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Instead of moving from the startValue to target, consider reversing the operations. Start from the target and attempt to reach the startValue. If the target is even, divide it by 2. If it is odd, add 1 to it. Continue until the target is less than or equal to the startValue. This allows you to efficiently compute the minimum steps by potentially making large reductions through division by 2.
Time Complexity: O(log n), where n is the target due to halving operation.
Space Complexity: O(1) as no extra space is used.
1public class BrokenCalculator {
2 public static int brokenCalc(int startValue, int target) {
3 int operations = 0;
4 while (target > startValue) {
5 operations++;
6 if (target % 2 == 1) {
7 target++;
8 } else {
9 target /= 2;
10 }
11 }
12 return operations + (startValue - target);
13 }
14
15 public static void main(String[] args) {
16 int startValue = 3, target = 10;
17 System.out.println(brokenCalc(startValue, target));
18 }
19}This Java program executes the same logic: reducing the target with divide by 2 or plus 1 operations till it becomes less than or equal to startValue. Extra decrements are added if the adjusted target is still greater than startValue.
Using a greedy approach, always attempt the operation that brings the target closer to the startValue significantly, which is multiplication by 2 in the reverse. Only when all doubling possibilities are exhausted, should we use subtraction/addition operations.
Time Complexity: O(log n) due to division operations.
Space Complexity: O(1).
1public
This Java solution performs reverse operations until the target is reduced adequately compared to the startValue, enhancing speed through halving.