Created
June 29, 2019 18:47
-
-
Save ajinkyajawale14499/567aa0823ddad23ba0ca7ac08ccaa836 to your computer and use it in GitHub Desktop.
Largest Histogram using dynamic programming
This file contains hidden or bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
public static int largestRectangleArea(int[] height) { | |
if (height == null || height.length == 0) { | |
return 0; | |
} | |
int[] lessFromLeft = new int[height.length]; // idx of the first bar the left that is lower than current | |
int[] lessFromRight = new int[height.length]; // idx of the first bar the right that is lower than current | |
lessFromRight[height.length - 1] = height.length; | |
lessFromLeft[0] = -1; | |
for (int i = 1; i < height.length; i++) { | |
int p = i - 1; | |
while (p >= 0 && height[p] >= height[i]) { | |
p = lessFromLeft[p]; | |
} | |
lessFromLeft[i] = p; | |
} | |
for (int i = height.length - 2; i >= 0; i--) { | |
int p = i + 1; | |
while (p < height.length && height[p] >= height[i]) { | |
p = lessFromRight[p]; | |
} | |
lessFromRight[i] = p; | |
} | |
int maxArea = 0; | |
for (int i = 0; i < height.length; i++) { | |
maxArea = Math.max(maxArea, height[i] * (lessFromRight[i] - lessFromLeft[i] - 1)); | |
} | |
return maxArea; | |
} |
Sign up for free
to join this conversation on GitHub.
Already have an account?
Sign in to comment