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LargestRectangleInHistogram.java
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LargestRectangleInHistogram.java
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package Leetcode;
import java.util.ArrayDeque;
import java.util.Arrays;
import java.util.Deque;
/**
* @author kalpak
*
* Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.
*
* Example 1:
* Input: heights = [2,1,5,6,2,3]
* Output: 10
*
* Explanation: The above is a histogram where width of each bar is 1.
* The largest rectangle is shown in the red area, which has an area = 10 units.
*
* Example 2:
* Input: heights = [2,4]
* Output: 4
*
* Constraints:
*
* 1 <= heights.length <= 105
* 0 <= heights[i] <= 104
*
*/
public class LargestRectangleInHistogram {
public static int largestRectangleArea(int[] heights) {
int[] firstSmallerToLeft = new int[heights.length];
int[] firstSmallerToRight = new int[heights.length];
Deque<Integer> stack = new ArrayDeque<>();
int result = 0;
Arrays.fill(firstSmallerToLeft, -1);
Arrays.fill(firstSmallerToRight, heights.length);
for (int i = 0; i < heights.length; i++) {
while (!stack.isEmpty() && (heights[stack.peek()] >= heights[i])) {
firstSmallerToRight[stack.poll()] = i;
}
if(!stack.isEmpty())
firstSmallerToLeft[i] = stack.peek();
stack.push(i);
}
for (int i = 0; i < heights.length; i++) {
result = Math.max(result, heights[i] * (firstSmallerToRight[i] - firstSmallerToLeft[i] - 1));
}
return result;
}
public static void main(String[] args) {
int[] arr = new int[]{2,1,5,6,2,3};
System.out.println(largestRectangleArea(arr));
arr = new int[]{2, 4};
System.out.println(largestRectangleArea(arr));
}
}