Given three integers A, B, and C, the task is to find the minimum possible value of |A – X| + |B – Y| + |C – Z| such that X * Y = Z.
Example:
Input: A = 19, B = 28, C = 522
Output: 2
Explanation: The most optimal choice of X, Y, and Z for the given A, B, and C are X = 18, Y = 29, and Z = 522. The equation X * Y = Z holds true and the value of |A – X| + |B – Y| + |C – Z| = 2 which is minimum possible.Input: A = 11, B = 11, C = 121
Output: 0
Explanation: The given values of A, B, and C satisfies A * B = C. Therefore the most optimal choice is X = A, Y = B, and Z = C.
Approach: The above problem can be solved using the following observations:
- The maximum value of |A – X| + |B – Y| + |C – Z| can be A + B + C for X, Y, and Z equal to 0.
- Based on the above observation, iterating over all the values of i * j such that i * j <= 2 * C and choosing the best value is the optimal choice.
Therefore, iterate over all values of i in the range [1, 2*C], and for every i, iterate over all values of j such that i * j <= 2 * C and keep track of the minimum possible value of |A – i| + |B – j| + |C – i * j|.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the minimum possible// value of |A - X| + |B - Y| + |C - Z|// such that X * Y = Z for given A, B and Cint minimizeCost(int A, int B, int C){ // Stores the minimum value of // |A - X| + |B - Y| + |C - Z| // such that X * Y = Z int ans = A + B + C; // Iterate over all values of i // in the range [1, 2*C] for (int i = 1; i <= 2 * C; i++) { int j = 0; // Iterate over all values of // j such that i*j <= 2*c while (i * j <= 2 * C) { // Update the value of ans ans = min(ans, abs(A - i) + abs(B - j) + abs(i * j - C)); j++; } } // Return answer return ans;}// Driver Codeint main(){ int A = 19, B = 28, C = 522; cout << minimizeCost(A, B, C); return 0;} |
Java
// Java program for the above approachclass GFG{// Function to find the minimum possible// value of |A - X| + |B - Y| + |C - Z|// such that X * Y = Z for given A, B and Cpublic static int minimizeCost(int A, int B, int C){ // Stores the minimum value of // |A - X| + |B - Y| + |C - Z| // such that X * Y = Z int ans = A + B + C; // Iterate over all values of i // in the range [1, 2*C] for (int i = 1; i <= 2 * C; i++) { int j = 0; // Iterate over all values of // j such that i*j <= 2*c while (i * j <= 2 * C) { // Update the value of ans ans = Math.min(ans, Math.abs(A - i) + Math.abs(B - j) + Math.abs(i * j - C)); j++; } } // Return answer return ans;}// Driver Codepublic static void main(String args[]){ int A = 19, B = 28, C = 522; System.out.print(minimizeCost(A, B, C));}}// This code is contributed by gfgking. |
Python3
# Python Program to implement# the above approach# Function to find the minimum possible# value of |A - X| + |B - Y| + |C - Z|# such that X * Y = Z for given A, B and Cdef minimizeCost(A, B, C): # Stores the minimum value of # |A - X| + |B - Y| + |C - Z| # such that X * Y = Z ans = A + B + C # Iterate over all values of i # in the range [1, 2*C] for i in range(1, 2 * C + 1): j = 0 # Iterate over all values of # j such that i*j <= 2*c while (i * j <= 2 * C): # Update the value of ans ans = min(ans, abs(A - i) + abs(B - j) + abs(i * j - C)) j += 1 # Return answer return ans# Driver CodeA = 19B = 28C = 522print(minimizeCost(A, B, C))# This code is contributed by Saurabh Jaiswal |
C#
// C# program for the above approachusing System;class GFG{// Function to find the minimum possible// value of |A - X| + |B - Y| + |C - Z|// such that X * Y = Z for given A, B and Cpublic static int minimizeCost(int A, int B, int C){ // Stores the minimum value of // |A - X| + |B - Y| + |C - Z| // such that X * Y = Z int ans = A + B + C; // Iterate over all values of i // in the range [1, 2*C] for (int i = 1; i <= 2 * C; i++) { int j = 0; // Iterate over all values of // j such that i*j <= 2*c while (i * j <= 2 * C) { // Update the value of ans ans = Math.Min(ans, Math.Abs(A - i) + Math.Abs(B - j) + Math.Abs(i * j - C)); j++; } } // Return answer return ans;}// Driver Codepublic static void Main(String []args){ int A = 19, B = 28, C = 522; Console.Write(minimizeCost(A, B, C));}}// This code is contributed by shivanisinghss2110 |
Javascript
<script> // JavaScript Program to implement // the above approach // Function to find the minimum possible // value of |A - X| + |B - Y| + |C - Z| // such that X * Y = Z for given A, B and C function minimizeCost(A, B, C) { // Stores the minimum value of // |A - X| + |B - Y| + |C - Z| // such that X * Y = Z let ans = A + B + C; // Iterate over all values of i // in the range [1, 2*C] for (let i = 1; i <= 2 * C; i++) { let j = 0; // Iterate over all values of // j such that i*j <= 2*c while (i * j <= 2 * C) { // Update the value of ans ans = Math.min(ans, Math.abs(A - i) + Math.abs(B - j) + Math.abs(i * j - C)); j++; } } // Return answer return ans; } // Driver Code let A = 19, B = 28, C = 522; document.write(minimizeCost(A, B, C)); // This code is contributed by Potta Lokesh </script> |
Output:
2
Time Complexity: O(C*log C)
Auxiliary Space: O(1)
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