LeetCode 640: Solve the Equation Solution in Python – A Step-by-Step Guide

What Is LeetCode 640: Solve the Equation?

In LeetCode 640: Solve the Equation, you’re given a string equation representing a linear equation in the form "ax + b = cx + d", where a, b, c, d are integers, and x is the variable. Your task is to solve for x and return:

• "x=value" if there’s a unique solution (e.g., "x=2").
• "infinite solutions" if the equation holds for all x (e.g., 2x + 3 = 2x + 3).
• "no solution" if no x satisfies it (e.g., 0x + 1 = 0x + 2).

For example, "x+5-3+x=6+x-2" simplifies to 2x + 2 = x + 4, yielding x = 2. The equation may include multiple x terms, constants, and signs, requiring careful parsing. This problem tests your ability to manipulate strings and solve linear algebra programmatically.

Problem Statement

• Input:
• equation: A string in the form "expression1=expression2".
• Output: A string—"x=value", "infinite solutions", or "no solution".
• Rules:
• Parse left and right sides of '='.
• Extract coefficients of x and constants.
• Solve ax + b = cx + d for x.

Constraints

• 3 ≤ equation.length ≤ 1000.
• Contains only digits, 'x', '+', '-', '='.
• Valid equation with one '='.
• Integers may have multiple digits, no leading zeros unless 0 itself.

Examples

• Input: equation = "x+5-3+x=6+x-2"
• Output: "x=2"
• Input: equation = "x=x"
• Output: "infinite solutions"
• Input: equation = "2x=x"
• Output: "x=0"
• Input: equation = "0x+1=0x+2"
• Output: "no solution"

These examples set the equation—let’s solve it!

Understanding the Problem: Solving Equations

To solve LeetCode 640: Solve the Equation in Python, we need to parse a string equation, extract the coefficients of x and constants from both sides, simplify to ax = b, and determine x or special cases. A naive approach—manually splitting and checking each char—could work but risks errors with multi-digit numbers or signs. Since it’s a linear equation, we can optimize by parsing systematically. We’ll use:

• Best Solution (Parsing with Coefficient Extraction): O(n) time, O(1) space—fast and clear (n = string length).
• Alternative Solution (Regex-Based Parsing): O(n) time, O(1) space—concise but regex-heavy.

Let’s start with the parsing solution, breaking it down for beginners.

Best Solution: Parsing with Coefficient Extraction

Why This Is the Best Solution

The parsing with coefficient extraction approach is the top choice for LeetCode 640 because it’s efficient—O(n) time with O(1) space—and intuitive, breaking the equation into left and right sides, then systematically extracting x coefficients and constants without relying on complex tools like regex. It fits constraints (n ≤ 1000) and is easy to follow, making it ideal for learning. It’s like simplifying an equation on paper, step-by-step!

How It Works

Think of this as an equation simplifier:

• Step 1: Split Equation:
• Divide at '=' into left and right expressions.
• Step 2: Parse Expression:
• For each side:
• Scan char by char.
• Track sign (+/-), build numbers, count x’s and constants.
• Step 3: Extract Coefficients:
• Left: a (x coeff), b (constant).
• Right: c (x coeff), d (constant).
• Step 4: Simplify and Solve:
• Move terms: (a - c)x = d - b.
• If a - c ≠ 0: x = (d - b) / (a - c).
• If a - c = 0: Check if d - b = 0 (infinite) or not (no solution).
• Step 5: Return Result:
• Format as "x=value" or special case.

It’s like an equation balancer—parse and solve!

Step-by-Step Example

Example: equation = "x+5-3+x=6+x-2"

• Step 1: Split:
• Left = "x+5-3+x", Right = "6+x-2".
• Step 2: Parse Left:
• "x": x_coeff += 1 (sign +).
• "+5": const += 5.
• "-3": const -= 3.
• "+x": x_coeff += 1.
• Result: a = 2, b = 2.
• Step 3: Parse Right:
• "6": const += 6.
• "+x": x_coeff += 1.
• "-2": const -= 2.
• Result: c = 1, d = 4.
• Step 4: Solve:
• (a - c)x = d - b → (2 - 1)x = 4 - 2 → x = 2 / 1 = 2.
• Step 5: Return "x=2".
• Output: "x=2".

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained clearly:

Copy codeclass Solution:
    def solveEquation(self, equation: str) -> str:
        # Step 1: Split into left and right
        left, right = equation.split('=')

        # Step 2: Helper to parse an expression
        def parse(expr):
            x_coeff = 0  # Coefficient of x
            const = 0    # Constant term
            i = 0
            sign = 1     # Current sign (+1 or -1)
            num = 0      # Current number being built

            while i < len(expr):
                if expr[i] == '+':
                    sign = 1
                    i += 1
                elif expr[i] == '-':
                    sign = -1
                    i += 1
                elif expr[i].isdigit():
                    num = 0
                    while i < len(expr) and expr[i].isdigit():
                        num = num * 10 + int(expr[i])
                        i += 1
                    if i < len(expr) and expr[i] == 'x':
                        x_coeff += sign * (num if num else 1)
                        i += 1
                    else:
                        const += sign * num
                else:  # 'x'
                    x_coeff += sign
                    i += 1
            return x_coeff, const

        # Step 3: Parse both sides
        a, b = parse(left)   # Left: ax + b
        c, d = parse(right)  # Right: cx + d

        # Step 4: Solve (a - c)x = d - b
        coeff = a - c
        const = d - b

        if coeff == 0:
            return "infinite solutions" if const == 0 else "no solution"
        x = const // coeff

        # Step 5: Return result
        return f"x={x}"

• Line 4: Split at '='.
• Lines 7-32: parse:
• Track sign, build numbers, update x_coeff/const.
• Lines 35-36: Extract coefficients.
• Lines 39-44: Solve:
• Check coeff = 0 cases, compute x otherwise.
• Time Complexity: O(n)—single pass.
• Space Complexity: O(1)—few variables.

This is like an equation parser—clear and quick!

Alternative Solution: Regular Expression-Based Parsing

Why an Alternative Approach?

The regex-based approach uses regular expressions to extract terms—O(n) time, O(1) space. It’s more concise, leveraging pattern matching, but less intuitive for beginners due to regex complexity. It’s like a pattern scanner for equations!

How It Works

Picture this as an equation scanner:

• Step 1: Split at '='.
• Step 2: Use regex to match x terms and constants.
• Step 3: Sum coefficients and constants.
• Step 4: Solve as above.
• Step 5: Return result.

It’s like a regex solver—short but tricky!

Step-by-Step Example

Example: Same as above

• Step 1: Left = "x+5-3+x", Right = "6+x-2".
• Step 2: Regex:
• Left: Matches ["x", "+5", "-3", "+x"] → a = 2, b = 2.
• Right: Matches ["6", "+x", "-2"] → c = 1, d = 4.
• Step 3: Solve: x = 2.
• Output: "x=2".

Code for Regex Approach

Copy codeimport re

class Solution:
    def solveEquation(self, equation: str) -> str:
        left, right = equation.split('=')

        def parse(expr):
            # Match terms: optional sign, optional digits, optional 'x'
            terms = re.findall(r'[+-]?\d*x|[+-]?\d+', expr)
            x_coeff = 0
            const = 0
            for term in terms:
                if 'x' in term:
                    coeff = term.rstrip('x')
                    x_coeff += int(coeff) if coeff not in ['+', '-'] else (1 if coeff == '+' else -1)
                else:
                    const += int(term)
            return x_coeff, const

        a, b = parse(left)
        c, d = parse(right)

        coeff = a - c
        const = d - b

        if coeff == 0:
            return "infinite solutions" if const == 0 else "no solution"
        x = const // coeff
        return f"x={x}"

• Line 1: Import re for regex.
• Lines 7-17: parse: Use regex to extract terms.
• Time Complexity: O(n)—regex parsing.
• Space Complexity: O(1)—few variables.

It’s a regex cracker—short but dense!

Comparing the Two Solutions

• Parsing (Best):
• Pros: O(n) time, O(1) space, clear and manual.
• Cons: More code.
• Regex (Alternative):
• Pros: O(n) time, O(1) space, concise.
• Cons: Regex learning curve.

Parsing wins for clarity.

Additional Examples and Edge Cases

• Input: "x=2x"
• Output: "x=0".
• Input: "0x=0x"
• Output: "infinite solutions".

Both handle these well.

Complexity Breakdown

• Parsing: Time O(n), Space O(1).
• Regex: Time O(n), Space O(1).

Parsing is optimal for understanding.

Key Takeaways

• Parsing: Step-by-step—smart!
• Regex: Pattern power—clear!
• Equations: Solving is fun.
• Python Tip: Strings rock—see Python Basics.

Final Thoughts: Solve That Equation

LeetCode 640: Solve the Equation in Python is a neat parsing challenge. Parsing with coefficient extraction offers clarity and control, while regex provides a sleek alternative. Want more? Try LeetCode 67: Add Binary or LeetCode 415: Add Strings. Ready to balance? Head to Solve LeetCode 640 on LeetCode and solve that equation today!