2443. Sum of Number and Its Reverse
Explanation:
To solve this problem, we need to iterate through all possible values up to half of the given number and check if the sum of the number and its reverse equals the given number. We can reverse a number by using modulo and division operations.
- Iterate from 0 to num/2.
- For each iteration, calculate the reverse of the current number.
- Check if the sum of the number and its reverse equals the given number.
- If a match is found, return true. Otherwise, return false at the end.
Time complexity: O(log(num)) - number of digits in num Space complexity: O(1) :
class Solution {
public boolean isSumOfNumberAndReverse(int num) {
for (int i = 0; i <= num / 2; i++) {
int reverse = 0;
int n = i;
while (n > 0) {
reverse = reverse * 10 + n % 10;
n = n / 10;
}
if (i + reverse == num) {
return true;
}
}
return false;
}
}Code Editor (Testing phase)
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