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.
Examples:
Input: X = 5, Y= 6, N = 4
Output: 6
Explanation:
Decrease the value of X by 4, X = 5 – 4 = 1 and Y = 6.
Therefore, the minimized product = X * Y = 1 * 6 = 6Input: X = 49, Y = 4256, N = 10
Output: 165984
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.Mathematical Proof:
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:
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to minimize// the product of two numbersint minProd(int X, int Y, int N){ if (X <= Y) { if (N < X) // Reducing X, N times, // minimizes the product return (X - N) * Y; else { // Reduce X to 1 and reduce // remaining N from Y return max(Y - (N - X + 1), 1); } } if (Y >= N) // Reducing Y, N times, // minimizes the product return (Y - N) * X; // Reduce Y to 1 and reduce // remaining N from X return max(X - (N - Y + 1), 1); ;}// Driver Codeint main(){ int X = 47, Y = 42, N = 167; cout << minProd(X, Y, N);} |
Java
// Java program to implement// the above approachimport java.io.*;class GFG{ // Function to minimize // the product of two numbers static int minProd(int X, int Y, int N) { if (X <= Y) { if (N < X) // Reducing X, N times, // minimizes the product return (X - N) * Y; else { // Reduce X to 1 and reduce // remaining N from Y return Math.max(Y - (N - X + 1), 1); } } if (Y >= N) // Reducing Y, N times, // minimizes the product return (Y - N) * X; // Reduce Y to 1 and reduce // remaining N from X return Math.max(X - (N - Y + 1), 1); } // Driver Code public static void main (String[] args){ int X = 47, Y = 42, N = 167; System.out.println(minProd(X, Y, N)); }}// This code is contributed by code_hunt |
Python3
# Python3 program to implement# the above approach# Function to minimize# the product of two numbersdef minProd(X, Y, N): if (X <= Y): if (N < X): # Reducing X, N times, # minimizes the product return (X - N) * Y else: # Reduce X to 1 and reduce # remaining N from Y return max(Y - (N - X + 1), 1) if (Y >= N): # Reducing Y, N times, # minimizes the product return (Y - N) * X # Reduce Y to 1 and reduce # remaining N from X return max(X - (N - Y + 1), 1)# Driver Codeif __name__ == "__main__": X = 47 Y = 42 N = 167 print (minProd(X, Y, N))# This code is contributed by Chitranayal |
C#
// C# program to implement// the above approachusing System;class GFG{ // Function to minimize // the product of two numbers static int minProd(int X, int Y, int N) { if (X <= Y) { if (N < X) // Reducing X, N times, // minimizes the product return (X - N) * Y; else { // Reduce X to 1 and reduce // remaining N from Y return Math.Max(Y - (N - X + 1), 1); } } if (Y >= N) // Reducing Y, N times, // minimizes the product return (Y - N) * X; // Reduce Y to 1 and reduce // remaining N from X return Math.Max(X - (N - Y + 1), 1); } // Driver Code public static void Main(String[] args){ int X = 47, Y = 42, N = 167; Console.WriteLine(minProd(X, Y, N)); }}// This code is contributed by Rajput-Ji |
1
Time Complexity: O(1)
Auxiliary Space: O(1)
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