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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.
1#include <stdio.h>
2
3int brokenCalc(int startValue, int target) {
4 int operations = 0;
5 while (target > startValue) {
6 operations++;
7 if (target % 2 == 1) {
8 target++;
9 } else {
10 target /= 2;
11 }
12 }
13 return operations + (startValue - target);
14}
15
16int main() {
17 int startValue = 3, target = 10;
18 printf("%d\n", brokenCalc(startValue, target));
19 return 0;
20}This C program uses a loop to apply reverse operations to reduce the target to a value no larger than the startValue. If target is even, it is divided by 2; if odd, it is incremented by 1. Lastly, the remaining difference between startValue and target is added to the operations, as this represents the needed decrements.
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.