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
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.
// Program to find the amount of water in j-th glass // of i-th row #include <stdio.h> #include <stdlib.h> #include <string.h> // Returns the amount of water in jth glass of ith row float 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 function int 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;
} |
// Program to find the amount /// of water in j-th glass // of i-th row import java.lang.*;
class GFG
{ // Returns the amount of water // in jth glass of ith row static float 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)
{ 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 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[( int )(i * (i - 1 ) /
2 + j - 1 )];
} // Driver Code public 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 |
# 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 row def 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 = [ 0 ] * 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 Code if __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 |
// Program to find the amount // of water in j-th glass // of i-th row using System;
class GFG
{ // Returns the amount of water // in jth glass of ith row static float 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)
{ 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 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[( int )(i * (i - 1) /
2 + j - 1)];
} // Driver Code static 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 // 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 row function findWater( $i , $j , $X )
{ // A row number i has maximum
// i columns. So input column
// number must be less than i
if ( $j > $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 water
echo "Amount of water in jth " ,
"glass of ith row is: " .
str_pad (findWater( $i , $j ,
$X ), 8, '0' );
// This Code is contributed by mits ?> |
<script> // 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 row function findWater(i , j, X)
{ // A row number i has maximum i // columns. So input column // number must be less than i if (j > i)
{ document.write( "Incorrect Input" );
} // There will be i*(i+1)/2 glasses // till ith row (including ith row) var ll = Math.round((i * (i + 1) ));
glass = Array.from({length: ll+2}, (_, i) => 0.0); // Put all water in first glass var 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] += 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[parseInt((i * (i - 1) /
2 + j - 1))];
} // Driver Code var i = 2, j = 2;
var X = 2.0; // Total amount of water
document.write( "Amount of water in jth " +
"glass of ith row is: " +
findWater(i, j, X));
// This code is contributed by 29AjayKumar </script> |
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)
This article is compiled by Rahul and reviewed by GeeksforGeeks team.
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.
// CPP program for above approach #include<bits/stdc++.h> using namespace std;
// Program to find the amount // of water in j-th glass // of i-th row void 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<i+1;n++)
{
for ( int m=0;m<(2*i+1);m++)
{
dp[n][m] = 0;
vis[n][m] = false ;
}
}
// store Triplet i.e curr-row , curr-col
queue<pair< int , int >>q;
dp[0][i] = k;
// we take the center of the first row for
// the location of the first glass
q.push({0,i});
vis[0][i] = true ;
// this while loop runs unless the queue is not empty
while (!q.empty())
{
// First we remove the pair from the queue
pair< int , int >temp = 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 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<<fixed<<setprecision(6)<<dp[i-1][2*j-1]<<endl;
} // Driver Code int main()
{ //k->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
|
// Program to find the amount /// of water in j-th glass // of i-th row import java.io.*;
import java.util.*;
// class Triplet which stores curr row // curr col and curr rem-water // of every glass class 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<Triplet> 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 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 total
// 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 total water is not equal to
// zero then we add left and right glass of the next row
// and gives half capacity of total 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
|
# 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 glass class 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 row def 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 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 total
# 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 total water is not equal to
# zero then we add left and right glass of the next row
# and gives half capacity of total 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 , 2
totalWater = 2 # Total amount of water
print ( "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. |
// Program to find the amount /// of water in j-th glass using 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<Triplet> queue = new List<Triplet>();
// 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[0];
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 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 total
// 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 total water is not equal to
// zero then we add left and right glass of the next row
// and gives half capacity of total 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. |
<script> // 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 glass class Triplet { constructor(row,col,rem_water)
{
this .row=row;
this .col=col;
this .rem_water=rem_water;
}
} // Returns the amount of water // in jth glass of ith row
function findWater(i,j,totalWater)
{ // stores how much amount of water present in every glass
let dp = new Array(i+1);
for (let k=0;k<dp.length;k++)
{
dp[k]= new Array((2*i)+1);
for (let l=0;l<dp[k].length;l++)
{
dp[k][l]=0;
}
}
// store Triplet i.e curr-row , curr-col, rem-water
let queue = [];
// we take the center of the first row for
// the location of the first glass
queue.push( new Triplet(0,i,totalWater));
// this while loop runs unless the queue is not empty
while (queue.length!=0)
{
// First we remove the Triplet from the queue
let curr = queue.shift();
// 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 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
let 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 total
// 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 total water is not equal to
// zero then we add left and right glass of the next row
// and gives half capacity of total water to both the
// glass
if (curr.rem_water != 0)
{
queue.push( new Triplet(curr.row + 1,curr.col -
1,(curr.rem_water/2)));
queue.push( new 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];
} // Driver Code let i = 2, j = 2; let totalWater = 2; // Total amount of water
document.write( "Amount of water in jth glass of ith row is:" );
document.write(findWater(i, j, totalWater).toFixed(6)); // This code is contributed by rag2127 </script> |
Amount of water in jth glass of ith row is:0.500000
Time Complexity: O(i^2)
Auxiliary Space: O(i^2)