DP 1

LC198_DP_house_robber.py

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+#  Time Complexity : O(n)
+#  Space Complexity : O(n)
+#  Did this code successfully run on Leetcode : yes
+#  Any problem you faced while coding this : No
+
+# Approach: built a dp[i] as the minimum coins needed to make amount i, checking every coin for each amount
+
+
+def rob(nums: list[int]) -> int:
+    n = len(nums)
+    if n == 1:
+        return nums[0]
+    prev = nums[0]
+    curr = max(nums[0], nums[1])
+
+    for i in range(2, n):
+        temp = curr
+        curr = max(temp, nums[i] + prev)
+        prev = temp
+
+    return curr

LC322_DP_coin_change.py

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+#  Time Complexity : O(n)
+#  Space Complexity : O(1)
+#  Did this code successfully run on Leetcode : yes
+#  Any problem you faced while coding this : No
+
+# Approach: at each house, choosing max(skip current, rob current + money from i-2) while keeping only the previous two DP values
+
+
+def coinChange(coins: list[int], amount: int) -> int:
+    m = len(coins)
+    n = amount
+    # m rows and n cols
+    dp = [0] * (n + 1)
+
+    for j in range(1, n + 1):
+        dp[j] = 99999
+
+    for i in range(1, m + 1):
+        for j in range(n + 1):
+            # choose case
+            if j >= coins[i - 1]:
+                dp[j] = min(dp[j], dp[j - coins[i - 1]] + 1)
+
+    if dp[n] == 99999:
+        return -1
+    return dp[n]