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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.
1function brokenCalc(startValue, target) {
2 let operations = 0;
3 while (target > startValue) {
4 operations++;
5 if (target % 2 === 1) {
6 target++;
7 } else {
8 target /= 2;
9 }
10 }
11 return operations + (startValue - target);
12}
13
14const startValue = 3;
15const target = 10;
16console.log(brokenCalc(startValue, target));This JavaScript code similarly uses reverse operations, efficiently minimizing the target with each step until it's below or equal to the 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).
1using System;
class BrokenCalculator {
public static int BrokenCalc(int startValue, int target) {
int operations = 0;
while (target > startValue) {
if (target % 2 == 1) {
target++;
} else {
target /= 2;
}
operations++;
}
return operations + (startValue - target);
}
static void Main() {
int startValue = 3, target = 10;
Console.WriteLine(BrokenCalc(startValue, target));
}
}This C# approach efficiently tracks operations to reduce target down to the start value using division and intentional incrementation.