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2139. Minimum Moves to Reach Target Score

Description

You are playing a game with integers. You start with the integer 1 and you want to reach the integer target.

In one move, you can either:

  • Increment the current integer by one (i.e., x = x + 1).
  • Double the current integer (i.e., x = 2 * x).

You can use the increment operation any number of times, however, you can only use the double operation at most maxDoubles times.

Given the two integers target and maxDoubles, return the minimum number of moves needed to reach target starting with 1.

 

Example 1:

Input: target = 5, maxDoubles = 0
Output: 4
Explanation: Keep incrementing by 1 until you reach target.

Example 2:

Input: target = 19, maxDoubles = 2
Output: 7
Explanation: Initially, x = 1
Increment 3 times so x = 4
Double once so x = 8
Increment once so x = 9
Double again so x = 18
Increment once so x = 19

Example 3:

Input: target = 10, maxDoubles = 4
Output: 4
Explanation: Initially, x = 1
Increment once so x = 2
Double once so x = 4
Increment once so x = 5
Double again so x = 10

 

Constraints:

  • 1 <= target <= 109
  • 0 <= maxDoubles <= 100

Solutions

Solution 1: Backtracking + Greedy

Let’s start by backtracking from the final state. Assuming the final state is $target$, we can get the previous state of $target$ as $target - 1$ or $target / 2$, depending on the parity of $target$ and the value of $maxDoubles$.

If $target=1$, no operation is needed, and we can return $0$ directly.

If $maxDoubles=0$, we can only use the increment operation, so we need $target-1$ operations.

If $target$ is even and $maxDoubles>0$, we can use the doubling operation, so we need $1$ operation, and then recursively solve $target/2$ and $maxDoubles-1$.

If $target$ is odd, we can only use the increment operation, so we need $1$ operation, and then recursively solve $target-1$ and $maxDoubles$.

The time complexity is $O(\min(\log target, maxDoubles))$, and the space complexity is $O(\min(\log target, maxDoubles))$.

We can also change the above process to an iterative way to avoid the space overhead of recursion.

  • class Solution {
    public int minMoves(int target, int maxDoubles) {
        if (target == 1) {
            return 0;
        }
        if (maxDoubles == 0) {
            return target - 1;
        }
        if (target % 2 == 0 && maxDoubles > 0) {
            return 1 + minMoves(target >> 1, maxDoubles - 1);
        }
        return 1 + minMoves(target - 1, maxDoubles);
    }
}

  • class Solution {
public:
    int minMoves(int target, int maxDoubles) {
        if (target == 1) {
            return 0;
        }
        if (maxDoubles == 0) {
            return target - 1;
        }
        if (target % 2 == 0 && maxDoubles > 0) {
            return 1 + minMoves(target >> 1, maxDoubles - 1);
        }
        return 1 + minMoves(target - 1, maxDoubles);
    }
};

  • class Solution:
    def minMoves(self, target: int, maxDoubles: int) -> int:
        if target == 1:
            return 0
        if maxDoubles == 0:
            return target - 1
        if target % 2 == 0 and maxDoubles:
            return 1 + self.minMoves(target >> 1, maxDoubles - 1)
        return 1 + self.minMoves(target - 1, maxDoubles)


  • func minMoves(target int, maxDoubles int) int {
	if target == 1 {
		return 0
	}
	if maxDoubles == 0 {
		return target - 1
	}
	if target%2 == 0 && maxDoubles > 0 {
		return 1 + minMoves(target>>1, maxDoubles-1)
	}
	return 1 + minMoves(target-1, maxDoubles)
}

  • function minMoves(target: number, maxDoubles: number): number {
    if (target === 1) {
        return 0;
    }
    if (maxDoubles === 0) {
        return target - 1;
    }
    if (target % 2 === 0 && maxDoubles) {
        return 1 + minMoves(target >> 1, maxDoubles - 1);
    }
    return 1 + minMoves(target - 1, maxDoubles);
}

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