Add Neighborhood Burglary

Author: marttp

kotlin/Dynamic Programming/NeighborhoodBurglarOptimized.kt

@@ -0,0 +1,18 @@
+fun neighborhoodBurglaryOptimized(houses: List<Int>): Int {
+    if (houses.isEmpty()) {
+        return 0
+    }
+    if (houses.size == 1) {
+        return houses[0]
+    }
+    // Initialize the variables with the base cases.
+    var prevMaxProfit = maxOf(houses[0], houses[1])
+    var prevPrevMaxProfit = houses[0]
+    for (i in 2 until houses.size) {
+        val currMaxProfit = maxOf(prevMaxProfit, houses[i] + prevPrevMaxProfit)
+        // Update the values for the next iteration.
+        prevPrevMaxProfit = prevMaxProfit
+        prevMaxProfit = currMaxProfit
+    }
+    return prevMaxProfit
+}

kotlin/Dynamic Programming/NeighborhoodBurglary.kt

@@ -0,0 +1,22 @@
+fun neighborhoodBurglary(houses: List<Int>): Int {
+    // Handle the cases when the array is less than the size of 2 to
+    // avoid out-of-bounds errors when assigning the base case values.
+    if (houses.isEmpty()) {
+        return 0
+    }
+    if (houses.size == 1) {
+        return houses[0]
+    }
+    val dp = IntArray(houses.size)
+    // Base case: when there's only one house, rob that house.
+    dp[0] = houses[0]
+    // Base case: when there are two houses, rob the one with the most
+    // money.
+    dp[1] = maxOf(houses[0], houses[1])
+    // Fill in the rest of the DP array.
+    for (i in 2 until houses.size) {
+        // 'dp[i]' = max(profit if we skip house 'i', profit if we rob house 'i').
+        dp[i] = maxOf(dp[i - 1], houses[i] + dp[i - 2])
+    }
+    return dp[houses.size - 1]
+}