If someone wants a short summary: The idea is to calculate the area between the lines at the two pointers, update the maximum area if the current area is larger, and then move the pointer pointing to the shorter line. By moving the shorter line, you potentially increase the height of the container, as increasing the width alone won't give a better result if one side is short.
One line sums up it all: If you have the max width between two bars, then the only way to increase the area is to increase the height. That is, the current area is the maximum for the area for the solution set containing the current minimum height bar.
better approaches int area; int maxarea=0; int start=0; int end=height.size()-1; while(start < end){ * we going to find the area of container* area=min(height[start], height[end])* (end-start); * after finding the area we check that if area is grater than maxarea we place the area value into maxarea* if(area > maxarea){ maxarea= area; } *after that we check that start value is less that end value than increment the start or else decrement the end index value* if(height [start] < height[end] ){ start++; } else { end--; } } return maxarea; please comment if any query is there?? thank you
MY approach is way better btw bcuz we dont need to mutate the min height and idx simple and better int maxArea(vector& height) { // Write your code here. int ans = 0, n = height.size(); int i = 0, j = n-1; while(iheight[j]){ ans = max(ans, (j-i)*height[j]); j--; } else{ ans = max(ans, (j-i)*height[i]); i++; } } return ans; }
int maxArea(vector& height) { int n=height.size(); int i=0,j=n-1; int area1=0; int area; int width; int hmin; while(iheight[j]) { hmin=height[j]; j--; } else { hmin=height[i]; i++; } area=width*hmin; if(area>area1) { area1=area; } } return area1; }
@@quant-prep2843 LMAO what's wrong with you ? Bhai doesnt mean what you have assumed to be. I hope you know that it is just a way to address someone politely,
Your explaination is good but your demo code is quite long. We just need to keep track left_height and right_height (with only 1 while loop)...anyway, thank you very much for your video.
is that a correct answer ? #include #include #include using namespace std; int sum; unsigned int maxArea; int area; int i, j; int minArea = 213213; int main() { maxArea > 0; area > 0; valarraya(9); a = { 1,8,6,2,5,4,8,3,7 }; for (int i = 0; i < 9; i++) { for (int j = 1; i+j < 9; j++) { if (a[i] >= a[i + j]) { area = j * a[i + j]; //cout area) { minArea = area; } //cout
If someone wants a short summary: The idea is to calculate the area between the lines at the two pointers, update the maximum area if the current area is larger, and then move the pointer pointing to the shorter line. By moving the shorter line, you potentially increase the height of the container, as increasing the width alone won't give a better result if one side is short.
Nice explanation. we can Generalize as below.
int maxArea(vector& height) {
int maxarea =0;
int minIndex = 0;
int maxIndex = height.size()-1;
while(minIndex!=maxIndex)
{
if(height[minIndex]>height[maxIndex])
swap(minIndex,maxIndex);
int newarea = height[minIndex] * (abs(minIndex-maxIndex));
if(newarea>maxarea)
maxarea = newarea;
if(minIndex > maxIndex)
minIndex--;
else
minIndex++;
}
return maxarea;
}
class Solution {
public:
int maxArea(vector& height) {
int i = 0, j = height.size()-1;
int m = 0,m1;
while(i
One line sums up it all: If you have the max width between two bars, then the only way to increase the area is to increase the height.
That is, the current area is the maximum for the area for the solution set containing the current minimum height bar.
U made my bulb glow just with one line Thanks!
better approaches
int area;
int maxarea=0;
int start=0;
int end=height.size()-1;
while(start < end){
* we going to find the area of container*
area=min(height[start], height[end])* (end-start);
* after finding the area we check that if area is grater than maxarea we place the area value into maxarea*
if(area > maxarea){
maxarea= area;
}
*after that we check that start value is less that end value than increment the start or else decrement the end index value*
if(height [start] < height[end] ){
start++;
}
else {
end--;
}
}
return maxarea;
please comment if any query is there??
thank you
nice explanation,first problem i seen algorithm and solved my own.
Well done
I watch your videos for intuition. Then take time to code that logic.
remember during the interview nobody will give you the intuititon, lol
@@quant-prep2843 working on it 😅
MY approach is way better btw bcuz we dont need to mutate the min height and idx
simple and better
int maxArea(vector& height) {
// Write your code here.
int ans = 0, n = height.size();
int i = 0, j = n-1;
while(iheight[j]){
ans = max(ans, (j-i)*height[j]);
j--;
}
else{
ans = max(ans, (j-i)*height[i]);
i++;
}
}
return ans;
}
class Solution {
public:
int maxArea(vector& height) {
int left=0;
int right=height.size()-1;
int max_area=0;
while(left
What kind of hardware you use for the pen ? Please share the amazon link if you have of the pen and digital slate you are using.
Hi. This is the pen. amzn.to/3tOVEP3
int maxArea(vector& height)
{
int n=height.size();
int i=0,j=n-1;
int area1=0;
int area;
int width;
int hmin;
while(iheight[j])
{
hmin=height[j];
j--;
}
else
{
hmin=height[i];
i++;
}
area=width*hmin;
if(area>area1)
{
area1=area;
}
}
return area1;
}
code quality is not good
simple solution
class Solution {
public:
int maxArea(vector& height) {
int maxArea = 0;
int i = 0, j = height.size() - 1;
while(i < j)
{
maxArea = max(maxArea, min(height[i], height[j]) * (j - i));
if(height[i] < height[j])
i++;
else
j--;
}
return maxArea;
}
};
Bhai do you also plan to make videos in Hindi ?
there is no bhai's,this is for all dont be a discriminant and gender based inequality
@@quant-prep2843 LMAO what's wrong with you ? Bhai doesnt mean what you have assumed to be. I hope you know that it is just a way to address someone politely,
Amazing explanation !!
Thanks.
Wow! Great Solution!
Glad you like it!
What is the difference b/w this Q and *Largest Rectangle in Histogram* ?
Nice explanation sir....
Thanks for liking
Your explaination is good but your demo code is quite long. We just need to keep track left_height and right_height (with only 1 while loop)...anyway, thank you very much for your video.
Noted
@@KnowledgeCenter this video is perfect 👍 as long as its not over 30 mins long🤣!! thanks for creating this C++ playlist
its a good explaination . but u are just telling the thing but u r not concernerd about the viewers understanding . be bold on your voice .
Weiting fore premiere
question ka link to provide kara dena chaiye
is that a correct answer ?
#include
#include
#include
using namespace std;
int sum;
unsigned int maxArea;
int area;
int i, j;
int minArea = 213213;
int main() {
maxArea > 0;
area > 0;
valarraya(9);
a = { 1,8,6,2,5,4,8,3,7 };
for (int i = 0; i < 9; i++) {
for (int j = 1; i+j < 9; j++) {
if (a[i] >= a[i + j]) {
area = j * a[i + j];
//cout area) {
minArea = area;
}
//cout
This code is quite big, this can be done simply with only 1 while loop/
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