# Javascript solution with some explaination

• ``````/**
* @param {number[]} heights
* @return {number}
*/
var largestRectangleArea = function(heights) {
var maxarea = 0
var stack = []

/*
* return the top of the stack, or null
*/
Array.prototype.top = function() {
return this.length && this[this.length - 1] || null
}

// left boundary of the entire problem is always 0
stack.push({leftbdforthenextpoint: 0, val: 0})

for (var i = 0; i < heights.length + 1; i++) {
if (i < heights.length && heights[i] > stack.top().val) {
// get the left boundaries here
//
// when the height is increasing, we push those heights
// into the stack, so any point in the stack can have the
// left boundary determined by the previous point in the stack
// to calculate the maximum area

stack.push({leftbdforthenextpoint: i + 1, val: heights[i]})
} else {
// get the right boundaries here
//
// let's say the current point index is i
// when the height is decreasing, we know a right
// boundary for point i - 1 is found (it's i)
// combind with the last step you can get the rectangle with
// height[i] and hence the area

while (stack.top() && stack.top().val > (heights[i] || 0)) {
var popped = stack.pop()
var leftbdforthenextpoint = stack.top() && stack.top().leftbdforthenextpoint || 0
maxarea = Math.max(maxarea, (i - leftbdforthenextpoint) * popped.val)
}
stack.push({leftbdforthenextpoint: i + 1, val: heights[i]})
}
}
return maxarea
};
``````

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