Coin Change

Problem1.py

@@ -0,0 +1,19 @@
+# Time Complexity --> O(m*n) where m is the amount and n is the number of coins
+# Space Complexity --> O(m)
+# Approach --> Still going over all the possibilities but optimizing the search by reusing the result for repeated subproblems. This is the Tabulation DP, a bottom up approach. The space complexity is optimized to use only a single array of length m+1 since the result at each node only depends on its curent value or the value before that position in array. 
+class Solution:
+    def coinChange(self, coins: List[int], amount: int) -> int:
+        n = len(coins)
+        m = amount
+        dp = [0 for i in range(m+1)]
+        for i in range(1,m+1):
+            dp[i] = amount+1
+
+        for i in range(1,n+1):
+            for j in range(1,m+1):
+                if j>=coins[i-1]:
+                    dp[j] = min(dp[j], 1+dp[j-coins[i-1]])
+        #print(dp)
+        if dp[m]>amount:
+            return -1
+        return dp[m]