What is the brute force solution for Minimize Result by Adding Parentheses to Expression?

The brute force approach for Minimize Result by Adding Parentheses to Expression is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try adding parentheses at every possible position around the '+' sign and calculate the result for each configuration. This will help us find the minimum value by brute force. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimize Result by Adding Parentheses to Expression?

Minimize Result by Adding Parentheses to Expression uses the following concepts: String, Enumeration. The recommended patterns to study are: Enumeration, Dynamic Programming.

What are the interview tips for Minimize Result by Adding Parentheses to Expression?

Always clarify the problem statement and constraints before jumping into coding. Think aloud to communicate your thought process and reasoning to the interviewer. Consider edge cases and test your solution with them to ensure robustness.

What are common mistakes when solving Minimize Result by Adding Parentheses to Expression?

Failing to account for cases where num1 or num2 could be empty, leading to incorrect calculations. Not considering the edge cases where the left or right part of the expression might be '1'.

#2232

Minimize Result by Adding Parentheses to Expression

Medium

StringEnumerationEnumerationDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

Instead of checking every possible position, we can calculate the minimum value directly by iterating through the expression and keeping track of potential splits around the '+' sign. This reduces unnecessary calculations.

⚙️

Algorithm

4 steps

1Step 1: Split the expression into num1 and num2 at the '+' sign.
2Step 2: Initialize variables to track the minimum value and the best expression.
3Step 3: Iterate through possible split points in num1 and num2, calculating the potential minimum value directly.
4Step 4: Update the minimum value and expression whenever a smaller value is found.

solution.py14 lines

1def minimizeResult(expression):
2    num1, num2 = expression.split('+')
3    min_value = float('inf')
4    result_expression = ''
5    for i in range(len(num1) + 1):
6        for j in range(len(num2)):
7            left = num1[:i] or '1'
8            right = num2[j:]
9            middle = num1[i:] + num2[:j]
10            value = int(left) * (int(middle) + int(right))
11            if value < min_value:
12                min_value = value
13                result_expression = f'{left}({middle}+{right})'
14    return result_expression

ℹ

Complexity note: The time complexity is O(n) because we only iterate through the string once to find the minimum value. The space complexity is O(1) since we are using a constant amount of space.

1The placement of parentheses significantly affects the result of the expression.
2Understanding how multiplication distributes over addition is key to minimizing the result.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.