Given three integers X, Y, and N, the task is to find the minimum possible positive product of X and Y that can be obtained by decreasing either the value of X or Y by 1 at most N times.
Input: X = 5, Y= 6, N = 4
Decrease the value of X by 4, X = 5 – 4 = 1 and Y = 6.
Therefore, the minimized product = X * Y = 1 * 6 = 6
Input: X = 49, Y = 4256, N = 10
Approach: The given problem can be solved based on the following observations:
If X ≤ Y: Reducing X minimizes the product.
If Y ≤ X: Reducing Y minimizes the product.
If (X – 2) * Y < (X – 1) * (Y – 1)
=> X * Y – 2 * Y < X * Y – X – Y + 1
=> – 2 × Y < -X – Y + 1
=> Y > X – 1
Follow the steps below to solve the problem:
- If X ≤ Y: Follow the steps below:
- If N < X: Print Y * (X – N) as the answer as reducing X minimizes the product.
- Otherwise, reduce X to 1 and reduce the remaining N from Y to minimize the product. Therefore, print Y – max(1, N – X + 1)) as the required minimized product.
- Otherwise, if N < Y, print X * (Y – N) as the minimized product. If N ≥ Y, reduce Y to 1 and print max(X – (N – Y + 1), 1) as the minimized product.
Below is the implementation of the above approach:
Time Complexity: O(1)
Auxiliary Space: O(1)
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