DP 55. Maximum Rectangle Area with all 1's | DP on Rectangles

Please watch the video at 1.25x for a better experience.

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In this video, we solve the Maximum Rectangle Area with all 1's problem.

[sunavaroy]

I am not sure whether you would be seeing this or not but I want to thank you for all the playlists (dp, graph, tree, recursion, placement series and the sde sheet) every content I have covered and just in 2 weeks literally just after giving proper 2 weeks to all the problems you have created I was able to clear all the amazon coding interviews. Not sure whether I would be able to clear the manegerial round but I have come so far because of the content you created. Not only amazon, Qualcomm, Vmware every where I am able to clear the rounds because solely of your content. Waiting for more content. Keep the great work going.

[varungurnani]

Special thanks to you for making a clarification on why this problem is in DP playlist.

[salmanahmedkhan]

This problem came in my Internship test. Unfortunately, i was unable to solve it and scared from this question. My bad, i haven't look youtube at that time. Gladly, i found a mentor to teach me this.

[devbhattdev]

man am feeling proud of myself i was so consitent in this series of the dp. i watched i code every single question. thank you striver bhaiya...

[naveensaicremsiyadlapalli]

Those who are getting wrong answer in Leetcode, In the function they gave the 2d character vector(matrix) not integer vector, so instead of writing matrix[i][j]==1 write matrix[i][j]=='1' then everything will be fine

[RamyasriMedagam]

Understood!! Thank you Striver for this amazing playlist and kudos to all the hard work you've put in to bring us this amazing content!!

[sauravchandra]

Oh nice building up on previous concepts. Understood, thanks!

[dashsights]

Thanks striver for giving line of thinking that's the most difficult thing for me! Leap of faith🎉

[ChewyPopcorn]

this blew my mind, on god, hats off man!

[vishalsinghrajpurohit]

The time complexity will be O(n* (m+m)) instead of O(n*(m+n)). because the row size=m and size of height vector=m

[gandhijainamgunvantkumar]

I think the time complexity would be O(n*m) + O(n*O(m))
O(n*m) because there are 2 loops and O(n* m) is for finding max area in histogram for each row and there are total n rows.

[prabhpreetsingh]

at 6:06 shouldn’t it be 3? there are 3 1’s together so 3*1=3

[cinime]

Understood! Thank you sooo much for your great work!!

[ronakslibrary]

You have explained it very well sir, thanks for such a wonderful video

[daranip]

Hard Problems are many times easy. But, with a difficult approach to come up with.

[vinayjangra]

In case, if you are curious to solve using Recursive Solution:: Here is the memoized recursive solution(in JAVA):-
(I HAVE ABSTRACT OUT THE MAX HISTOGRAM FUNCTION) :-)
public int maximalRectangle(char[][] matrix) {


int[]dp = new int[matrix.length];
for(int i = 0; i < dp.length; i++)
{
dp[i] = -1;
}
int[]heights = new int[matrix[0].length];

return maxRec(0, matrix, heights, dp);
}

public int maxRec(int i, char[][] matrix, int[]heights, int[]dp)
{
if(i == matrix.length)
{

return 0;
}

if(dp[i] != -1)
{
return dp[i];
}
int[] temp = new int[heights.length];
for(int j = 0; j < heights.length; j++)
{
if(matrix[i][j] == '1')
{
heights[j]++;
// System.out.println("+1");
}
else
{
heights[j] = 0;
}
}

for(int k = 0; k < temp.length; k++)
{
temp[k] = heights[k];
}

int rec = maxRec(i + 1, matrix, heights, dp);

for(int k = 0; k < temp.length; k++)
{
heights[k] = temp[k];
}
int height =

int ans = Math.max(height, rec);

dp[i] = ans;
return ans;
}

[punkstAr]

max area of second row rect should be 3 not 2

[kanakbanerjee]

Time complexity is O(n*m) not O(n*(n+m))

[aizadiqbal]

for max area of row 2...why won't it be 1x3 ??

[lazymethisside]

Today I came to know why all my seniors and batchmates, say about Striver . Such a beautiful explanation 🫡