Given an empty glass, this glass has to be filled with water and the task is to find the maximum amount of water that the glass has held at any moment.
Given conditions:
The program takes an input N denoting the number of steps. Each step consists of two inputs T and X where T is the flag condition denoting whether the X ml of water has to be poured or drinked based on below conditions: 1. if T = 0, Pour X ml (millilitres) of water in the glass 2. if T = 1, Drink X ml of water from the glass
Examples:
Input: N = 4,
0 1
0 1
0 1
1 3
Output: 3
Explanation:
The glass initially has 0 ml of water.
The maximum value is obtained
after the first 3 operations.
Input: N = 2
1 15
1 24
Output: 39
Approach THe approach can be simply put as adding up the volumes whenever T is 0 and finding the maximum of this volume. Implementation:
C++
// C++ implementation of the approach #include<iostream> using namespace std; // returns the largest volume int largest_volume(int n, int *t, int *x) { // arbitrarily large value int minValue = 100000000; // stores the maximum int maxValue = 0; // Current Volume of the glass int c = 0; for (int i = 0; i < n; i++) { // if 1st operation is performed if (t[i] == 0) { // increment with current x c += x[i]; // take current max maxValue = max(maxValue, c); } // if 2nd operation is performed else { // decrement with current x c -= x[i]; // take current min minValue = min(minValue, c); } } // returns the largest difference return maxValue - minValue; } // Driver code int main() { int n = 4; int t[4] = {0}; int x[4] = {0}; t[0] = 0; x[0] = 1; t[1] = 0; x[1] = 1; t[2] = 0; x[2] = 1; t[3] = 1; x[3] = 3; // Find the largest volume int ans = largest_volume(n, t, x); // Print the largest volume cout<< ans; return 0; } // This code is contributed by PrinciRaj1992 |
Java
// Java implementation of the approach class GFG { // returns the largest volume static int largest_volume(int n, int[] t, int[] x) { // arbitrarily large value int min = 100000000; // stores the maximum int max = 0; // Current Volume of the glass int c = 0; for (int i = 0; i < n; i++) { // if 1st operation is performed if (t[i] == 0) { // increment with current x c += x[i]; // take current max max = Math.max(max, c); } // if 2nd operation is performed else { // decrement with current x c -= x[i]; // take current min min = Math.min(min, c); } } // returns the largest difference return max - min; } // Driver code public static void main(String[] args) { int n = 4; int[] t = new int[4]; int[] x = new int[4]; t[0] = 0; x[0] = 1; t[1] = 0; x[1] = 1; t[2] = 0; x[2] = 1; t[3] = 1; x[3] = 3; // Find the largest volume int ans = largest_volume(n, t, x); // Print the largest volume System.out.println(ans); } } |
Python3
# Python3 implementation of the approach # Returns the largest volume def largest_volume(n, t, x): # Arbitrarily large value Min = 100000000 # Stores the maximum Max = 0 # Current Volume of the glass c = 0 for i in range(0, n): # If 1st operation is performed if t[i] == 0: # Increment with current x c += x[i] # Take current max Max = max(Max, c) # If 2nd operation is performed else: # Decrement with current x c -= x[i] # Take current min Min = min(Min, c) # Returns the largest difference return Max - Min # Driver code if __name__ == "__main__": n = 4 t = [0, 0, 0, 1] x = [1, 1, 1, 3] # Find the largest volume ans = largest_volume(n, t, x) # Print the largest volume print(ans) # This code is contributed by Rituraj Jain |
C#
// C# implementation of the approach using System; class GFG { // returns the largest volume static int largest_volume(int n, int[] t, int[] x) { // arbitrarily large value int min = 100000000; // stores the maximum int max = 0; // Current Volume of the glass int c = 0; for (int i = 0; i < n; i++) { // if 1st operation is performed if (t[i] == 0) { // increment with current x c += x[i]; // take current max max = Math.Max(max, c); } // if 2nd operation is performed else { // decrement with current x c -= x[i]; // take current min min = Math.Min(min, c); } } // returns the largest difference return max - min; } // Driver code public static void Main() { int n = 4; int[] t = new int[4]; int[] x = new int[4]; t[0] = 0; x[0] = 1; t[1] = 0; x[1] = 1; t[2] = 0; x[2] = 1; t[3] = 1; x[3] = 3; // Find the largest volume int ans = largest_volume(n, t, x); // Print the largest volume Console.WriteLine(ans); } } // This code is contributed by // tufan_gupta2000 |
Javascript
<script> // Javascript implementation of the approach // returns the largest volume function largest_volume(n, t, x) { // arbitrarily large value let minValue = 100000000; // stores the maximum let maxValue = 0; // Current Volume of the glass let c = 0; for (let i = 0; i < n; i++) { // if 1st operation is performed if (t[i] == 0) { // increment with current x c += x[i]; // take current max maxValue = Math.max(maxValue, c); } // if 2nd operation is performed else { // decrement with current x c -= x[i]; // take current min minValue = Math.min(minValue, c); } } // returns the largest difference return maxValue - minValue; } // Driver code const n = 4; const t = Array(n).fill(0); const x = Array(n).fill(0); t[0] = 0; x[0] = 1; t[1] = 0; x[1] = 1; t[2] = 0; x[2] = 1; t[3] = 1; x[3] = 3; // Find the largest volume let ans = largest_volume(n, t, x); // Print the largest volume document.write(ans); // This code is contributed by Pushpesh Raj. </script> |
Output:
3
Time Complexity: O(n)
Auxiliary Space: O(1)
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