }     // The index of jth glass    // in ith row will    // be i*(i-1)/2 + j - 1    return \$glass[(int)(\$i * (\$i - 1) /                          2 + \$j - 1)];} // Driver Code\$i = 2;\$j = 2;\$X = 2.0; // Total amount of waterecho "Amount of water in jth " ,        "glass of ith row is: ".       str_pad(findWater(\$i, \$j,                   \$X), 8, '0'); // This Code is contributed by mits?>

Javascript



Output:

Amount of water in jth glass of ith row is: 0.500000

Time Complexity: O(i*(i+1)/2) or O(i^2)
Auxiliary Space: O(i*(i+1)/2) or O(i^2)

Method 2 (Using BFS Traversal)

we will discuss another approach to this problem. First, we add a Triplet(row, col,rem-water) in the queue which indicates the starting value of the first element and fills 1-litre water.  Then we simply apply bfs i.e. and we add left(row+1, col-1,rem-water) Triplet and right(row+1, col+1,rem-water) Triplet into the queue with half of the remaining water in first Triplet and another half into the next Triplet.

Following is the implementation of this solution.

C++

 // CPP program for above approach#includeusing namespace std; // Program to find the amount // of water in j-th glass// of i-th rowvoid findWater(float k, int i, int j){         // stores how much amount of water    // present in every glass     float dp[i+1][2*i + 1];    bool vis[i+1][2*i + 1];         // initialize dp and visited arrays    for(int n=0;n>q;    dp[i] = k;         // we take the center of the first row for    // the location of the first glass    q.push({0,i});    vis[i] = true;         // this while loop runs unless the queue is not empty    while(!q.empty())    {        // First we remove the pair from the queue        pairtemp = q.front();        q.pop();        int n = temp.first;        int m = temp.second;                          // as we know we have  to calculate the        // amount of water in jth glass        // of ith row . so first we check if we find solutions        // for the the every glass of i'th row.        // so, if we have calculated the result        // then we break the loop        // and return our answer        if(i == n)            break;                                  float x = dp[n][m];        if(float((x-1.0)/2.0) < 0)        {            dp[n+1][m-1] += 0;            dp[n+1][m+1] += 0;        }        else        {            dp[n+1][m-1] += float((x-1.0)/2.0);            dp[n+1][m+1] += float((x-1.0)/2.0);           }        if(vis[n+1][m-1]==false)        {            q.push({n+1,m-1});            vis[n+1][m-1] = true;        }        if(vis[n+1][m+1]==false)        {            q.push({n+1,m+1});            vis[n+1][m+1] = true;        }    }        if(dp[i-1][2*j-1]>1)        dp[i-1][2*j-1] = 1.0;             cout<<"Amount of water in jth glass of ith row is: ";         // print the result for jth glass of ith row    cout<water in litres    float k;     cin>>k;       //i->rows  j->cols    int i,j;    cin>>i>>j;        // Function Call     findWater(k,i,j);    return 0;}  // This code is contributed by Sagar Jangra and Naresh Saharan

Java

 // Program to find the amount /// of water in j-th glass// of i-th rowimport java.io.*;import java.util.*; // class Triplet which stores curr row// curr col and curr rem-water// of every glassclass Triplet{  int row;  int col;  double rem_water;  Triplet(int row,int col,double rem_water)  {    this.row=row;this.col=col;this.rem_water=rem_water;  }}  class GFG{  // Returns the amount of water  // in jth glass of ith row  public static double findWater(int i,int j,int totalWater)  {     // stores how much amount of water present in every glass      double dp[][] = new double[i+1][2*i+1];     // store Triplet i.e curr-row , curr-col, rem-water    Queue queue = new LinkedList<>();     // we take the center of the first row for    // the location of the first glass    queue.add(new Triplet(0,i,totalWater));      // this while loop runs unless the queue is not empty    while(!queue.isEmpty())    {      // First we remove the Triplet from the queue      Triplet curr = queue.remove();       // as we know we have  to calculate the      // amount of water in jth glass      // of ith row . so first we check if we find solutions      // for the the every glass of i'th row.      // so, if we have calculated the result      // then we break the loop      // and return our answer      if(curr.row == i)        break;       // As we know maximum capacity of every glass      // is 1 unit. so first we check the capacity      // of curr glass is full or not.      if(dp[curr.row][curr.col] != 1.0)      {        // calculate the remaining capacity of water for curr glass        double rem_water = 1-dp[curr.row][curr.col];         // if the remaining capacity of water for curr glass        // is greater than then the remaining capacity of tatal        // water then we put all remaining water into curr glass        if(rem_water > curr.rem_water)        {          dp[curr.row][curr.col] += curr.rem_water;          curr.rem_water = 0;        }        else        {          dp[curr.row][curr.col] += rem_water;          curr.rem_water -= rem_water;        }      }       // if remaining capacity of tatal water is not equal to      // zero then we add left and right glass of the next row      // and gives half capacity of tatal water to both the      // glass      if(curr.rem_water != 0)      {        queue.add(new Triplet(curr.row + 1,curr.col -                                  1,curr.rem_water/2.0));        queue.add(new Triplet(curr.row + 1,curr.col +                                  1,curr.rem_water/2.0));      }    }     // return the result for jth glass of ith row    return dp[i-1][2*j-1];  }   // Driver Code  public static void main (String[] args)  {    int i = 2, j = 2;    int totalWater = 2; // Total amount of water    System.out.print("Amount of water in jth glass of ith row is:");    System.out.format("%.6f", findWater(i, j, totalWater));    } }  // this code is contributed by  Naresh Saharan and Sagar Jangra

Python3

 # Program to find the amount# of water in j-th glass of i-th row # class Triplet which stores curr row# curr col and curr rem-water# of every glassclass Triplet:    def __init__(self, row, col, rem_water):        self.row = row        self.col = col        self.rem_water = rem_water # Returns the amount of water# in jth glass of ith rowdef findWater(i, j, totalWater):       # stores how much amount of water present in every glass     dp = [[0.0 for i in range(2*i+1)] for j in range(i+1)]      # store Triplet i.e curr-row , curr-col, rem-water    queue = []      # we take the center of the first row for    # the location of the first glass    queue.append(Triplet(0,i,totalWater))      # this while loop runs unless the queue is not empty    while len(queue) != 0:      # First we remove the Triplet from the queue      curr = queue.pop(0)      # as we know we have  to calculate the      # amount of water in jth glass      # of ith row . so first we check if we find solutions      # for the the every glass of i'th row.      # so, if we have calculated the result      # then we break the loop      # and return our answer      if curr.row == i:        break        # As we know maximum capacity of every glass      # is 1 unit. so first we check the capacity      # of curr glass is full or not.      if dp[curr.row][curr.col] != 1.0:                 # calculate the remaining capacity of water for curr glass        rem_water = 1-dp[curr.row][curr.col]          # if the remaining capacity of water for curr glass        # is greater than then the remaining capacity of tatal        # water then we put all remaining water into curr glass        if rem_water > curr.rem_water:          dp[curr.row][curr.col] += curr.rem_water          curr.rem_water = 0        else:          dp[curr.row][curr.col] += rem_water          curr.rem_water -= rem_water        # if remaining capacity of tatal water is not equal to      # zero then we add left and right glass of the next row      # and gives half capacity of tatal water to both the      # glass      if curr.rem_water != 0:        queue.append(Triplet(curr.row + 1,curr.col - 1,(curr.rem_water/2)))        queue.append(Triplet(curr.row + 1,curr.col + 1,(curr.rem_water/2)))      # return the result for jth glass of ith row    return dp[i - 1][2 * j - 1] i, j = 2, 2totalWater = 2 # Total amount of waterprint("Amount of water in jth glass of ith row is:", end = "")print("{0:.6f}".format(findWater(i, j, totalWater))) # This code is contributed by decode2207.

C#

 // Program to find the amount/// of water in j-th glassusing System;using System.Collections.Generic;class GFG {         // class Triplet which stores curr row    // curr col and curr rem-water    // of every glass    class Triplet {                public int row, col;        public double rem_water;                public Triplet(int row, int col, double rem_water)        {            this.row = row;            this.col = col;            this.rem_water = rem_water;        }    }       // Returns the amount of water  // in jth glass of ith row  public static double findWater(int i, int j, int totalWater)  {      // stores how much amount of water present in every glass     double[,] dp = new double[i+1,2*i+1];      // store Triplet i.e curr-row , curr-col, rem-water    List queue = new List();      // we take the center of the first row for    // the location of the first glass    queue.Add(new Triplet(0,i,totalWater));        // this while loop runs unless the queue is not empty    while(queue.Count > 0)    {      // First we remove the Triplet from the queue      Triplet curr = queue;      queue.RemoveAt(0);        // as we know we have  to calculate the      // amount of water in jth glass      // of ith row . so first we check if we find solutions      // for the the every glass of i'th row.      // so, if we have calculated the result      // then we break the loop      // and return our answer      if(curr.row == i)        break;        // As we know maximum capacity of every glass      // is 1 unit. so first we check the capacity      // of curr glass is full or not.      if(dp[curr.row,curr.col] != 1.0)      {        // calculate the remaining capacity of water for curr glass        double rem_water = 1-dp[curr.row,curr.col];          // if the remaining capacity of water for curr glass        // is greater than then the remaining capacity of tatal        // water then we put all remaining water into curr glass        if(rem_water > curr.rem_water)        {          dp[curr.row,curr.col] += curr.rem_water;          curr.rem_water = 0;        }        else        {          dp[curr.row,curr.col] += rem_water;          curr.rem_water -= rem_water;        }      }        // if remaining capacity of tatal water is not equal to      // zero then we add left and right glass of the next row      // and gives half capacity of tatal water to both the      // glass      if(curr.rem_water != 0)      {        queue.Add(new Triplet(curr.row + 1,curr.col -                                  1,curr.rem_water/2.0));        queue.Add(new Triplet(curr.row + 1,curr.col +                                  1,curr.rem_water/2.0));      }    }      // return the result for jth glass of ith row    return dp[i - 1, 2 * j - 1];  }   static void Main() {    int i = 2, j = 2;    int totalWater = 2; // Total amount of water    Console.Write("Amount of water in jth glass of ith row is:");    Console.Write(findWater(i, j, totalWater).ToString("0.000000"));   }} // This code is contributed by divyeshrabadiya07.

Javascript


Output
Amount of water in jth glass of ith row is:0.500000

Time Complexity:  O(2^i)

Space Complexity:  O(i^2)

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