# Broken Calculator Problem

## Description

LeetCode Problem 991.

There is a broken calculator that has the integer startValue on its display initially. In one operation, you can:

• multiply the number on display by 2, or
• subtract 1 from the number on display.

Given two integers startValue and target, return the minimum number of operations needed to display target on the calculator.

Example 1:

``````1
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3
Input: startValue = 2, target = 3
Output: 2
Explanation: Use double operation and then decrement operation {2 -> 4 -> 3}.
``````

Example 2:

``````1
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3
Input: startValue = 5, target = 8
Output: 2
Explanation: Use decrement and then double {5 -> 4 -> 8}.
``````

Example 3:

``````1
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3
Input: startValue = 3, target = 10
Output: 3
Explanation: Use double, decrement and double {3 -> 6 -> 5 -> 10}.
``````

Example 4:

``````1
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3
Input: startValue = 1024, target = 1
Output: 1023
Explanation: Use decrement operations 1023 times.
``````

Constraints:

• 1 <= x, y <= 10^9

## Sample C++ Code

``````1
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class Solution {
public:
// 1. reduce target value as much as possible
// because this question can be done by greedy
// 2. if divisible by 2 divide by 2
// 3. else increase by 1
// 4. if target becomes less than startValue then only option left is to increasse
int brokenCalc(int startValue, int target) {
long long ans = 0;
long long tar = target;
while (startValue != tar) {
if (tar % 2 == 0 and startValue < tar) {
tar /= 2;
}
else if (tar > startValue) {
tar += 1;
}
else{
ans += abs(tar - startValue) - 1;
tar = startValue;
}
ans++;
}
return ans;
}
};
``````