Python Solution for Coin Change Problem in DP

Author: dsrao711

Dynamic Programming/CoinChange.py

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+# https://leetcode.com/problems/coin-change/
+
+#### In the given problem , we are supposed to find minimum number of coins to get the sum equal to the amount given
+
+#### For given example
+# - Input :
+
+# 1. coins = [7 , 5 , 1]
+# 2. amount = 18
+
+# - O/p :
+#     4
+
+# We can achieve 18 by 
+# - 2 Coins of 5
+# - 1 coin of 7
+# - 1 coin of 1
+# - Sum = 4
+
+# 18 can be achieved by various combinations of 7 , 5 , 1
+
+# https://ibb.co/TtcVRYx
+
+# Refer the image above
+
+# Lets call 11 , 13 , 17 , 6 , 12 , 10 ... as subSums
+
+# As shown in the image above , 6 has been calculated previously and later on 6 is required in latter part .
+# Instead of computing it again , we can store the value for 6 i.e 2 (1 coin of 5 and 1 coin of 1)
+
+# As we need to store the value of recursion ie. Subsums in a memory , hence we choose DP here
+
+# Consider a list op = [] with size equal to the amount
+
+# op[] stores the number of coins required for each number from  1, 2 ,...amount 
+# We are supposed to return op[amount] 
+
+# In this case op[18] returns the minimum coins required to produce the sum 18
+
+
+class Solution(object):
+    def coinChange(self, coins, amount):
+        """
+        :type coins: List[int]
+        :type amount: int
+        :rtype: int
+        """
+        op = [amount+1] * (amount+1)
+        op[0] = 0
+        for i in xrange(1, amount+1):
+            for c in coins:
+                if i >= c:
+                    op[i] = min(op[i], op[i-c] + 1)
+                    
+        if op[amount] >= amount+1:
+             return -1
+        return op[amount]
+    
+