Container with Most Water

Given n non-negative integers $a_1, a_2, ..., a_n$ where each represents a point at coordinate $(i, a_i)$ . ‘ n ‘ vertical lines are drawn such that the two endpoints of line i is at $(i, a_i)$ and $(i, 0)$ .
Find two lines, which together with x-axis forms a container, such that the container contains the most water.

The program should return an integer which corresponds to the maximum area of water that can be contained ( maximum area instead of maximum volume sounds weird but this is 2D plane we are working with for simplicity ).

Note : You may not slant the container.

Examples :

Input : [1, 5, 4, 3]
Output : 6
Explanation : 
5 and 3 are distance 2 apart. 
So the size of the base = 2. 
Height of container = min(5, 3) = 3. 
So total area = 3 * 2 = 6

Input : [3, 1, 2, 4, 5]
Output : 12
Explanation : 
5 and 3 are distance 4 apart. 
So the size of the base = 4. 
Height of container = min(5, 3) = 3. 
So total area = 4 * 3 = 12

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach :

Note 1 : When you consider a1 and aN, then the area is (N-1) * min(a1, aN).
Note 2 : The base (N-1) is the maximum possible.

This implies that if there was a better solution possible, it will definitely have the Height greater than min(a₁, a_N).
$Base * Height > (N-1) * min(a_1, a_N)$
We know that, Base min(a₁, a_N)
This means that we can discard min(a₁, a_N) from our set and look to solve this problem again from the start.
If a₁ < a_N, then the problem reduces to solving the same thing for a₂, a_N.
Else, it reduces to solving the same thing for a₁, a_N-1

C++

// C++ code for Max  
// Water Container 
#include<iostream> 
using namespace std; 
  
int maxArea(int A[], int len) 
{ 
    int l = 0; 
    int r = len -1; 
    int area = 0; 
      
    while (l < r) 
    { 
        // Calculating the max area 
        area = max(area, min(A[l], 
                        A[r]) * (r - l)); 
                          
            if (A[l] < A[r]) 
                l += 1; 
                  
            else
                r -= 1; 
    } 
    return area; 
} 
  
// Driver code 
int main() 
{ 
    int a[] = {1, 5, 4, 3}; 
    int b[] = {3, 1, 2, 4, 5}; 
      
    int len1 = sizeof(a) / sizeof(a[0]); 
    cout << maxArea(a, len1); 
      
    int len2 = sizeof(b) / sizeof(b[0]); 
    cout << endl << maxArea(b, len2); 
} 
  
// This code is contributed by Smitha Dinesh Semwal 

Java

// Java code for Max  
// Water Container 
import java.util.*; 
  
class Area{ 
  
    public static int maxArea(int A[], int len) 
    { 
        int l = 0; 
        int r = len -1; 
        int area = 0; 
      
        while (l < r) 
        { 
            // Calculating the max area 
            area = Math.max(area,  
                        Math.min(A[l], A[r]) * (r - l)); 
                          
            if (A[l] < A[r]) 
                l += 1; 
                  
            else
                r -= 1; 
        } 
        return area; 
    } 
      
    public static void main(String[] args) 
    { 
        int a[] = {1, 5, 4, 3}; 
        int b[] = {3, 1, 2, 4, 5}; 
      
        int len1 = 4; 
        System.out.print( maxArea(a, len1)+"\n" ); 
      
        int len2 = 5; 
        System.out.print( maxArea(b, len2) ); 
    } 
} 
  
// This code is contributed by rishabh_jain 

Python

# Python code for Max  
# Water Container 
def maxArea( A): 
    l = 0
    r = len(A) -1
    area = 0
      
    while l < r: 
        # Calculating the max area 
        area = max(area, min(A[l],  
                        A[r]) * (r - l)) 
      
        if A[l] < A[r]: 
            l += 1
        else: 
            r -= 1
    return area 
  
# Driver code 
a = [1, 5, 4, 3] 
b = [3, 1, 2, 4, 5] 
  
print(maxArea(a)) 
print(maxArea(b)) 

C#

// C# code for Max  
// Water Container 
using System; 
  
class Area{ 
  
    public static int maxArea(int []A, int len) 
    { 
        int l = 0; 
        int r = len -1; 
        int area = 0; 
      
        while (l < r) 
        { 
            // Calculating the max area 
            area = Math.Max(area,  
                        Math.Min(A[l], A[r]) * (r - l)); 
                          
            if (A[l] < A[r]) 
                l += 1; 
                  
            else
                r -= 1; 
        } 
        return area; 
    } 
      
    // Driver code 
    public static void Main() 
    { 
        int []a = {1, 5, 4, 3}; 
        int []b = {3, 1, 2, 4, 5}; 
      
        int len1 = 4; 
        Console.WriteLine( maxArea(a, len1)); 
      
        int len2 = 5; 
        Console.WriteLine( maxArea(b, len2) ); 
    } 
} 
  
// This code is contributed by Vt_M 

PHP

<?php 
// PHP code for Max  
// Water Container 
function maxArea($A, $len) 
{ 
    $l = 0; 
    $r = $len -1; 
    $area = 0; 
      
    while ($l < $r) 
    { 
        // Calculating the max area 
        $area = max($area, min($A[$l], 
                    $A[$r]) * ($r - $l)); 
                          
            if ($A[$l] < $A[$r]) 
                $l += 1; 
                  
            else
                $r -= 1; 
    } 
    return $area; 
} 
  
// Driver code 
$a = array(1, 5, 4, 3); 
$b = array(3, 1, 2, 4, 5); 
  
$len1 = sizeof($a) / sizeof($a[0]); 
echo maxArea($a, $len1). "\n"; 
  
$len2 = sizeof($b) / sizeof($b[0]); 
echo maxArea($b, $len2); 
      
// This code is contributed by mits 
?> 

Output :

6
12

My Personal Notes

Suggest a Topic

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python

C#

PHP

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