LeetCode 991: Broken Calculator Solution

Problem Explanation

Explanation:

To solve this problem, we can work backwards from the target to the startValue. At each step, we choose either to divide by 2 or add 1 based on whether the current number is even or odd. By doing this, we can reach the startValue from the target. The number of operations needed will be the minimum steps taken.

Algorithm:

Initialize a variable result to 0.
Iterate while target is greater than startValue:
- If target is even, divide it by 2.
- If target is odd, increment it by 1.
- Increment result by 1.
Add the difference between startValue and target to result.

Time Complexity:

The time complexity of this algorithm is O(log(target)).

Space Complexity:

The space complexity is O(1).

Solution Code

class Solution {
    public int brokenCalc(int startValue, int target) {
        int result = 0;
        while (target > startValue) {
            if (target % 2 == 0) {
                target /= 2;
            } else {
                target++;
            }
            result++;
        }
        return result + startValue - target;
    }
}

Frequently Asked Questions

How to solve LeetCode 991 (Broken Calculator)?

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What is the time complexity of LeetCode 991 (Broken Calculator)?

The time complexity for LeetCode 991 (Broken Calculator) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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991. Broken Calculator