2843. Count Symmetric Integers

Description

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

Solutions

Solution 1: Enumeration

We enumerate each integer \(x\) in the range \([low, high]\), and check whether it is a palindromic number. If it is, then the answer \(ans\) is increased by \(1\).

The time complexity is \(O(n \times \log m)\), and the space complexity is \(O(\log m)\). Here, \(n\) is the number of integers in the range \([low, high]\), and \(m\) is the maximum integer given in the problem.

Python3JavaC++GoTypeScript

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))

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;
    }
}

class Solution {
public:
    int countSymmetricIntegers(int low, int high) {
        int ans = 0;
        auto f = [](int x) {
            string s = to_string(x);
            int n = s.size();
            if (n & 1) {
                return 0;
            }
            int a = 0, b = 0;
            for (int i = 0; i < n / 2; ++i) {
                a += s[i] - '0';
                b += s[n / 2 + i] - '0';
            }
            return a == b ? 1 : 0;
        };
        for (int x = low; x <= high; ++x) {
            ans += f(x);
        }
        return ans;
    }
};

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
}

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;
}

2843. Count Symmetric Integers

Description

Solutions

Solution 1: Enumeration

Comments