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You are given a string that represents an equation of the form ax + b = cx + d where:

Your task is to solve for x.

If there is a solution for x, return it as a string in the form "x=#value", where #value is an integer. If there are infinite solutions, return "Infinite solutions". If there is no solution, return "No solution".

Example 1:

Input: "x+5-3+x=6+x-2"
Output: "x=2"

Example 2:

Input: "x=x"
Output: "Infinite solutions"

Example 3:

Input: "2x=x"
Output: "x=0"

Example 4:

Input: "2x+3x-6x=x+2"
Output: "x=-1"

Example 5:

Input: "x=x+2"
Output: "No solution"

Clarifying Questions

  1. Are all the constants and coefficients in the equation integers?
  2. Can the input string always be assumed to be a valid equation format?

Strategy

  1. Parse the Equation: Split the equation at the = sign to separate it into the left-hand side (LHS) and the right-hand side (RHS).
  2. Coefficient Extraction: Define a helper function to extract the coefficient of x and the constant term from the string.
  3. Combine Coefficients: Sum up the coefficients of x and constants for both sides of the equation.
  4. Solve Equation:
    • If the coefficients of x on both sides are the same, check the constants:
      • If constants are also the same, return "Infinite solutions".
      • Otherwise, return "No solution".
    • If the coefficients are different, solve for x.

Code

Here’s a complete Python code implementing the strategy:

def solveEquation(equation: str) -> str:
    def parse_expr(expr):
        # Initialize variables to hold the sum of x coefficients and constant term
        x_coeff = 0
        const = 0
        num = 0
        sign = 1  # 1 means positive sign, -1 means negative sign
        i = 0
        n = len(expr)
        while i < n:
            if expr[i] == '+':
                sign = 1
                i += 1
            elif expr[i] == '-':
                sign = -1
                i += 1
            elif expr[i] == 'x':
                # Check if 'x' is preceded by a number
                if i == 0 or not expr[i-1].isdigit():
                    # The coefficient before 'x' is either 1 or -1
                    x_coeff += sign
                else:
                    x_coeff += sign * num
                num = 0
                i += 1
            else:
                num = 0
                while i < n and expr[i].isdigit():
                    num = num * 10 + int(expr[i])
                    i += 1
                if i < n and expr[i] == 'x':
                    x_coeff += sign * num
                    i += 1
                else:
                    const += sign * num
                    num = 0
        return x_coeff, const

    # Split the equation into LHS and RHS
    lhs, rhs = equation.split('=')
    lhs_x_coeff, lhs_const = parse_expr(lhs)
    rhs_x_coeff, rhs_const = parse_expr(rhs)

    # Calculate the net coefficients and constant term
    net_x_coeff = lhs_x_coeff - rhs_x_coeff
    net_const = rhs_const - lhs_const

    if net_x_coeff == 0:
        if net_const == 0:
            return "Infinite solutions"
        else:
            return "No solution"
    else:
        x_value = net_const // net_x_coeff
        return f"x={x_value}"

# Test the function with a sample input
equation = "x+5-3+x=6+x-2"
print(solveEquation(equation))  # Output: "x=2"

Time Complexity

The time complexity of this solution is O(n), where n is the length of the input string. This is because we are processing each character in the string exactly once during the parsing steps.

Let me know if you have any further questions or need additional clarifications!

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