LeetCode #2843 — EASY

Count Symmetric Integers

Build confidence with an intuition-first walkthrough focused on math fundamentals.

The Problem

Problem Statement

You are given two positive integers low and high.

An integer x consisting of 2 * n digits is symmetric if the sum of the first n digits of x is equal to the sum of the last n digits of x. Numbers with an odd number of digits are never symmetric.

Return the number of symmetric integers in the range [low, high].

Example 1:

Input: low = 1, high = 100
Output: 9
Explanation: There are 9 symmetric integers between 1 and 100: 11, 22, 33, 44, 55, 66, 77, 88, and 99.

Example 2:

Input: low = 1200, high = 1230
Output: 4
Explanation: There are 4 symmetric integers between 1200 and 1230: 1203, 1212, 1221, and 1230.

Constraints:

1 <= low <= high <= 10⁴

Step 01

Brute Force Baseline

Problem summary: You are given two positive integers low and high. An integer x consisting of 2 * n digits is symmetric if the sum of the first n digits of x is equal to the sum of the last n digits of x. Numbers with an odd number of digits are never symmetric. Return the number of symmetric integers in the range [low, high].

Baseline thinking

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

Pattern signal: Math

Example 1

1
100

Example 2

1200
1230

Core Insight

What unlocks the optimal approach

<div class="_1l1MA">Iterate over all numbers from <code>low</code> to <code>high</code></div>
<div class="_1l1MA">Convert each number to a string and compare the sum of the first half with that of the second.</div>

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 #2843:   Count Symmetric Integers
class Solution {
    public int countSymmetricIntegers(int low, int high) {
        int ans = 0;
        for (int x = low; x <= high; ++x) {
            ans += f(x);
        }
        return ans;
    }

    private int f(int x) {
        String s = "" + x;
        int n = s.length();
        if (n % 2 == 1) {
            return 0;
        }
        int a = 0, b = 0;
        for (int i = 0; i < n / 2; ++i) {
            a += s.charAt(i) - '0';
        }
        for (int i = n / 2; i < n; ++i) {
            b += s.charAt(i) - '0';
        }
        return a == b ? 1 : 0;
    }
}

// Accepted solution for LeetCode #2843:   Count Symmetric Integers
func countSymmetricIntegers(low int, high int) (ans int) {
	f := func(x int) int {
		s := strconv.Itoa(x)
		n := len(s)
		if n&1 == 1 {
			return 0
		}
		a, b := 0, 0
		for i := 0; i < n/2; i++ {
			a += int(s[i] - '0')
			b += int(s[n/2+i] - '0')
		}
		if a == b {
			return 1
		}
		return 0
	}
	for x := low; x <= high; x++ {
		ans += f(x)
	}
	return
}

# Accepted solution for LeetCode #2843:   Count Symmetric Integers
class Solution:
    def countSymmetricIntegers(self, low: int, high: int) -> int:
        def f(x: int) -> bool:
            s = str(x)
            if len(s) & 1:
                return False
            n = len(s) // 2
            return sum(map(int, s[:n])) == sum(map(int, s[n:]))

        return sum(f(x) for x in range(low, high + 1))

// Accepted solution for LeetCode #2843:   Count Symmetric Integers
impl Solution {
    pub fn count_symmetric_integers(low: i32, high: i32) -> i32 {
        let mut ans = 0;
        for x in low..=high {
            ans += Self::f(x);
        }
        ans
    }

    fn f(x: i32) -> i32 {
        let s = x.to_string();
        let n = s.len();
        if n % 2 == 1 {
            return 0;
        }
        let bytes = s.as_bytes();
        let mut a = 0;
        let mut b = 0;
        for i in 0..n / 2 {
            a += (bytes[i] - b'0') as i32;
        }
        for i in n / 2..n {
            b += (bytes[i] - b'0') as i32;
        }
        if a == b {
            1
        } else {
            0
        }
    }
}

// Accepted solution for LeetCode #2843:   Count Symmetric Integers
function countSymmetricIntegers(low: number, high: number): number {
    let ans = 0;
    const f = (x: number): number => {
        const s = x.toString();
        const n = s.length;
        if (n & 1) {
            return 0;
        }
        let a = 0;
        let b = 0;
        for (let i = 0; i < n >> 1; ++i) {
            a += Number(s[i]);
            b += Number(s[(n >> 1) + i]);
        }
        return a === b ? 1 : 0;
    };
    for (let x = low; x <= high; ++x) {
        ans += f(x);
    }
    return ans;
}

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(n × log m)

Space

O(log m)

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.