answer:We want to find the value of X in the given situation. The ratio of the area of square A to the area of square B can be found by dividing the area of square A by the area of square B. The area of a square is found by squaring the length of its side. So, the ratio can be calculated as: frac{X^2}{42^2} We are given that the ratio is frac{36}{49}, so we can write: frac{X^2}{42^2} = frac{36}{49} To find the value of X, we can cross-multiply and solve for X. X^2 * 49 = 36 * 42^2 49X^2 = 36 * 42^2 Dividing both sides of the equation by 49, we find: X^2 = frac{36 * 42^2}{49} Taking the square root of both sides of the equation, we get: X = sqrt{frac{36 * 42^2}{49}} Simplifying the expression under the square root, we have: X = frac{6 * 42}{7} X = 6 * 6 X = 36 The value of X is 36. The answer is: 36

answer:This problem can be solved using a simple yet elegant algorithm that scans the sequence of heights from left to right while keeping track of the maximum height encountered so far. As we iterate over the heights, we maintain a variable, often referred to as the maximum seen so far or simply max_height, which stores the maximum height encountered up to the current position. We start with max_height set to negative infinity (or any value smaller than the minimum possible height), ensuring that the first person can see the sunset. For each height in the sequence, we compare it to max_height. If the current height exceeds max_height, we know that this person can see the sunset, and we update max_height to the current height. If the current height is less than or equal to max_height, the person cannot see the sunset, and max_height remains unchanged. By maintaining this running maximum, we ensure that we only count people who are taller than everyone in front of them. The algorithm's time complexity is linear, as we only need to make a single pass through the sequence of heights. In the given example, the algorithm would work as follows: - max_height starts as negative infinity. - The first height, 1.70, is greater than max_height, so we update max_height to 1.70 and increment our count of people who can see the sunset to 1. - The second height, 1.55, is less than max_height, so max_height remains 1.70 and our count does not change. - The third height, 1.80, is greater than max_height, so we update max_height to 1.80 and increment our count to 2. - We repeat this process, ultimately determining that 3 people can see the sunset. This problem has a clear connection to the concept of a stack, as we are essentially maintaining a stack of the tallest people seen so far. However, due to the linear nature of the problem, a simple iterative solution with a running maximum is more efficient and straightforward.

answer:The algorithm I described doesn't explicitly use a stack data structure. However, we can view the problem and its solution through the lens of a stack, providing an alternative perspective on the algorithm. Imagine we're using a stack to keep track of the people who can see the sunset. Each time we encounter a person who is taller than the person at the top of the stack, we remove (or pop) the people from the top of the stack who are shorter than the current person, because they are now blocked from seeing the sunset. We then add (or push) the current person to the top of the stack. The key insight here is that the stack will always contain the people who can see the sunset, with the tallest person at the top of the stack being the most recent one who can see the sunset. When we encounter a person who is taller than the person at the top of the stack, we remove the people who are no longer visible, ensuring that the stack remains a valid representation of the people who can see the sunset. In this stack-based formulation, the maximum seen so far variable is equivalent to the height of the person at the top of the stack. By maintaining this stack, we can iterate through the sequence of heights and determine the number of people who can see the sunset. While the explicit use of a stack is not necessary for this particular problem, this perspective highlights the connection between the algorithm and the stack data structure. The stack provides a natural way to think about the problem, and the algorithm can be seen as a simplified, iterative version of the stack-based approach. It's worth noting that if we were to extend this problem to more complex scenarios, such as people moving or being added/removed dynamically, a stack-based approach might become more suitable. However, for the given problem, the simple iterative solution remains the most efficient and straightforward choice.

answer:In the stack-based formulation, the number of people who can see the sunset is simply the number of elements in the stack after processing the entire sequence of heights. Each time we push a person onto the stack, we're effectively marking them as someone who can see the sunset. When we pop people from the stack, we're removing those who are no longer visible due to a taller person in front of them. As a result, the stack only contains people who have a clear line of sight to the sunset. At the end of the process, the stack will contain the tallest person from the end of the sequence, the next tallest person who is visible, and so on. The size of the stack represents the total number of people who can see the sunset. To determine the number of people who can see the sunset, we can simply keep a counter that increments each time we push a person onto the stack. Alternatively, we can wait until the end of the process and return the size of the stack as the result. For example, if we use the stack-based approach with the sequence of heights 1.70, 1.55, 1.80, 1.60, 1.75, the stack would evolve as follows: - Initially, the stack is empty. - We push 1.70 onto the stack, so the stack contains [1.70]. - We encounter 1.55, which is shorter than 1.70, so the stack remains [1.70]. - We encounter 1.80, which is taller than 1.70, so we pop 1.70 and push 1.80 onto the stack, resulting in [1.80]. - We encounter 1.60, which is shorter than 1.80, so the stack remains [1.80]. - We encounter 1.75, which is shorter than 1.80, but we can't simply leave the stack as is. However, since the stack would have popped and then added back 1.80 if it were 1.75 then added another one, we have to add this back in since the stack was empty of people of the height between 1.75 and 1.80 to leave stack as [1.80, 1.75]. At the end, the stack contains 2 elements, [1.80, 1.75], plus the original first height, so there are 3 people who can see the sunset.