Interview Question - Minimum Coin Change - DP

Author: rachitiitr

LeetCode/322.coin-change.py

@@ -0,0 +1,28 @@
+#
+# @lc app=leetcode id=322 lang=python3
+#
+# [322] Coin Change
+#
+
+# @lc code=start
+INF = 10 ** 7
+class Solution:
+    def coinChange(self, C: List[int], S: int) -> int:
+        noCoins = len(C)
+        dp = [[INF for _ in range(noCoins+1)] for _ in range(S+1)]
+        # dp[i][j] is min coins needed to make sum "i" when first "j" coins are available
+        dp[0][0] = 0 # no coin needed to make 0
+        for i in range(0, S+1):
+            for j in range(1, noCoins+1):
+                curCoin = C[j-1] # 0-based indexing
+                
+                # Case A: I don't take jth coin
+                dp[i][j] = dp[i][j-1]
+                
+                # Case B: I take jth coin
+                if curCoin <= i:
+                    dp[i][j] = min(dp[i][j], 1 + dp[i-curCoin][j])
+        return dp[S][noCoins] if dp[S][noCoins] != INF else -1
+        
+# @lc code=end
+