LeetCode Entry
2843. Count Symmetric Integers
Numbers with equal half' sums
2843. Count Symmetric Integers easy
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Problem TLDR
Numbers with equal half’ sums #easy
Intuition
Brute. Force.
Approach
- convert to string and don’t
Complexity
-
Time complexity: \(O((h - l)lg(h))\)
-
Space complexity: \(O(1)\)
Code
fun countSymmetricIntegers(low: Int, high: Int) =
(low..high).count {
val s = "$it".map { it - '0' }; val n = s.size
n % 2 < 1 && s.sum() == s.take(n / 2).sum() * 2
}
pub fn count_symmetric_integers(low: i32, high: i32) -> i32 {
(low..=high).filter(|&x| {
let (mut a, mut b, mut s, mut h, mut c) = (x, x, 0, 0, 0);
while a > 0 { s += a % 10; a /= 10; c += 1;
if c % 2 > 0 { h += b % 10; b /= 10 }}
c & 1 < 1 && s == h * 2
}).count() as _
}
int countSymmetricIntegers(int low, int high) {
int r = 0;
for (int x = low; x <= high; ++x) {
int a = x, b = x, s = 0, h = 0, c = 0;
while (a) {
s += a % 10; a /= 10; ++c;
if (c % 2) { h += b % 10; b /= 10; }
}
r += c % 2 < 1 && s == h * 2;
}
return r;
}