Solve the Equation

Solve the Equation - Problem

Solve a given equation and return the value of 'x' in the form of a string "x=#value".

The equation contains only '+' and '-' operations, the variable 'x' and its coefficient.

Return cases:

"No solution" if there is no solution for the equation
"Infinite solutions" if there are infinite solutions for the equation
"x=#value" if there is exactly one solution (where #value is an integer)

Input & Output

Example 1 — Basic Case

$ Input: equation = "x+5-3+x=6+x-2"

› Output: "x=2"

💡 Note: Left side: x + x + 5 - 3 = 2x + 2. Right side: 6 + x - 2 = x + 4. Moving terms: 2x - x = 4 - 2, so x = 2.

Example 2 — No Solution

$ Input: equation = "x=x"

› Output: "Infinite solutions"

💡 Note: Both sides are identical (x = x), so any value of x satisfies the equation.

Example 3 — Contradictory

$ Input: equation = "2x=x"

› Output: "x=0"

💡 Note: Moving terms: 2x - x = 0, so x = 0.

Constraints

The equation contains only '+', '-' operation, the variable 'x' and its coefficient
There is exactly one '=' in the equation
The equation is valid and follows the format described

Visualization

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Asked in

G Google 15 a Amazon 12

The key insight is to parse both sides of the equation and collect coefficients of x and constants separately. Move all x terms to the left side and constants to the right side to form ax = b, then solve based on the value of coefficient a. Best approach uses string parsing with O(n) time and O(1) space.

Common Approaches

✓ Character-by-Character Parsing

⏱️ Time: O(n) Space: O(1)

Iterate through each character of the equation, building numbers and handling signs. Track x coefficients and constants separately for left and right sides of the equation.

String Parsing with Sign Tracking

⏱️ Time: O(n) Space: O(1)

Split the equation into tokens, then process each token to determine if it's a coefficient of x or a constant. Use string manipulation to handle multi-digit numbers and implicit coefficients.

Character-by-Character Parsing — Algorithm Steps

Parse left side character by character
Parse right side character by character
Move all x terms to left, constants to right
Solve ax = b

Visualization

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Step-by-Step Walkthrough

Parse Left Side

x + 5 - 3 + x → x_coeff=2, const=2

Parse Right Side

6 + x - 2 → x_coeff=1, const=4

Solve

2x + 2 = x + 4 → x = 2

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void parseSide(char* side, int* xCoeff, int* constant) {
    *xCoeff = 0;
    *constant = 0;
    int i = 0;
    int sign = 1;
    int len = strlen(side);
    
    while (i < len) {
        if (side[i] == '+' || side[i] == '-') {
            sign = side[i] == '+' ? 1 : -1;
            i++;
        } else {
            int num = 0;
            int hasNum = 0;
            while (i < len && side[i] >= '0' && side[i] <= '9') {
                num = num * 10 + (side[i] - '0');
                hasNum = 1;
                i++;
            }
            
            if (i < len && side[i] == 'x') {
                // This is x coefficient
                if (!hasNum) {
                    *xCoeff += sign * 1;
                } else {
                    *xCoeff += sign * num;
                }
                i++;
            } else {
                // This is constant
                *constant += sign * num;
            }
        }
    }
}

char* solution(char* equation) {
    static char result[100];
    char* equalPos = strchr(equation, '=');
    *equalPos = '\0';
    
    char* left = equation;
    char* right = equalPos + 1;
    
    int leftX, leftConst, rightX, rightConst;
    parseSide(left, &leftX, &leftConst);
    parseSide(right, &rightX, &rightConst);
    
    // Move all x to left, constants to right: ax = b
    int a = leftX - rightX;
    int b = rightConst - leftConst;
    
    if (a == 0) {
        if (b == 0) {
            strcpy(result, "Infinite solutions");
        } else {
            strcpy(result, "No solution");
        }
    } else {
        sprintf(result, "x=%d", b / a);
    }
    
    return result;
}

int main() {
    char equation[1000];
    fgets(equation, sizeof(equation), stdin);
    equation[strcspn(equation, "\n")] = 0;
    printf("%s\n", solution(equation));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through equation string of length n

✓ Linear Growth

Space Complexity

O(1)

Only storing coefficients and constants as integers

✓ Linear Space

23.4K Views

Medium Frequency

~15 min Avg. Time

856 Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Character-by-Character Parsing — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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