def maxArea(height):
max_area = 0
left = 0
right = len(height) - 1
while left < right:
area = min(height[left], height[right]) * (right - left)
max_area = max(max_area, area)
if height[left] < height[right]:
left += 1
else:
right -= 1
return max_area
height = [1, 8, 6, 2, 5, 4, 8, 3, 7]
print(maxArea(height))
Explanation:
The "Container With Most Water" problem asks to find two lines in an array that form a container, such that the container encloses the most water.
The Python, Java, and C++ solutions follow a similar two-pointer approach to find the maximum area: br
Initialize two pointers, left and right, pointing to the start and end of the array.
Calculate the area between the two pointers using the formula min(height[left], height[right]) * (right - left).
Update the maximum area if the current area is greater.
Move the pointer that corresponds to the smaller height towards the center.
If height[left] < height[right], increment left.
If height[left] >= height[right], decrement right.
Repeat the above steps until the pointers meet or cross each other.
The time complexity of the solution is O(n), where n is the number of elements in the array.
I hope this explanation helps! Let me know if you have any further questions.
class Solution {
public int maxArea(int[] height) {
int maxArea = 0;
int left = 0;
int right = height.length - 1;
while (left < right) {
int area = Math.min(height[left], height[right]) * (right - left);
maxArea = Math.max(maxArea, area);
if (height[left] < height[right]) {
left++;
} else {
right--;
}
}
return maxArea;
}
public static void main(String[] args) {
int[] height = {1, 8, 6, 2, 5, 4, 8, 3, 7};
Solution solution = new Solution();
System.out.println(solution.maxArea(height));
}
}
Explanation:
The "Container With Most Water" problem asks to find two lines in an array that form a container, such that the container encloses the most water.
The Python, Java, and C++ solutions follow a similar two-pointer approach to find the maximum area: br
Initialize two pointers, left and right, pointing to the start and end of the array.
Calculate the area between the two pointers using the formula min(height[left], height[right]) * (right - left).
Update the maximum area if the current area is greater.
Move the pointer that corresponds to the smaller height towards the center.
If height[left] < height[right], increment left.
If height[left] >= height[right], decrement right.
Repeat the above steps until the pointers meet or cross each other.
The time complexity of the solution is O(n), where n is the number of elements in the array.
I hope this explanation helps! Let me know if you have any further questions.
#include <iostream>
#include <vector>
using namespace std;
int maxArea(vector& height) {
int maxArea = 0;
int left = 0;
int right = height.size() - 1;
while (left < right) {
int area = min(height[left], height[right]) * (right - left);
maxArea = max(maxArea, area);
if (height[left] < height[right]) {
left++;
} else {
right--;
}
}
return maxArea;
}
int main() {
vector height = {1, 8, 6, 2, 5, 4, 8, 3, 7};
cout << maxArea(height) << endl;
return 0;
}
Explanation:
The "Container With Most Water" problem asks to find two lines in an array that form a container, such that the container encloses the most water.
The Python, Java, and C++ solutions follow a similar two-pointer approach to find the maximum area: br
Initialize two pointers, left and right, pointing to the start and end of the array.
Calculate the area between the two pointers using the formula min(height[left], height[right]) * (right - left).
Update the maximum area if the current area is greater.
Move the pointer that corresponds to the smaller height towards the center.
If height[left] < height[right], increment left.
If height[left] >= height[right], decrement right.
Repeat the above steps until the pointers meet or cross each other.
The time complexity of the solution is O(n), where n is the number of elements in the array.
I hope this explanation helps! Let me know if you have any further questions.