Count Symmetric Integers - Practice Coding Problems

Count Symmetric Integers - Problem

Math Enumeration Easy

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].

Input & Output

Example 1 — Small Range

$ Input: low = 1, high = 100

› Output: 9

💡 Note: Symmetric numbers: 11 (1=1), 22 (2=2), 33 (3=3), 44 (4=4), 55 (5=5), 66 (6=6), 77 (7=7), 88 (8=8), 99 (9=9). All single digits and odd-length numbers are not symmetric.

Example 2 — Four Digit Range

$ Input: low = 1200, high = 1230

› Output: 1

💡 Note: Only 1221 is symmetric in this range: left sum = 1+2 = 3, right sum = 2+1 = 3. Numbers like 1234 have left sum = 1+2 = 3, right sum = 3+4 = 7.

Example 3 — No Symmetric Numbers

$ Input: low = 1, high = 10

› Output: 0

💡 Note: No symmetric numbers exist. Single digit numbers (1-9) have odd length, and 10 has left sum = 1, right sum = 0.

Constraints

1 ≤ low ≤ high ≤ 10⁹
Numbers with odd digits are never symmetric
Only even-length numbers can be symmetric

Visualization

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Understanding the Visualization

Input Range

Given range [1, 100] to check for symmetric numbers

Check Symmetry

Split even-length numbers and compare digit sums

Count Results

Count numbers where left sum = right sum

Key Takeaway

🎯 Key Insight: Only even-length numbers can be symmetric - split in half and compare digit sums

Asked in

G Google 25 M Microsoft 20 a Amazon 15

The key insight is that only numbers with even digit count can be symmetric, where left half digit sum equals right half digit sum. Brute force checks every number in range O(n×d). Mathematical approach organizes by digit length for better range filtering. Time: O(n×d), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force Check Every Number	O(n × d)	O(1)	Iterate through all numbers and check if each one is symmetric
Mathematical Enumeration	O(10^k × k)	O(1)	Generate symmetric numbers by digit length patterns

Brute Force Check Every Number — Algorithm Steps

Step 1: Iterate through each number from low to high
Step 2: For each number, check if it has even number of digits
Step 3: Split digits in half and compare sums of left and right parts
Step 4: Count numbers where left sum equals right sum

Visualization

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Step-by-Step Walkthrough

Check Each Number

Iterate through range checking digit sums

Split and Compare

Split digits in half and compare sums

Count Matches

Count numbers with equal left/right sums

Code -

solution.c — C

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

int solution(int low, int high) {
    int count = 0;
    for (int num = low; num <= high; num++) {
        char s[20];
        sprintf(s, "%d", num);
        int len = strlen(s);
        if (len % 2 == 0) {  // Only even length numbers can be symmetric
            int mid = len / 2;
            int leftSum = 0, rightSum = 0;
            for (int i = 0; i < mid; i++) {
                leftSum += s[i] - '0';
                rightSum += s[mid + i] - '0';
            }
            if (leftSum == rightSum) {
                count++;
            }
        }
    }
    return count;
}

int main() {
    int low, high;
    scanf("%d", &low);
    scanf("%d", &high);
    int result = solution(low, high);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × d)

n numbers in range, d average digits per number for sum calculation

✓ Linear Growth

Space Complexity

O(1)

Only using variables for current number processing

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Check Every Number — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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