Completed DP2

Solutions.py

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+# Problem1(https://leetcode.com/problems/paint-house/)
+# Time Complexity : O(N)
+# Space Complexity : O(N)
+# Did this code successfully run on Leetcode :yes
+# Any problem you faced while coding this :no
+
+## no adjacent house can be painted with the same colour
+## starting with the greedy selecting smallest value first, can work for some test cases but wont work for many 
+## going with the exaustive approach of creating 3 trees and deciding which one will be better. 
+## the time complexity will be exponential in that case, can I do any better? 
+## trying to convert exuastive to memoization by storing the results of already completed nodes
+## We have 3 cases to choose from, and each step we can calculate the cost, by choosing that and one of either two remaining
+## storing the results in memo, as we calculate
+## finally returning the min of three trees
+
+def helper(i,j,costs,memo):
+    ## base case
+    if i>=len(costs):
+        return 0
+    if memo[i][j] != -1:
+        return memo[i][j]
+    ## case 1 : choose red
+    if j ==0:
+        res= costs[i][j] + min(helper(i+1,1,costs,memo), helper(i+1,2,costs,memo))
+        memo[i][j] = res
+        return res
+
+    ## case 2 : choose blue 
+    if j ==1:
+        res= costs[i][j] + min(helper(i+1,0,costs,memo), helper(i+1,2,costs,memo))
+        memo[i][j] = res
+        return res
+    ## case 3 : choose green
+    if j==2:
+        res= costs[i][j] + min(helper(i+1,0,costs,memo), helper(i+1,1,costs,memo))
+        memo[i][j] = res
+        return res
+
+rows = len(costs)
+cols = 3 ## 3 colors
+memo = [[-1]*(cols+1) for _ in range(rows+1) ]
+
+colr = helper(0,0,costs,memo)
+colb = helper(0,1,costs,memo)
+colg = helper(0,2,costs,memo)
+
+return min(colr,min(colb,colg))
+
+
+
+# Problem2 (https://leetcode.com/problems/coin-change-2/)
+# Time Complexity : O(N)
+# Space Complexity : O(N)
+# Did this code successfully run on Leetcode :yes
+# Any problem you faced while coding this :no
+
+## starting with the greedy approach again to choose the biggest coin and start decreasing the amount until amount is 0 
+## the greedy approach fails for the non conical denomination coins. 
+## we can start with exaustive approach to select and not select each coin and build a tree
+## the exaustive approach will be O(2^n) because of the depth\
+## converting to memoization to save the some complexity
+
+
+def helper(amount,coins,idx):
+    ## base case
+
+    if amount ==0:
+        return 1
+    if amount <0 or idx >= len(coins):
+        return 0
+    if memo[amount][idx] != -1:
+        return memo[amount][idx]
+    ## choose the coin
+    case1 = helper(amount-coins[idx],coins,idx)
+    ## dont choose the coin
+    case2 = helper(amount,coins,idx+1)
+    res = case1+case2
+    memo[amount][idx] = res
+    return res
+
+memo = [[-1]*len(coins) for _ in range(amount+1)]
+res = helper(amount,coins,0)
+return res