LeetCode 2119: A Number After a Double Reversal Solution
Master LeetCode problem 2119 (A Number After a Double Reversal), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
2119. A Number After a Double Reversal
Problem Explanation
Explanation
To solve this problem, we need to reverse the given integer twice and check if the final result is equal to the original number. We can achieve this by implementing a function to reverse a number, applying it twice, and then comparing the result with the original number.
- Reverse the given number.
- Reverse the result obtained from step 1.
- Compare the final result with the original number to determine if they are equal.
Time Complexity
The time complexity of this solution is O(log n), where n is the value of the input number. This is because we iterate through the digits of the number in each step of the reversal process.
Space Complexity
The space complexity of this solution is O(1) as we are not using any extra space that grows with the input size.
Solution Code
class Solution {
public boolean isNumberAfterDoubleReversal(int num) {
int reversed1 = reverseNumber(num);
int reversed2 = reverseNumber(reversed1);
return reversed2 == num;
}
private int reverseNumber(int num) {
int reversed = 0;
while (num != 0) {
reversed = reversed * 10 + num % 10;
num /= 10;
}
return reversed;
}
}Try It Yourself
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How to solve LeetCode 2119 (A Number After a Double Reversal)?
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What is the time complexity of LeetCode 2119 (A Number After a Double Reversal)?
The time complexity for LeetCode 2119 (A Number After a Double Reversal) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.
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