How do you solve Integer Replacement on LeetCode?

Integer Replacement (LeetCode #397) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log n) time complexity and O(1) space complexity. Key insight: For odd numbers, the choice between incrementing or decrementing can significantly affect the number of operations needed.

What is the brute force solution for Integer Replacement?

The brute force approach for Integer Replacement is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach involves recursively trying all possible operations on the number until we reach 1. It explores both options for odd numbers and halves even numbers, counting the steps taken. The optimal Optimal Solution approach improves this to O(log n).

What algorithm or data structure is used to solve Integer Replacement?

Integer Replacement uses the following concepts: Dynamic Programming, Greedy, Bit Manipulation, Memoization. The recommended patterns to study are: Bit Manipulation, Greedy Algorithms.

What are the interview tips for Integer Replacement?

Always explain your thought process clearly when choosing between operations. Consider edge cases, especially for small values of n. Practice recognizing patterns in number manipulation problems.

What are common mistakes when solving Integer Replacement?

Not considering the implications of choosing n + 1 vs n - 1 for odd numbers. Overlooking the efficiency gained from halving even numbers.

#397

Integer Replacement

Medium

Dynamic ProgrammingGreedyBit ManipulationMemoizationBit ManipulationGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(log n)
Space	O(n)	O(1)

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Intuition

Time O(log n)Space O(1)

The optimal solution uses a greedy approach combined with bit manipulation. It reduces the number of operations by choosing the best operation for odd numbers based on their binary representation.

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Algorithm

3 steps

1Step 1: Initialize a counter for operations.
2Step 2: While n is greater than 1, check if n is even or odd.
3Step 3: If n is even, divide by 2. If odd, check the last two bits to decide whether to increment or decrement n, aiming to make it even.

solution.py12 lines

1def integerReplacement(n):
2    count = 0
3    while n > 1:
4        if n % 2 == 0:
5            n //= 2
6        else:
7            if n == 3 or (n & 2) == 0:
8                n -= 1
9            else:
10                n += 1
11        count += 1
12    return count

ℹ

Complexity note: The time complexity is O(log n) because each operation effectively reduces n by half in the case of even numbers, and the number of operations is logarithmic relative to the size of n.

1For odd numbers, the choice between incrementing or decrementing can significantly affect the number of operations needed.
2Understanding binary representation helps in determining the optimal path for odd numbers.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.