# Program to find amount of water in a given glass

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 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 top glass, how much water will be contained by jth glass in ith row?

Example. If you will put 2 litre 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 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 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 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; ` `} `

## Java

 `// 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 `

## 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 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 `

## C#

 `// 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

 ` ``\$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 ` `?> `

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)