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Container with Most Water
  • Difficulty Level : Medium
  • Last Updated : 04 Feb, 2021

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 the 2D plane we are working with for simplicity).

Note: You may not slant the container. 

Examples :  

Input: array = [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: array = [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

Naive Solution:  

  • Approach: The idea is quite simple and involves checking every pair of boundaries or array element and find out the maximum area under any pair of boundaries.
  • Algorithm: 
    1. Create a nested loop, the outer loop traverses the array from 0 to end (index of this loop is i).
    2. The inner loop traverses the array from i + 1 to end (index of this loop is j).
    3. Find the water that can be contained in the container with height of boundaries as array[i] and array[j], i.e area = (j – i)* min(array[i],array[j]), if this area is greater than current maximum, update the current maximum
    4. Print the current maximum.
  • Implementation:

C++14




// C++ code for Max
// Water Container
#include <iostream>
using namespace std;
 
int maxArea(int A[], int len)
{
    int area = 0;
    for (int i = 0; i < len; i++) {
        for (int j = i + 1; j < len; j++) {
            // Calculating the max area
            area = max(area, min(A[j], A[i]) * (j - i));
        }
    }
    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);
}


Java




// Java code for Max 
// Water Container
import java.io.*;
 
class GFG{
 
public static int maxArea(int[] a)
{
 
    int Area = 0;
     
    for(int i = 0; i < a.length; i++)
    {
        for(int j = i + 1; j < a.length; j++)
        {
            Area = Math.max(
                Area, Math.min(a[i], a[j]) *
                              (j - i));
        }
    }
    return Area;
}
 
// Driver code
public static void main(String[] args)
{
    int a[] = { 1, 5, 4, 3 };
    int b[] = { 3, 1, 2, 4, 5 };
 
    System.out.println(maxArea(a));
    System.out.println(maxArea(b));
}
}
 
// This code is contributed by mulchandanimanisha5


Python3




# Python3 code for Max
# Water Container
def maxArea(A, Len) :
    area = 0
    for i in range(Len) :
        for j in range(i + 1, Len) :
           
            # Calculating the max area
            area = max(area, min(A[j], A[i]) * (j - i))
    return area
 
# Driver code
a = [ 1, 5, 4, 3 ]
b = [ 3, 1, 2, 4, 5 ]
 
len1 = len(a)
print(maxArea(a, len1))
 
len2 = len(b)
print(maxArea(b, len2))
 
# This code is contributed by divyesh072019.


C#




// C# code for Max
// Water Container
using System;
class GFG
{
 
public static int maxArea(int[] a)
{
 
    int Area = 0;
     
    for(int i = 0; i < a.Length; i++)
    {
        for(int j = i + 1; j < a.Length; j++)
        {
            Area = Math.Max(Area, Math.Min(a[i], a[j]) *
                            (j - i));
        }
    }
    return Area;
}
 
// Driver code
public static void Main(String[] args)
{
    int []a = { 1, 5, 4, 3 };
    int []b = { 3, 1, 2, 4, 5 };
 
    Console.WriteLine(maxArea(a));
    Console.Write(maxArea(b));
}
}
 
// This code is contributed by shivanisinghss2110


Output: 



6
12

 

Complexity Analysis: 

  • Time Complexity: O(n^2). 
    As nested traversal of the array is required, so time complexity is O(n^2)
  • Space Complexity: O(1). 
    As no extra space is required, so space complexity is constant.

Efficient Solution:  

  • Approach: Given an array of heights of lines of boundaries of the container, find the maximum water that can be stored in a container. So start with the first and last element and check the amount of water that can be contained and store that container. Now the question arises is there any better boundaries or lines that can contain maximum water. So there is a clever way to find that. Initially, there are two indices the first and last index pointing to the first and the last element (which acts as boundaries of the container), if the value first index is less than the last index increase the first index else decrease the last index. As the increase in width is constant, by following the above process the optimal answer can be reached.
    The GIF below explains the approach.

  • Algorithm: 
    1. Keep two index, first = 0 and last = n-1 and a value max_area that stores the maximum area.
    2. Run a loop until first is less than the last.
    3. Update the max_area with maximum of max_area and min(array[first] , array[last])*(last-first)
    4. if the value at array[first] is greater the array[last] then update last as last – 1 else update first as first + 1
    5. Print the maximum area.
  • Implementation:

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


Python3




# Python3 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

Complexity Analysis: 

  • Time Complexity: O(n). 
    As only one traversal of the array is required, so time complexity is O(n).
  • Space Complexity: O(1). 
    No extra space is required, so space complexity is constant.

 

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