LeetCode #592 — MEDIUM

Fraction Addition and Subtraction

Move from brute-force thinking to an efficient approach using math strategy.

The Problem

Problem Statement

Given a string expression representing an expression of fraction addition and subtraction, return the calculation result in string format.

The final result should be an irreducible fraction. If your final result is an integer, change it to the format of a fraction that has a denominator 1. So in this case, 2 should be converted to 2/1.

Example 1:

Input: expression = "-1/2+1/2"
Output: "0/1"

Example 2:

Input: expression = "-1/2+1/2+1/3"
Output: "1/3"

Example 3:

Input: expression = "1/3-1/2"
Output: "-1/6"

Constraints:

The input string only contains '0' to '9', '/', '+' and '-'. So does the output.
Each fraction (input and output) has the format ±numerator/denominator. If the first input fraction or the output is positive, then '+' will be omitted.
The input only contains valid irreducible fractions, where the numerator and denominator of each fraction will always be in the range [1, 10]. If the denominator is 1, it means this fraction is actually an integer in a fraction format defined above.
The number of given fractions will be in the range [1, 10].
The numerator and denominator of the final result are guaranteed to be valid and in the range of 32-bit int.

Step 01

Brute Force Baseline

Problem summary: Given a string expression representing an expression of fraction addition and subtraction, return the calculation result in string format. The final result should be an irreducible fraction. If your final result is an integer, change it to the format of a fraction that has a denominator 1. So in this case, 2 should be converted to 2/1.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

"-1/2+1/2"

Example 2

"-1/2+1/2+1/3"

Example 3

"1/3-1/2"

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #592: Fraction Addition and Subtraction
class Solution {
    public String fractionAddition(String expression) {
        int x = 0, y = 6 * 7 * 8 * 9 * 10;
        if (Character.isDigit(expression.charAt(0))) {
            expression = "+" + expression;
        }
        int i = 0, n = expression.length();
        while (i < n) {
            int sign = expression.charAt(i) == '-' ? -1 : 1;
            ++i;
            int j = i;
            while (j < n && expression.charAt(j) != '+' && expression.charAt(j) != '-') {
                ++j;
            }
            String s = expression.substring(i, j);
            String[] t = s.split("/");
            int a = Integer.parseInt(t[0]), b = Integer.parseInt(t[1]);
            x += sign * a * y / b;
            i = j;
        }
        int z = gcd(Math.abs(x), y);
        x /= z;
        y /= z;
        return x + "/" + y;
    }

    private int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
}

// Accepted solution for LeetCode #592: Fraction Addition and Subtraction
func fractionAddition(expression string) string {
	x, y := 0, 6*7*8*9*10
	if unicode.IsDigit(rune(expression[0])) {
		expression = "+" + expression
	}
	i, n := 0, len(expression)
	for i < n {
		sign := 1
		if expression[i] == '-' {
			sign = -1
		}
		i++
		j := i
		for j < n && expression[j] != '+' && expression[j] != '-' {
			j++
		}
		s := expression[i:j]
		t := strings.Split(s, "/")
		a, _ := strconv.Atoi(t[0])
		b, _ := strconv.Atoi(t[1])
		x += sign * a * y / b
		i = j
	}
	z := gcd(abs(x), y)
	x /= z
	y /= z
	return fmt.Sprintf("%d/%d", x, y)
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

func gcd(a, b int) int {
	if b == 0 {
		return a
	}
	return gcd(b, a%b)
}

# Accepted solution for LeetCode #592: Fraction Addition and Subtraction
class Solution:
    def fractionAddition(self, expression: str) -> str:
        x, y = 0, 6 * 7 * 8 * 9 * 10
        if expression[0].isdigit():
            expression = '+' + expression
        i, n = 0, len(expression)
        while i < n:
            sign = -1 if expression[i] == '-' else 1
            i += 1
            j = i
            while j < n and expression[j] not in '+-':
                j += 1
            s = expression[i:j]
            a, b = s.split('/')
            x += sign * int(a) * y // int(b)
            i = j
        z = gcd(x, y)
        x //= z
        y //= z
        return f'{x}/{y}'

// Accepted solution for LeetCode #592: Fraction Addition and Subtraction
struct Solution;
use std::fmt::{Display, Formatter, Result};

enum Tok {
    Op(char),
    Num(i32),
}

struct Fraction {
    sign: i32,
    numerator: i32,
    denominator: i32,
}

impl Fraction {
    fn new(sign: i32, numerator: i32, denominator: i32) -> Self {
        Fraction {
            sign,
            numerator,
            denominator,
        }
    }
    fn add(self, other: Self) -> Self {
        let mut numerator = self.sign * self.numerator * other.denominator
            + other.sign * other.numerator * self.denominator;
        let denominator = self.denominator * other.denominator;
        let mut sign = 1;
        if numerator < 0 {
            sign *= -1;
            numerator *= -1;
        }
        Fraction {
            sign,
            numerator,
            denominator,
        }
    }
    fn reduce(self) -> Self {
        let sign = self.sign;
        let mut denominator = self.denominator;
        let mut numerator = self.numerator;
        let gcd = gcd(denominator, numerator);
        denominator /= gcd;
        numerator /= gcd;
        Fraction {
            sign,
            numerator,
            denominator,
        }
    }
}

impl Display for Fraction {
    fn fmt(&self, f: &mut Formatter<'_>) -> Result {
        if self.sign > 0 {
            write!(f, "{}/{}", self.numerator, self.denominator)
        } else {
            write!(f, "-{}/{}", self.numerator, self.denominator)
        }
    }
}

fn gcd(mut m: i32, mut n: i32) -> i32 {
    while m != 0 {
        let temp = m;
        m = n % temp;
        n = temp;
    }
    n.abs()
}

impl Solution {
    fn fraction_addition(expression: String) -> String {
        let mut toks: Vec<Tok> = vec![];
        let mut c_it = expression.chars().peekable();
        while let Some(c) = c_it.next() {
            match c {
                '-' | '+' | '/' => {
                    toks.push(Tok::Op(c));
                }
                _ => {
                    let mut val = (c as u8 - b'0') as i32;
                    while let Some(p) = c_it.peek() {
                        if p.is_digit(10) {
                            val *= 10;
                            val += (c_it.next().unwrap() as u8 - b'0') as i32;
                        } else {
                            break;
                        }
                    }
                    toks.push(Tok::Num(val));
                }
            }
        }
        let mut t_it = toks.into_iter().peekable();
        let mut fractions: Vec<Fraction> = vec![];
        while t_it.peek().is_some() {
            let mut sign = 1;
            let numerator;
            let denominator;
            if let Some(Tok::Op('-')) = t_it.peek() {
                t_it.next().unwrap();
                sign = -1;
            }
            if let Some(Tok::Op('+')) = t_it.peek() {
                t_it.next().unwrap();
            }
            if let Tok::Num(x) = t_it.next().unwrap() {
                numerator = x;
            } else {
                panic!();
            }
            if let Tok::Op('/') = t_it.next().unwrap() {
            } else {
                panic!();
            }
            if let Tok::Num(y) = t_it.next().unwrap() {
                denominator = y;
            } else {
                panic!();
            }
            fractions.push(Fraction::new(sign, numerator, denominator));
        }
        while fractions.len() > 1 {
            let a = fractions.pop().unwrap();
            let b = fractions.pop().unwrap();
            let c = a.add(b);
            fractions.push(c);
        }
        let mut res = fractions.pop().unwrap();
        res = res.reduce();
        res.to_string()
    }
}

#[test]
fn test() {
    let expression = "-1/2+1/2".to_string();
    let res = "0/1".to_string();
    assert_eq!(Solution::fraction_addition(expression), res);
    let expression = "-1/2+1/2+1/3".to_string();
    let res = "1/3".to_string();
    assert_eq!(Solution::fraction_addition(expression), res);
    let expression = "1/3-1/2".to_string();
    let res = "-1/6".to_string();
    assert_eq!(Solution::fraction_addition(expression), res);
    let expression = "5/3+1/3".to_string();
    let res = "2/1".to_string();
    assert_eq!(Solution::fraction_addition(expression), res);
}

// Accepted solution for LeetCode #592: Fraction Addition and Subtraction
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #592: Fraction Addition and Subtraction
// class Solution {
//     public String fractionAddition(String expression) {
//         int x = 0, y = 6 * 7 * 8 * 9 * 10;
//         if (Character.isDigit(expression.charAt(0))) {
//             expression = "+" + expression;
//         }
//         int i = 0, n = expression.length();
//         while (i < n) {
//             int sign = expression.charAt(i) == '-' ? -1 : 1;
//             ++i;
//             int j = i;
//             while (j < n && expression.charAt(j) != '+' && expression.charAt(j) != '-') {
//                 ++j;
//             }
//             String s = expression.substring(i, j);
//             String[] t = s.split("/");
//             int a = Integer.parseInt(t[0]), b = Integer.parseInt(t[1]);
//             x += sign * a * y / b;
//             i = j;
//         }
//         int z = gcd(Math.abs(x), y);
//         x /= z;
//         y /= z;
//         return x + "/" + y;
//     }
// 
//     private int gcd(int a, int b) {
//         return b == 0 ? a : gcd(b, a % b);
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.