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7 4. House 3 (Amount: 3): temp = 11 (previous currMax) currMax = max(11, 7 + 3) = 10 prevMax = 11 5. House 1 (Amount: 1): temp = 10 (previous currMax) currMax = max(10, 11 + 1) = 12 prevMax = 10 The final value of currMax is 12, which represents the maximum amount that can be robbed without robbing