Sponsored
Sponsored
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.
1using System;
2
3class BrokenCalculator {
4 public static int BrokenCalc(int startValue, int target) {
5 int operations = 0;
6 while (target > startValue) {
7 operations++;
8 if (target % 2 == 1) {
9 target++;
10 } else {
11 target /= 2;
12 }
13 }
14 return operations + (startValue - target);
15 }
16
17 static void Main() {
18 int startValue = 3, target = 10;
19 Console.WriteLine(BrokenCalc(startValue, target));
20 }
21}This C# solution cycles through reverse operations similar to previous languages, efficiently counting the steps needed to reduce the target.
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).
1function
This JavaScript code applies a choice strategy, maximizing the utility of division while fulfilling the role of additions when necessary, to achieve accurate operation counts.