Minimize Result by Adding Parentheses to Expression

Medium

Asked by:

13 views

Topics:

StringsArrays

You are given a 0-indexed string expression of the form "<num1>+<num2>" where <num1> and <num2> represent positive integers.

Add a pair of parentheses to expression such that after the addition of parentheses, expression is a valid mathematical expression and evaluates to the smallest possible value. The left parenthesis must be added to the left of '+' and the right parenthesis must be added to the right of '+'.

Return expression after adding a pair of parentheses such that expression evaluates to the smallest possible value. If there are multiple answers that yield the same result, return any of them.

The input has been generated such that the original value of expression, and the value of expression after adding any pair of parentheses that meets the requirements fits within a signed 32-bit integer.

Example 1:

Input: expression = "247+38"
Output: "2(47+38)"
Explanation: The expression evaluates to 2 * (47 + 38) = 2 * 85 = 170.
Note that "2(4)7+38" is invalid because the right parenthesis must be to the right of the '+'.
It can be shown that 170 is the smallest possible value.

Example 2:

Input: expression = "12+34"
Output: "1(2+3)4"
Explanation: The expression evaluates to 1 * (2 + 3) * 4 = 1 * 5 * 4 = 20.

Example 3:

Input: expression = "999+999"
Output: "(999+999)"
Explanation: The expression evaluates to 999 + 999 = 1998.

Constraints:

3 <= expression.length <= 10
expression consists of digits from '1' to '9' and '+'.
expression starts and ends with digits.
expression contains exactly one '+'.
The original value of expression, and the value of expression after adding any pair of parentheses that meets the requirements fits within a signed 32-bit integer.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the maximum possible values of the two integers in the expression?
Will the input expression always be in the format 'integer + integer' with no leading/trailing spaces or other characters?
If there are multiple parenthesis placements that result in the same minimum value, is any one of them acceptable?
Can you provide examples of edge cases or inputs with very small or very large numbers?
Is the '+' operator guaranteed to be the only operator present in the expression?

Brute Force Solution

Approach

The brute force strategy explores every single possible way to add parentheses to the expression. It's like trying every combination and checking which one gives you the smallest final number. We'll go through each combination, calculate the result, and keep track of the best one.

Here's how the algorithm would work step-by-step:

Imagine you can put a left parenthesis anywhere before the plus sign and a right parenthesis anywhere after the plus sign.
Start by putting the left parenthesis in the very first possible spot, right before the first number.
Then, try putting the right parenthesis in every spot after the plus sign, one at a time.
For each of these parenthesis combinations, calculate the result of the expression.
Now, move the left parenthesis to the next possible spot, and repeat the process of trying all the different spots for the right parenthesis.
Keep doing this until you've tried putting the left parenthesis in every possible location.
While you're doing all this, remember to keep track of the smallest result you've found so far, and the corresponding placement of the parentheses.
After you've tried all the possibilities, the smallest result you've kept track of will be your answer.

Code Implementation

def minimize_result(expression):
    minimum_result = float('inf')
    best_expression = ""

    plus_index = expression.find('+')

    # Iterate through all possible positions for the left parenthesis
    for left_index in range(plus_index):
        # Iterate through all possible positions for the right parenthesis
        for right_index in range(plus_index + 1, len(expression) + 1):

            # Extract the parts of the expression
            left_part = expression[:left_index]
            inside_part = expression[left_index:right_index]
            right_part = expression[right_index:]

            # Evaluate the expression within the parentheses
            inside_numbers = inside_part.split('+')
            inside_result = int(inside_numbers[0]) + int(inside_numbers[1])

            # Handle multiplication for the left and right parts
            left_value = 1
            if left_part:
                left_value = int(left_part)

            right_value = 1
            if right_part:
                right_value = int(right_part)

            current_result = left_value * inside_result * right_value

            # Update the minimum result and best expression if needed
            if current_result < minimum_result:
                minimum_result = current_result
                best_expression = left_part + '(' + inside_part + ')' + right_part

    return best_expression

Big(O) Analysis

Time Complexity

O(n³) – Let n be the length of the input string. The outer loop iterates through possible positions for the left parenthesis, which takes O(n) time. The inner loop iterates through possible positions for the right parenthesis, which also takes O(n) time. For each combination of parenthesis positions, evaluating the resulting expression involves string slicing and arithmetic operations, taking O(n) time. Thus, the overall time complexity is O(n * n * n) = O(n³).

Space Complexity

O(1) – The brute force approach described iterates through different combinations of parenthesis placements but does not explicitly create any auxiliary data structures whose size depends on the input expression. It keeps track of the smallest result found so far and the corresponding parenthesis placement, but these are stored in a few variables that take constant space regardless of the input string length N. Therefore, the auxiliary space complexity is constant.

Optimal Solution

Approach

The key to solving this problem efficiently is realizing we don't need to check every single place we could put the parentheses. We can focus on strategically placing the opening and closing parentheses to minimize the result of the expression.

Here's how the algorithm would work step-by-step:

Find the position of the plus sign in the string, as it separates the two numbers we need to work with.
Consider every possible place to put the opening parenthesis *before* the plus sign. This means starting from the beginning of the first number and moving towards the plus sign.
For each opening parenthesis position, consider every possible place to put the closing parenthesis *after* the plus sign. This means starting right after the plus sign and moving towards the end of the second number.
For each combination of opening and closing parenthesis positions, evaluate the expression. Remember you're trying to minimize the result of the entire expression.
Keep track of the minimum result you've found so far, and the corresponding expression with the parentheses.
After checking all possible parenthesis positions, return the expression that produced the minimum result.

Code Implementation

def minimize_result(expression):
    addition_index = expression.find('+')

    minimum_result = float('inf')
    best_expression = ""

    # Try all possible left parenthesis placements
    for left_parenthesis_index in range(addition_index):
        
        #Try all possible right parenthesis placements
        for right_parenthesis_index in range(addition_index + 1, len(expression) + 1):
            
            current_expression = (
                expression[:left_parenthesis_index]
                + '('
                + expression[left_parenthesis_index:right_parenthesis_index]
                + ')'
                + expression[right_parenthesis_index:]
            )

            # Evaluate the expression with parentheses
            left_operand_string = current_expression[:left_parenthesis_index]
            parenthesized_expression = current_expression[left_parenthesis_index:right_parenthesis_index+2]
            right_operand_string = current_expression[right_parenthesis_index+2:]

            plus_index_local = parenthesized_expression.find('+')
            
            first_number_string = parenthesized_expression[1:plus_index_local]
            second_number_string = parenthesized_expression[plus_index_local+1:-1]
            
            first_number = int(first_number_string)
            second_number = int(second_number_string)
            parenthesized_result = first_number + second_number

            # Handle empty strings for multiplication to work
            left_operand = int(left_operand_string) if left_operand_string else 1
            right_operand = int(right_operand_string) if right_operand_string else 1
            
            current_result = left_operand * parenthesized_result * right_operand

            # Track the expression for minimum result
            if current_result < minimum_result:
                # Need to update best found expression.
                minimum_result = current_result
                best_expression = (
                    expression[:left_parenthesis_index]
                    + '('
                    + expression[left_parenthesis_index:right_parenthesis_index]
                    + ')'
                    + expression[right_parenthesis_index:]
                )

    return best_expression

Big(O) Analysis

Time Complexity

O(n³) – The algorithm first finds the plus sign in the input string of length n, which takes O(n) time but doesn't dominate the complexity. The algorithm then iterates to place the opening parenthesis. The outer loop can iterate up to n times in the worst case. For each opening parenthesis placement, the inner loop iterates to place the closing parenthesis, which can also iterate up to n times. Inside these nested loops, the algorithm evaluates the expression formed by the current parenthesis placements. Evaluating the expression involves extracting substrings and performing arithmetic operations, which can take O(n) time in the worst case. Therefore, the overall time complexity is O(n * n * n), which simplifies to O(n³).

Space Complexity

O(1) – The algorithm primarily utilizes variables to store the position of the plus sign and indices for iterating through possible parenthesis placements. It also uses variables to store the minimum result found so far and the corresponding string, all of which consume constant space regardless of the input string's length (N). No auxiliary data structures, like lists or hash maps, are created that scale with the size of the input string, resulting in a constant auxiliary space complexity.

Edge Cases

Case	How to Handle
Null or empty input string	Return an empty string or throw an IllegalArgumentException as there is no expression to process.
Input string does not contain a '+' operator.	Return the original string, or throw an IllegalArgumentException as the input doesn't conform to the expected format.
Input string contains non-numeric characters other than '+'.	Filter non-numeric chars except '+' or throw an IllegalArgumentException for invalid input.
Integer overflow during multiplication or addition.	Use long data types for intermediate calculations or check for overflow before performing operations.
Input string contains leading zeros in the numbers.	Remove leading zeros before parsing the numbers to ensure correct arithmetic.
Multiple pairs of parentheses result in the same minimum value	Return the lexicographically smallest string among those with the minimum value; any minimal solution is acceptable as long as consistent.
The numbers in the string are very large (close to the maximum integer value).	Use long data types for calculations and compare to LONG_MAX/MIN before casting to prevent overflow.
String contains multiple '+' operators, or no number between the '+' operators.	Return empty string, throw error, or consider only the first valid 'number1 + number2' pattern.

Null or empty input string

How to Handle:

Return an empty string or throw an IllegalArgumentException as there is no expression to process.

Input string does not contain a '+' operator.

How to Handle:

Return the original string, or throw an IllegalArgumentException as the input doesn't conform to the expected format.

Input string contains non-numeric characters other than '+'.

How to Handle:

Filter non-numeric chars except '+' or throw an IllegalArgumentException for invalid input.

Integer overflow during multiplication or addition.

How to Handle:

Use long data types for intermediate calculations or check for overflow before performing operations.

Input string contains leading zeros in the numbers.

How to Handle:

Remove leading zeros before parsing the numbers to ensure correct arithmetic.

Multiple pairs of parentheses result in the same minimum value

How to Handle:

Return the lexicographically smallest string among those with the minimum value; any minimal solution is acceptable as long as consistent.

The numbers in the string are very large (close to the maximum integer value).

How to Handle:

Use long data types for calculations and compare to LONG_MAX/MIN before casting to prevent overflow.

String contains multiple '+' operators, or no number between the '+' operators.

How to Handle:

Return empty string, throw error, or consider only the first valid 'number1 + number2' pattern.