LeetCode 397: Integer Replacement Solution
Master LeetCode problem 397 (Integer Replacement), a medium 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.
397. Integer Replacement
Problem Explanation
Explanation
To solve this problem, we can use a recursive approach. At each step, we check if the number is even or odd. If it's even, we divide it by 2, if it's odd, we check the parity of the number after adding or subtracting 1, and choose the option that minimizes the number of operations. We keep track of the number of operations needed to reach 1 and return this count as the result.
Time Complexity
The time complexity of this approach is O(log n) because at each step, the number is halved.
Space Complexity
The space complexity is O(log n) due to the recursive calls on the call stack.
Solution Code
class Solution {
public int integerReplacement(int n) {
if (n == 1) {
return 0;
}
if (n % 2 == 0) {
return 1 + integerReplacement(n / 2);
} else {
return 1 + Math.min(integerReplacement(n + 1), integerReplacement(n - 1));
}
}
}Try It Yourself
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What is the time complexity of LeetCode 397 (Integer Replacement)?
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