diff --git a/CoinChange2.kt b/CoinChange2.kt
new file mode 100644
index 00000000..8ff1d4fe
--- /dev/null
+++ b/CoinChange2.kt
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+// In this problem, we use DP bottom up approach to calculate the number of ways to make the amount. We use a 1D array to store the number of ways. 
+// If the amount is less than the coin value, we carry forward the previous value. Otherwise, we add the number of ways to make the amount by including the coin.
+// Time Complexity: O(m*n) where m is the number of coins and n is the amount.
+// Space Complexity: O(n) where n is the amount.
+
+class Solution {
+    fun change(amount: Int, coins: IntArray): Int {
+    val m = coins.size
+    val n = amount
+
+    val dp = IntArray(n + 1)
+    dp[0] = 1
+
+    for(i in 1..m) {
+        for(j in 0..n) {
+            if(j < coins[i-1]) {
+                dp[j] = dp[j]
+            } else {
+                dp[j] = dp[j] + dp[j - coins[i - 1]]
+            }
+        }
+    }
+
+    return dp[n]
+}
+}
diff --git a/PaintHouse.kt b/PaintHouse.kt
new file mode 100644
index 00000000..a8a51ebe
--- /dev/null
+++ b/PaintHouse.kt
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+// In this problem, we have 3 variables for each color tracking the cost of painting the houses. we start from the last house and move to the first house,
+// updating the costs for each color based on the minimum cost of painting the next house with a different color.
+// Time Complexity: O(n) where n is the number of houses.
+// Space Complexity: O(1) constant extra space
+
+class Solution {
+    fun minCost(costs: Array<IntArray>): Int {
+        val n = costs.size
+        var cr = costs[n-1][0]
+        var cb = costs[n-1][1] 
+        var cg = costs[n-1][2]
+
+        for(i in n-2 downTo 0) {
+            val tmpR = cr
+            val tmpB = cb
+
+            cr = costs[i][0] + Math.min(cb, cg)
+            cb = costs[i][1] + Math.min(tmpR, cg)
+            cg = costs[i][2] + Math.min(tmpR, tmpB)
+        }
+
+        return Math.min(cr, Math.min(cb, cg))
+    }
+}
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