[Leetcode] Divide and conquer, DP--241. Different Ways to Add Parentheses

Given a string of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. The valid operators are +, - and *.

Example 1

Input: "2-1-1".

((2-1)-1) = 0
(2-(1-1)) = 2

Output: [0, 2]

Example 2

Input: "2*3-4*5"

(2*(3-(4*5))) = -34
((2*3)-(4*5)) = -14
((2*(3-4))*5) = -10
(2*((3-4)*5)) = -10
(((2*3)-4)*5) = 10

Output: [-34, -14, -10, -10, 10]

Solution:

1. 1st use divide and conquer   

        we can recursively divide the string into left and right part substring.

　　 then  recursively calculate the left and right part substring  result when there is only one numerical number and the other numberical number with a valid operator

　　 finally,  merge their results.

　    note:

　　how to divide?  iterate from the input string;  if there is a "+", "-", "*" operator,  we will divide the string into left and right substring.

　　when there is only one numerical number, directly add into the result list

 

 1  def calcMath(num1, num2, operator):           #python does not have switch/case
 2             res = 0
 3             if operator == "+":
 4                 res = num1 + num2
 5             elif operator == "-":
 6                 res = num1 - num2
 7             elif operator == "*":
 8                 res = num1 * num2
 9             return res
10             
11         result = []
12         for i, c in enumerate(input):
13             if  c == "+" or c == "-" or c == "*":
14                 leftSubStr  = input[0:i]
15                 rightSubStr = input[i+1::]
16                 
17                 leftLst = self.diffWaysToCompute(leftSubStr)
18                 rightLst = self.diffWaysToCompute(rightSubStr)
19                 #print ("ttttt  ", i, c, leftLst, rightLst)
20                 for ls in leftLst:
21                     for rs in rightLst:
22                         rs = calcMath(ls, rs, c)
23                         result.append(rs)
24                 #print ("ttttt  ", result)
25         if len(result) == 0:
26             result.append(int(input))
27         return result

   

　　2.  2nd optimization 1st method; (Dynamic programming) using hashmap to store previous subproblem's result; Dynamic programming
　　#in the previous calculations, there are repeated calculation of substring with same left substring and right substring calculation

 1 def calcMath(num1, num2, operator):           #python does not have switch/case
 2             res = 0
 3             if operator == "+":
 4                 res = num1 + num2
 5             elif operator == "-":
 6                 res = num1 - num2
 7             elif operator == "*":
 8                 res = num1 * num2
 9             return res
10             
11        
12             
13         def  diffWaysToComputeHelper(input, start, end):
14             
15             result = []
16             strKey = str(start) + "-" + str(end)
17             if strKey in dic:
18                 result = dic[strKey]
19                 return result
20             #print ("sssss  ", result, start, end)
21             for i, c in enumerate(input[start:end+1]):
22                 if  c == "+" or c == "-" or c == "*":
23                     leftLst = diffWaysToComputeHelper(input, start, start+i-1 )
24                     rightLst = diffWaysToComputeHelper(input, start+i+1, end)
25                     #print ("xxxxx  ", i, c, leftLst, rightLst, start, end)
26                     for ls in leftLst:
27                         for rs in rightLst:
28                             rs = calcMath(ls, rs, c)
29                             result.append(rs)
30                            
31             #print ("ttttt  ", result, start, end, input[start:end+1])
32             if len(result) == 0:
33                 result.append(int(input[start:end+1]))
34             dic[strKey] = result
35             return result
36         
37         dic = {}                #memoization
38         return diffWaysToComputeHelper(input, 0, len(input)-1)

--reference:

http://www.geeksforgeeks.org/all-ways-to-add-parenthesis-for-evaluation/