How do you solve Minimum Moves to Reach Target Score on LeetCode?

Minimum Moves to Reach Target Score (LeetCode #2139) 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: Using the doubling operation strategically can significantly reduce the number of moves needed.

What is the brute force solution for Minimum Moves to Reach Target Score?

The brute force approach for Minimum Moves to Reach Target Score is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves trying every possible combination of increments and doubles until we reach the target. This method is straightforward but inefficient, especially for larger targets. The optimal Optimal Solution approach improves this to O(log n).

What algorithm or data structure is used to solve Minimum Moves to Reach Target Score?

Minimum Moves to Reach Target Score uses the following concepts: Math, Greedy. The recommended patterns to study are: Greedy, Mathematical Optimization.

What are the interview tips for Minimum Moves to Reach Target Score?

Always clarify the constraints and limits of operations before diving into coding. Think about edge cases, such as when maxDoubles is zero. Practice explaining your thought process as you code; it helps solidify your understanding.

What are common mistakes when solving Minimum Moves to Reach Target Score?

Not considering the reverse approach, which can simplify the problem. Failing to account for the maximum number of doubles when making decisions.

#2139

Minimum Moves to Reach Target Score

Medium

MathGreedyGreedyMathematical Optimization

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(log n)Space O(1)

Instead of building up from 1 to the target, we can reverse the problem by starting at the target and working our way back to 1. This allows us to make more strategic decisions about when to double or decrement.

⚙️

Algorithm

4 steps

1Step 1: Initialize moves to 0.
2Step 2: While target is greater than 1, check if target is even and maxDoubles is available.
3Step 3: If both conditions are met, halve the target and decrement maxDoubles. Otherwise, decrement the target by 1.
4Step 4: Increment the moves counter until target reaches 1.

solution.py13 lines

1# Full working Python code
2
3def minMoves(target, maxDoubles):
4    moves = 0
5    while target > 1:
6        if maxDoubles > 0 and target % 2 == 0:
7            target //= 2
8            maxDoubles -= 1
9        else:
10            target -= 1
11        moves += 1
12    return moves
13

ℹ

Complexity note: The time complexity is O(log n) because each doubling operation effectively halves the target, leading to a logarithmic number of operations relative to the target value.

1Using the doubling operation strategically can significantly reduce the number of moves needed.
2Working backwards from the target allows for a more efficient approach, especially when the target is large.

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