Completed DP-2

problem-1.py

@@ -0,0 +1,46 @@
+# LEETCODE PROBLEM PAINT HOUSE
+# TIME COMPLEXITY:O(N) where N is the array length for the given number of colors present
+# SPACE COMPLEXITY: O(1)
+# Any problem you faced while coding this: None
+
+
+class Solution:
+    def minCost(self,costs):
+
+        #Approach 1
+        # m=len(costs)
+        # n=len(costs[0])
+
+        # dp=[[0]*n for _ in range(m)]
+
+        # dp[0][0]=costs[0][0]
+        # dp[0][1]=costs[0][1]
+        # dp[0][2]=costs[0][2]
+
+        # for i in range(1,m):
+        #     dp[i][0]=costs[i][0]+min(dp[i-1][1],dp[i-1][2])
+        #     dp[i][1]=costs[i][1]+min(dp[i-1][0],dp[i-1][2])
+        #     dp[i][2]=costs[i][2]+min(dp[i-1][0],dp[i-1][1])
+
+        # return min(dp[m-1][0],dp[m-1][1],dp[m-1][2])
+
+        #Approach 2
+        m=len(costs)
+        n=costs[0]
+        colorRed=costs[0][0]
+        colorBlue=costs[0][1]
+        colorGreen=costs[0][2]
+
+        for i in range(1,m):
+            tempRed=colorRed
+            tempBlue=colorBlue
+
+            colorRed=costs[i][0]+min(colorBlue,colorGreen)
+            colorBlue=costs[i][1]+min(tempRed,colorGreen)
+            colorGreen=costs[i][2]+min(tempRed,tempBlue)
+
+        return min(colorRed,colorBlue,colorGreen)
+
+costs=[[17,2,17],[16,16,5],[14,3,19]]
+sol=Solution()
+print(sol.minCost(costs))

problem-2.py

@@ -0,0 +1,44 @@
+# LEETCODE PROBLEM 518. COIN CHANGE II
+# TIME COMPLEXITY:O(M *N) where M is the number of coins and N is the amount
+# SPACE COMPLEXITY: O(N) where N amount
+# Any problem you faced while coding this: Had a few hiccup in like understanding also while 
+# writing it
+
+
+
+class Solution(object):
+    def change(self, amount, coins):
+        """
+        :type amount: int
+        :type coins: List[int]
+        :rtype: int
+        """
+        #Approach 1
+        # In this method a table is being made and the referencing is being done on the 
+        # basis of the value of j example if j is 6 and is greater than the coin we are
+        # referencing ie 5 we would add the value present in the row above the same
+        # col and in the current row we will go back to 6-5=1 column to get value and then
+        # add it
+        # m=len(coins)
+        # n=amount
+        # dp=[[0]*(n+1) for _ in range(m+1)]
+        # dp[0][0]=1
+        # for i in range(1,m+1):
+        #     for j in range(n+1):
+        #         if (j<coins[i-1]):
+        #             dp[i][j]=dp[i-1][j]
+        #         else:
+        #             dp[i][j]=dp[i-1][j]+dp[i][j-coins[i-1]]
+        # return dp[m][n]
+
+        #Approach 2
+        # The column is updated as per the coin value that we are iterating and on that
+        # basis we get a final answer which will be the last column of the given matrix
+        m=len(coins)
+        n=amount
+        dp=[0]*(n+1)
+        dp[0]=1
+        for i in range(1,m+1):
+            for j in range(coins[i-1],n+1):
+                dp[j]=dp[j]+dp[j-coins[i-1]]
+        return dp[n]