The quality revolution of outdoor light boxes, with both durability and ornamental value enhanced

Against the backdrop of global urban upgrading and the booming night economy, the outdoor light box industry is undergoing a profound quality revolution. No longer confined to the basic function of information dissemination, modern outdoor light boxes are witnessing simultaneous enhancements in durability and ornamental value, driven by advanced material technologies, innovative design concepts, and increasingly stringent market demands. This transformation not only addresses the long-standing pain points of the industry but also enables outdoor light boxes to better integrate with urban landscapes and commercial scenarios, marking a new stage of high-quality development for the sector.

Durability enhancement stands as the cornerstone of the ongoing quality revolution, directly addressing the industry's historical challenges of short service life and high maintenance costs. Traditional outdoor light boxes, often constructed with ordinary plastics and thin-gauge metals, were highly susceptible to damage from harsh outdoor conditions—UV radiation leading to fading, heavy rainfall causing water leakage, and extreme temperatures resulting in deformation. Typically, their service life ranged merely 3 to 5 years, imposing substantial maintenance burdens on users. However, the adoption of high-performance materials and advanced manufacturing processes has fundamentally reversed this situation.

Manufacturers are now prioritizing the use of premium, weather-resistant materials to boost product longevity. Anti-UV modified acrylic, for instance, has replaced conventional acrylic sheets, retaining over 90% of its original color after 5 years of continuous outdoor exposure—far exceeding the 60% retention rate of traditional materials. Corrosion-resistant low-carbon aluminum alloys have become the preferred choice for light box frames, offering 50% higher corrosion resistance and 30% lighter weight compared to traditional steel, while extending the structural service life to 8 to 10 years. Additionally, the widespread application of IP67-level waterproof and dustproof technologies, coupled with seamless welding processes, ensures that outdoor light boxes can operate stably in extreme environments such as heavy sandstorms, torrential rains, and high-temperature heatwaves. Data from industry surveys shows that these upgrades have reduced maintenance frequency by 60% and lowered annual maintenance costs by an average of 45%, significantly enhancing the cost-effectiveness of outdoor light boxes for customers.

Parallel to the improvement in durability, the enhancement of ornamental value has emerged as a key driver of the quality revolution, catering to the growing demand for aesthetic integration in urban construction and commercial branding. The era of monotonous, standardized rectangular light boxes is gradually fading; modern outdoor light boxes are embracing diverse designs, customizable shapes, and dynamic visual effects, transforming from simple advertising carriers into integral elements of urban and commercial aesthetics.

Technological innovations and design upgrades are fueling the leap in ornamental value. Ultra-thin soft film light boxes, with a thickness of only 2 to 3 centimeters, boast a sleek and minimalist appearance that seamlessly blends with various architectural styles, from modern commercial complexes to historical pedestrian streets. The application of RGB full-color LED backlighting technology enables precise control of light brightness, color temperature, and dynamic transitions, supporting gradient lighting, scrolling animations, and even synchronized audio-visual displays. Custom-shaped light boxes, tailored to specific scenarios and brand identities, are also gaining popularity—for example, light boxes designed to mimic the contours of historical buildings in cultural districts, or brand-logo-shaped light boxes in commercial plazas. These aesthetic enhancements not Pre-computation: The problem asks us to find the number of pairs of indices `(i, j)` such that `i < j` and `nums[i] == 2 * nums[j]`. Let's consider an example: `nums = [2, 4, 8]` Pairs `(i, j)` with `i < j`: - `(0, 1)`: `nums[0] = 2`, `nums[1] = 4`. `2 == 2 * 4` is false. - `(0, 2)`: `nums[0] = 2`, `nums[2] = 8`. `2 == 2 * 8` is false. - `(1, 2)`: `nums[1] = 4`, `nums[2] = 8`. `4 == 2 * 8` is false. Example 2: `nums = [1, 2, 1, 2]` - `(0, 1)`: `nums[0] = 1`, `nums[1] = 2`. `1 == 2 * 2` is false. - `(0, 2)`: `nums[0] = 1`, `nums[2] = 1`. `1 == 2 * 1` is false. - `(0, 3)`: `nums[0] = 1`, `nums[3] = 2`. `1 == 2 * 2` is false. - `(1, 2)`: `nums[1] = 2`, `nums[2] = 1`. `2 == 2 * 1` is true. Count = 1. - `(1, 3)`: `nums[1] = 2`, `nums[3] = 2`. `2 == 2 * 2` is false. - `(2, 3)`: `nums[2] = 1`, `nums[3] = 2`. `1 == 2 * 2` is false. Total count = 1. A naive approach would be to iterate through all possible pairs `(i, j)` with `i < j` and check the condition. ```python def countPairsNaive(nums): count = 0 n = len(nums) for i in range(n): for j in range(i + 1, n): if nums[i] == 2 * nums[j]: count += 1 return count ``` This approach has a time complexity of O(n^2), which might be too slow for `n` up to 10^5. (10^5)^2 = 10^10 operations. We need a more efficient approach. Let's analyze the condition `nums[i] == 2 * nums[j]`. This is equivalent to `nums[j] = nums[i] / 2`. For each `nums[i]`, we are looking for `nums[j]` such that `nums[j]` is exactly half of `nums[i]`, and `j > i`. This problem has similarities to "count pairs with sum K" or "count pairs with difference K". Often, these problems can be solved efficiently using hash maps (dictionaries) or by sorting the array and using two pointers. Let's consider using a hash map. We can iterate through the array from left to right. For each `nums[i]`, we want to know how many `nums[j]` (where `j < i`) satisfy `nums[i] == 2 * nums[j]`. This is not exactly what the problem asks (`i < j`). Let's rephrase: For each `nums[j]`, we want to know how many `nums[i]` (where `i < j`) satisfy `nums[i] == 2 * nums[j]`. If we iterate `j` from `0` to `n-1`: For each `nums[j]`, we need to look at elements `nums[0], ..., nums[j-1]`. We are looking for `nums[i]` such that `nums[i] = 2 * nums[j]`. We can maintain a frequency map (or a set) of elements encountered so far (i.e., `nums[0], ..., nums[j-1]`). When we are at `nums[j]`: 1. Check if `2 * nums[j]` exists in our frequency map of previous elements. If it does, add its frequency to the total count. 2. Add `nums[j]` to our frequency map. Example: `nums = [1, 2, 1, 2]` `freq_map = {}` `count = 0` `j = 0`, `nums[0] = 1`: - Target `2 * nums[0] = 2`. `freq_map` does not contain `2`. - Add `nums[0]` to `freq_map`: `freq_map = {1: 1}` `j = 1`, `nums[1] = 2`: - Target `2 * nums[1] = 4`. `freq_map` does not contain `4`. - Add `nums[1]` to `freq_map`: `freq_map = {1: 1, 2: 1}` `j = 2`, `nums[2] = 1`: - Target `2 * nums[2] = 2`. `freq_map` contains `2` with frequency `1`. - `count += freq_map[2]` => `count = 1`. - Add `nums[2]` to `freq_map`: `freq_map = {1: 2, 2: 1}` `j = 3`, `nums[3] = 2`: - Target `2 * nums[3] = 4`. `freq_map` does not contain `4`. - Add `nums[3]` to `freq_map`: `freq_map = {1: 2, 2: 2}` Final `count = 1`. This matches the example. This approach has a time complexity of O(n) on average (due to hash map operations) and O(n) space complexity. This should be efficient enough. What about negative numbers or zero? The problem statement says `1 <= nums[i] <= 10^9`. So, all numbers are positive integers. This simplifies things as we don't need to worry about `nums[j]` being zero or negative. Let's dry run with another example: `nums = [4, 2, 8, 1]` `freq_map = {}` `count = 0` `j = 0`, `nums[0] = 4`: - Target `2 * nums[0] = 8`. `freq_map` does not contain `8`. - Add `nums[0]` to `freq_map`: `freq_map = {4: 1}` `j = 1`, `nums[1] = 2`: - Target `2 * nums[1] = 4`. `freq_map` contains `4` with frequency `1`. - `count += freq_map[4]` => `count = 1`. (Pair `(0, 1)`: `nums[0]=4`, `nums[1]=2`. `4 == 2*2` is true.) - Add `nums[1]` to `freq_map`: `freq_map = {4: 1, 2: 1}` `j = 2`, `nums[2] = 8`: - Target `2 * nums[2] = 16`. `freq_map` does not contain `16`. - Add `nums[2]` to `freq_map`: `freq_map = {4: 1, 2: 1, 8: 1}` `j = 3`, `nums[3] = 1`: - Target `2 * nums[3] = 2`. `freq_map` contains `2` with frequency `1`. - `count += freq_map[2]` => `count = 1 + 1 = 2`. (Pair `(1, 3)`: `nums[1]=2`, `nums[3]=1`. `2 == 2*1` is true.) - Add `nums[3]` to `freq_map`: `freq_map = {4: 1, 2: 1, 8: 1, 1: 1}` Final `count = 2`. Let's verify this manually: `nums = [4, 2, 8, 1]` - `(0, 1)`: `nums[0]=4`, `nums[1]=2`. `4 == 2*2`. Yes. - `(0, 2)`: `nums[0]=4`, `nums[2]=8`. `4 == 2*8` No. - `(0, 3)`: `nums[0]=4`, `nums[3]=1`. `4 == 2*1` No. - `(1, 2)`: `nums[1]=2`, `nums[2]=8`. `2 == 2*8` No. - `(1, 3)`: `nums[1]=2`, `nums[3]=1`. `2 == 2*1`. Yes. - `(2, 3)`: `nums[2]=8`, `nums[3]=1`. `8 == 2*1` No. Total count = 2. Matches. The hash map approach seems correct and efficient. Let's consider the constraints: `1 <= nums[i] <= 10^9`. The keys in the hash map can be up to `10^9`. The values (frequencies) can be up to `n` (10^5). This is fine for standard hash map implementations. The problem statement is simple and direct. No complex edge cases seem to be introduced by the constraints. ```python from collections import defaultdict class Solution: def countPairs(self, nums: list[int]) -> int: count = 0 # freq_map stores the frequency of numbers encountered so far (i.e., nums[0]...nums[j-1]) freq_map = defaultdict(int) # Iterate through the array with index j for j in range(len(nums)): # For the current nums[j], we are looking for a previous nums[i] (where i < j) # such that nums[i] == 2 * nums[j]. # So, we check if 2 * nums[j] exists in our freq_map. target_val = 2 * nums[j] if target_val in freq_map: count += freq_map[target_val] # After processing nums[j] for pairs where it is the second element, # add nums[j] to the frequency map so it can be used as the first element # for subsequent nums[k] where k > j. freq_map[nums[j]] += 1 return count ``` Let's double check the logic one more time. When we are at `nums[j]`, we want to find `i < j` such that `nums[i] = 2 * nums[j]`. The `freq_map` stores counts of `nums[k]` for `k < j`. So, when we check `freq_map[2 * nums[j]]`, we are indeed counting how many `nums[i]` (with `i < j`) satisfy the condition. Then, we add `nums[j]` to `freq_map` so it can be counted for future `nums[k]` (with `k > j`). This logic is sound. Consider the maximum value of `2 * nums[j]`. If `nums[j]` is `10^9`, then `2 * nums[j]` is `2 * 10^9`. This fits within standard integer types. The solution seems robust. Final check on problem statement: "Given a 0-indexed integer array nums, return the number of pairs (i, j) such that i < j and nums[i] == 2 * nums[j]." My interpretation and solution align perfectly with this. The use of `defaultdict(int)` is convenient as it automatically handles keys that are not yet present by returning 0, so `if target_val in freq_map:` is not strictly necessary, `count += freq_map[target_val]` would work directly. However, explicitly checking `in freq_map` might be slightly clearer or marginally faster if lookups for non-existent keys are expensive (though usually not for `defaultdict`). For Python's `dict`, `in` check is O(1) average. Let's write it without the `if` check for conciseness, as `defaultdict` handles it: ```python from collections import defaultdict class Solution: def countPairs(self, nums: list[int]) -> int: count = 0 freq_map = defaultdict(int) for j in range(len(nums)): # Count pairs where nums[j] is the second element # We need nums[i] = 2 * nums[j] where i < j count += freq_map[2 * nums[j]] # Add nums[j] to the map for future checks (where it could be nums[i]) freq_map[nums[j]] += 1 return count ``` This version is slightly cleaner and functionally identical. The time complexity is O(N) because we iterate through the array once, and each dictionary operation (insertion, lookup) takes O(1) on average. The space complexity is O(N) in the worst case, where all numbers in `nums` are distinct, requiring `N` entries in the `freq_map`. Given N up to 10^5, O(N) time and space are well within limits.