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House Robber

Certification: Intermediate Level Accuracy: 0% Submissions: 0 Points: 10

Write a JavaScript program to find the maximum sum of non-adjacent elements in an array representing money in houses. You are a robber planning to rob houses along a street, but you cannot rob two adjacent houses because it will trigger an alarm system.

Example 1
    
      
  • Input: nums = [2, 1, 4, 9]
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  • Output: 11
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  • Explanation: 
    • We can rob house 0 (money = 2) and house 2 (money = 4) for total = 6. 
    • We can rob house 1 (money = 1) and house 3 (money = 9) for total = 10. 
    • We can rob house 0 (money = 2) and house 3 (money = 9) for total = 11. 
    • The maximum sum is 11 by robbing houses 0 and 3.
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Example 2
    
      
  • Input: nums = [2, 7, 9, 3, 1]
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  • Output: 12
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  • Explanation: 
    • We can rob house 0 (money = 2), house 2 (money = 9), and house 4 (money = 1) for total = 12. 
    • We can rob house 1 (money = 7) and house 3 (money = 3) for total = 10. 
    • Other combinations yield lower sums. 
    • The maximum sum is 12.
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Constraints
    
      
  • 1 ≤ nums.length ≤ 100
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  • 0 ≤ nums[i] ≤ 400
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  • You cannot rob two adjacent houses
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  • Time Complexity: O(n)
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  • Space Complexity: O(1)
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ArraysDynamic Programming PwCD. E. Shaw
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Hint
Solution

Solution Hints

    
      
  • Use dynamic programming approach to solve this problem
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  • For each house, decide whether to rob it or not based on maximum profit
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  • If you rob current house, add previous non-adjacent maximum to current value
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  • If you don't rob current house, take the maximum up to previous house
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  • Keep track of maximum sum including and excluding current house
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Steps to solve by this approach:

 Step 1: Handle base cases - if array is empty return 0, if single element return that element
 Step 2: Initialize two variables: prevMax (max sum excluding previous) and currMax (max sum so far)
 Step 3: Iterate through each house in the array
 Step 4: For current house, calculate maximum between (prevMax + current house value) and currMax
 Step 5: Update prevMax to store the previous currMax value
 Step 6: Update currMax with the calculated maximum from step 4
 Step 7: Return currMax as the final maximum sum of non-adjacent elements

Submitted Code :