LeetCode-in-Java
3918. Sum of Primes Between Number and Its Reverse
Medium
You are given an integer n.
Let r be the integer formed by reversing the digits of n.
Return the sum of all prime numbers between min(n, r) and max(n, r), inclusive.
Example 1:
Input: n = 13
Output: 132
Explanation:
- The reverse of 13 is 31. Thus, the range is
[13, 31]. - The prime numbers in this range are 13, 17, 19, 23, 29, and 31.
- The sum of these prime numbers is
13 + 17 + 19 + 23 + 29 + 31 = 132.
Example 2:
Input: n = 10
Output: 17
Explanation:
- The reverse of 10 is 1. Thus, the range is
[1, 10]. - The prime numbers in this range are 2, 3, 5, and 7.
- The sum of these prime numbers is
2 + 3 + 5 + 7 = 17.
Example 3:
Input: n = 8
Output: 0
Explanation:
- The reverse of 8 is 8. Thus, the range is
[8, 8]. - There are no prime numbers in this range, so the sum is 0.
Constraints:
1 <= n <= 1000
Solution
public class Solution {
private boolean isPrime(int x) {
if (x <= 1) {
return false;
}
if (x == 2) {
return true;
}
if (x % 2 == 0) {
return false;
}
for (int i = 3; i * i <= x; i += 2) {
if (x % i == 0) {
return false;
}
}
return true;
}
private int reverseNum(int n) {
int r = 0;
while (n > 0) {
r = r * 10 + (n % 10);
n /= 10;
}
return r;
}
public int sumOfPrimesInRange(int n) {
int r = reverseNum(n);
int low = Math.min(n, r);
int high = Math.max(n, r);
int sum = 0;
for (int i = low; i <= high; i++) {
if (isPrime(i)) {
sum += i;
}
}
return sum;
}
}