LeetCode 84: Largest Rectangle in Histogram Solution
Master LeetCode problem 84 (Largest Rectangle in Histogram), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
84. Largest Rectangle in Histogram
Problem Explanation
Explanation
To solve this problem, we can use a stack to keep track of the indices of the bars in the histogram. We iterate through the histogram and for each bar, we check if its height is greater than the height of the bar at the index on top of the stack. If it is, we push the current index onto the stack. If not, we keep popping indices from the stack and calculate the area of the rectangle formed by the popped bar as the smallest bar and the current bar. We keep track of the maximum area found so far.
At the end of the iteration, we may still have some bars left in the stack. We continue popping bars from the stack and calculate the area until the stack is empty.
Time complexity: O(n) where n is the number of bars in the histogram. Space complexity: O(n) where n is the number of bars in the histogram.
Solution Code
class Solution {
public int largestRectangleArea(int[] heights) {
Stack<Integer> stack = new Stack<>();
int maxArea = 0;
for (int i = 0; i <= heights.length; i++) {
int currHeight = (i == heights.length) ? 0 : heights[i];
while (!stack.isEmpty() && currHeight < heights[stack.peek()]) {
int height = heights[stack.pop()];
int width = stack.isEmpty() ? i : i - stack.peek() - 1;
maxArea = Math.max(maxArea, height * width);
}
stack.push(i);
}
return maxArea;
}
}Try It Yourself
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