# largest rectangle hackerrank solution java

The first and only line of input contains two space separated integers denoting the width and height of the rectangle. The stack is empty so we push i = 3 and then increment i to i== 4; Line 8. We pop the top of the stack which holds top = 2 and compute the area of the rectangle area = height[2] * 3 which produces area = 2 * 3 = 6. Line 7. Save the source file in the corresponding folder in your forked repo. The initial idea is to take the first rectangle (height [0] == 4) and set the current maxArea = 4. Determine if a set of points coincides with the edges of a non-degenerate rectangle. The largest rectangle is shown in the shaded area, which has area = 10 unit. The area is based on the height * length. Given the array, nums= [2,3,6,6,5] we see that the largest value in the array is 6 and the second largest value is 5. The maxArea is not updated. Implemented the code and gave it a try. If you have comments or questions regarding this entry or any other entry in this blog, please send me a message via email. The area formed is . Recommended: Please try your approach on first, before moving on to the solution. Minimum Absolute Difference In An Array Hackerrank Solution In Java. If you join K adjacent buildings, they will form a solid rectangle of area K * min(h, … , h). © 2020 The Poor Coder | Hackerrank Solutions - In this challenge, we practice creating objects. Line 6. Largest Rectangle solution. Line 9. We then go to the second rectangle (height[1] == 3). Perhaps Java is not fast enough when compared to C or C++. The idea is to use Dynamic Programming to solve this problem. The area = 2 * (9 – 3 – 1) = 2 * 5 = 10. Solutions of more than 380 problems of Hackerrank across several domains. Your task is to rearrange them according to their CGPA in decreasing order. At this point the area from the first two rectangles is 3 * 2 = 6. Rectangle The Rectangle class should have two data fields-width and height of int types. For the first 2 buildings the common area is determined by the min(height[0], height[1]) * 2. Line 16. The largest rectangle is shown in the shaded area, which has area = 10 unit. Output Formateval(ez_write_tag([[300,250],'thepoorcoder_com-box-3','ezslot_4',102,'0','0'])); The output should consist of exactly two lines: In the first line, print the width and height of the rectangle separated by space. Note that the stack now holds the indices 3 and 4 to height[3] == 1 and height[4] == 2. Equal Stacks, here is my solution in java which can pass this testcase too.. static int equalStacks(int[] h1, int[] h2, int[] h3) { Stack

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