algoadvance

Given a string expression 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"
Output: [0, 2]
Explanation: ((2-1)-1) = 0 
             (2-(1-1)) = 2

Example 2:

Input: "2*3-4*5"
Output: [-34, -14, -10, -10, 10]
Explanation: 
(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

Clarifying Questions

1. What is the length range of the input string?
   • Typically, we’ll assume it to be at most 20 characters for this problem.
2. Are there any parentheses in the input string?
   • No, the input string will only contain digits (0-9) and operators (+, -, *).
3. Are inputs guaranteed to be valid (e.g., no division by zero, no invalid characters)?
   • Yes, inputs are guaranteed to be valid.

Strategy

1. Divide and Conquer:
   • We’ll use a recursive approach which divides the problem at each operator.
   • For each operator in the input string, split the expression into left and right sub-expressions.
   • Recursively evaluate each sub-expression.
2. Combine Results:
   • For every result from the left sub-expression and every result from the right sub-expression, combine them using the current operator.
   • Store and return the combined results.
3. Base Case:
   • When the input string is a single number, return it as the base case.

Code

def diffWaysToCompute(expression):
    def compute(left, right, op):
        results = []
        for l in left:
            for r in right:
                if op == '+':
                    results.append(l + r)
                elif op == '-':
                    results.append(l - r)
                elif op == '*':
                    results.append(l * r)
        return results

    if expression.isdigit():
        return [int(expression)]

    results = []
    for i, ch in enumerate(expression):
        if ch in "+-*":
            left = diffWaysToCompute(expression[:i])
            right = diffWaysToCompute(expression[i+1:])
            results.extend(compute(left, right, ch))
    return results

# Example usages
print(diffWaysToCompute("2-1-1"))  # Output: [0, 2]
print(diffWaysToCompute("2*3-4*5"))  # Output: [-34, -14, -10, -10, 10]

Time Complexity

• Base Case: When a single number is the input, it takes constant time O(1) to process.
• Recursive Case: Let’s denote the number of operations in the expression as m. Each operation splits the problem into sub-problems.
• For each operator, the function divides the expression into two parts and solves each recursively. This results in a Catalan number which has a time complexity of O(4^n). However, the complexity can be simplified to an exponential function due to overlapping subproblems, making it approximately O(2^n).

Thus, while the exact number of operations is hard to pinpoint, it grows exponentially with the length of the input expression due to the nature of recursive splitting.

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