How do you solve Count Symmetric Integers on LeetCode?

Count Symmetric Integers (LeetCode #2843) can be solved using Brute Force. The optimal approach is Brute Force with undefined time complexity and undefined space complexity. Key insight: The brute force approach checks each number in the range from low to high. For each number, it converts it to a string, splits it into two halves, and compares the sums of the two halves to determine if it's symmetric.

What is the time complexity of Count Symmetric Integers?

The optimal solution for Count Symmetric Integers runs in undefined time and uses undefined space. undefined

What algorithm or data structure is used to solve Count Symmetric Integers?

Count Symmetric Integers uses the following concepts: Math, Enumeration. The recommended patterns to study are: .

#2843

Count Symmetric Integers

Easy

MathEnumeration

LeetCode ↗

Approaches

💡

Intuition

Time Space

The brute force approach checks each number in the range from low to high. For each number, it converts it to a string, splits it into two halves, and compares the sums of the two halves to determine if it's symmetric.

⚙️

Algorithm

6 steps

1Step 1: Initialize a counter to zero.
2Step 2: Loop through each number from low to high.
3Step 3: Convert the number to a string and check if its length is even.
4Step 4: Split the string into two halves and calculate the sum of digits for both halves.
5Step 5: If the sums are equal, increment the counter.
6Step 6: Return the counter.

solution.py12 lines

1# Full working Python code
2low, high = 1, 100
3count = 0
4for num in range(low, high + 1):
5    s = str(num)
6    if len(s) % 2 == 0:
7        mid = len(s) // 2
8        left_sum = sum(int(d) for d in s[:mid])
9        right_sum = sum(int(d) for d in s[mid:])
10        if left_sum == right_sum:
11            count += 1
12print(count)

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.