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Ads placement on the Zillow platform is the best way to maximize revenue. By analyzing user behavior and ad performance data, we can determine the optimal spots for ads. This means understanding how users interact with the platform and which ads perform the best in different locations. With this information, we can place ads where they are most likely to be seen and clicked on, ensuring we get the most out of our advertising space.

Design Technique: Dynamic Programming for Ad Placement

Dynamic programming approach helps in systematically calculating the maximum revenue by considering all possible placements of ads. By building up the solution incrementally, from considering no ads to all ads and placements, we ensure that we find the optimal solution. Each step (decision to place or not place an ad) is based on maximizing the revenue considering the previous decisions and their outcomes.

Implementation

1. Define State:

dp[i][j] represents the maximum revenue achievable using the first i ads and j placements.

2. Transition:

For each ad i and each possible placement j, we have two choices:
- Do not place the ad i at position j: Revenue remains as with the first i-1 ads, dp[i-1][j].
- Place the ad i at position j: Revenue is updated to include placing ad i at position j plus the maximum revenue achievable with i-1 ads and j-1 placements, dp[i-1][j-1] + revenue[i][j].

3. Initialization:

Initialize dp[0][0] = 0: Represents no ads placed and no placements, hence zero revenue.

4. Final State:

The maximum revenue achievable will be stored in dp[n][m], where n is the number of ads and m is the number of placements.

Algorithm

function optimizeAdPlacement(ads, placements):
    n = length(ads)          // Number of ads
    m = length(placements)   // Number of placement slots
    
    // Initializing dp array
    dp = array of size (n+1, m+1) with all values 0
    

    for i from 1 to n:
        for j from 1 to m:
            if placements[j-1] >= cost[i-1]:  
                dp[i][j] = max(dp[i-1][j], dp[i-1][j-1] + revenue[i-1])
            else:
                dp[i][j] = dp[i-1][j]
    
    // return maximum revenue
    return dp[n][m]

Knapsack Problem

The above dynamic problem approach resembles the knapsack 0/1 problem which also concentrates on maximizing the profit while having capacity constraints. In the knapsack problem, items have weights and values, and the goal is to maximize the total value without exceeding a weight limit. Similarly, in the revenue maximization problem, ads have revenues associated with specific placements, and the challenge is to maximize revenue within the constraints of available ad slots.

Time and Space Complexity

Time Complexity: O(N*logN) N:number of ads(items)
Space Complexity: O(N)

Zillow's Ad Placement Algorithm for maximizing revenue