Introduction

In this blog post, we will explore how to solve the LeetCode problem "282" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given a string num that contains only digits and an integer target, return all possibilities to insert the binary operators '+', '-', and/or '' between the digits of num so that the resultant expression evaluates to the target value. Note that operands in the returned expressions should not contain leading zeros. Example 1: Input: num = "123", target = 6 Output: ["123","1+2+3"] Explanation: Both "123" and "1+2+3" evaluate to 6. Example 2: Input: num = "232", target = 8 Output: ["23+2","2+32"] Explanation: Both "23+2" and "2+3*2" evaluate to 8. Example 3: Input: num = "3456237490", target = 9191 Output: [] Explanation: There are no expressions that can be created from "3456237490" to evaluate to 9191. Constraints: 1 <= num.length <= 10 num consists of only digits. -231 <= target <= 231 - 1

Explanation

Here's the breakdown of the solution:

• Backtracking: The solution employs a backtracking algorithm to explore all possible combinations of operators and operands.
• Recursive Exploration: The core logic resides within a recursive function that iterates through the digits of the input number. At each step, it considers adding '+', '-', or '*' operators or no operator (appending to the current number).

• Evaluation and Pruning: The recursive function maintains the current expression's value and the value of the last operand. When it reaches the end of the number, it checks if the expression's value matches the target. If not, the path is pruned. Leading zeros are also avoided during number formation.

• Runtime Complexity: O(4^N), where N is the length of the input string num. Each digit has 4 possibilities: '+', '-', '*', or no operator (concatenation).

• Storage Complexity: O(N), where N is the length of the input string num. This is primarily due to the depth of the recursion stack and the space used to store the intermediate expressions.

Code

    def addOperators(num: str, target: int) -> list[str]:
    """
    Finds all possible ways to insert binary operators ('+', '-', '*') between the digits of num
    so that the resultant expression evaluates to the target value.

    Args:
        num: A string containing only digits.
        target: An integer target value.

    Returns:
        A list of strings, where each string is a valid expression that evaluates to the target.
    """

    def helper(index: int, current_expression: str, current_value: int, previous_operand: int) -> None:
        """
        Recursive helper function to explore all possible combinations of operators and operands.

        Args:
            index: The current index in the num string.
            current_expression: The current expression string being built.
            current_value: The current evaluated value of the expression.
            previous_operand: The value of the last operand added to the expression.
        """
        if index == len(num):
            if current_value == target:
                result.append(current_expression)
            return

        for i in range(index, len(num)):
            # Avoid leading zeros
            if i > index and num[index] == '0':
                break

            current_num = int(num[index:i + 1])

            if index == 0:
                # First number in the expression
                helper(i + 1, str(current_num), current_num, current_num)
            else:
                # Addition
                helper(i + 1, current_expression + "+" + str(current_num), current_value + current_num, current_num)

                # Subtraction
                helper(i + 1, current_expression + "-" + str(current_num), current_value - current_num, -current_num)

                # Multiplication
                helper(i + 1, current_expression + "*" + str(current_num), current_value - previous_operand + previous_operand * current_num, previous_operand * current_num)

    result: list[str] = []
    helper(0, "", 0, 0)
    return result