2843. Count Symmetric Integers LeetCode Solution

In this guide, you will get 2843. Count Symmetric Integers LeetCode Solution with the best time and space complexity. The solution to Count Symmetric Integers problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Count Symmetric Integers solution in C++
Count Symmetric Integers solution in Java
Count Symmetric Integers solution in Python
Additional Resources

2843. Count Symmetric Integers LeetCode Solution image

Problem Statement of Count Symmetric Integers

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

Complexity Analysis

Time Complexity: O(\texttt{high})
Space Complexity: O(1)

2843. Count Symmetric Integers LeetCode Solution in C++

class Solution {
 public:
  int countSymmetricIntegers(int low, int high) {
    int ans = 0;

    for (int num = low; num <= high; ++num)
      if (isSymmetricInteger(num))
        ++ans;

    return ans;
  }

 private:
  bool isSymmetricInteger(int num) {
    if (num >= 10 && num <= 99)
      return num / 10 == num % 10;
    if (num >= 1000 && num <= 9999) {
      const int left = num / 100;
      const int right = num % 100;
      return left / 10 + left % 10 == right / 10 + right % 10;
    }
    return false;
  }
};
/* code provided by PROGIEZ */

2843. Count Symmetric Integers LeetCode Solution in Java

class Solution {
  public int countSymmetricIntegers(int low, int high) {
    int ans = 0;

    for (int num = low; num <= high; ++num)
      if (isSymmetricInteger(num))
        ++ans;

    return ans;
  }

  private boolean isSymmetricInteger(int num) {
    if (num >= 10 && num <= 99)
      return num / 10 == num % 10;
    if (num >= 1000 && num <= 9999) {
      final int left = num / 100;
      final int right = num % 100;
      return left / 10 + left % 10 == right / 10 + right % 10;
    }
    return false;
  }
}
// code provided by PROGIEZ

2843. Count Symmetric Integers LeetCode Solution in Python

class Solution:
  def countSymmetricIntegers(self, low: int, high: int) -> int:
    def isSymmetricInteger(num: int) -> bool:
      if num >= 10 and num <= 99:
        return num // 10 == num % 10
      if num >= 1000 and num <= 9999:
        left = num // 100
        right = num % 100
        return left // 10 + left % 10 == right // 10 + right % 10
      return False

    return sum(isSymmetricInteger(num) for num in range(low, high + 1))
 # code by PROGIEZ