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

Source: Amazon Interview | Set 12

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.

// 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; }

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. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above