Introduction

In this blog post, we will explore how to solve the LeetCode problem "640" 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

Solve a given equation and return the value of 'x' in the form of a string "x=#value". The equation contains only '+', '-' operation, the variable 'x' and its coefficient. You should return "No solution" if there is no solution for the equation, or "Infinite solutions" if there are infinite solutions for the equation. If there is exactly one solution for the equation, we ensure that the value of 'x' is an integer. Example 1: Input: equation = "x+5-3+x=6+x-2" Output: "x=2" Example 2: Input: equation = "x=x" Output: "Infinite solutions" Example 3: Input: equation = "2x=x" Output: "x=0" Constraints: 3 <= equation.length <= 1000 equation has exactly one '='. equation consists of integers with an absolute value in the range [0, 100] without any leading zeros, and the variable 'x'. The input is generated that if there is a single solution, it will be an integer.

Explanation

Here's the breakdown of the approach, complexities, and the Python code:

• High-Level Approach:

  • Parse the equation string, splitting it into left and right sides based on the '=' sign.
  • Process each side to collect the coefficient of 'x' and the constant term.
  • Solve for 'x' based on the collected coefficients and constants, handling special cases like no solution and infinite solutions.

• Complexity:

  • Runtime Complexity: O(n), where n is the length of the equation string.
  • Storage Complexity: O(1)

Code

    def solveEquation(equation: str) -> str:
    """Solves an equation and returns the value of 'x'.

    Args:
        equation: The equation string.

    Returns:
        The value of 'x' in the form of a string "x=#value",
        "No solution" if there is no solution, or
        "Infinite solutions" if there are infinite solutions.
    """

    def parse_side(side: str) -> tuple[int, int]:
        """Parses one side of the equation and returns the coefficient of x and the constant term."""
        x_coeff = 0
        const = 0
        i = 0
        sign = 1  # 1 for positive, -1 for negative

        while i < len(side):
            if side[i] == '+':
                sign = 1
                i += 1
            elif side[i] == '-':
                sign = -1
                i += 1
            else:
                j = i
                while j < len(side) and side[j].isdigit():
                    j += 1

                num_str = side[i:j]

                if j < len(side) and side[j] == 'x':
                    if num_str == "":
                        x_coeff += sign * 1
                    else:
                        x_coeff += sign * int(num_str)
                    j += 1
                else:
                    const += sign * int(num_str)

                i = j

        return x_coeff, const

    left, right = equation.split('=')
    x_coeff_left, const_left = parse_side(left)
    x_coeff_right, const_right = parse_side(right)

    x_coeff_total = x_coeff_left - x_coeff_right
    const_total = const_right - const_left

    if x_coeff_total == 0:
        if const_total == 0:
            return "Infinite solutions"
        else:
            return "No solution"
    else:
        x_value = const_total // x_coeff_total
        return "x=" + str(x_value)