Given an array arr[] consisting of N integers, the following three operations can be performed on any element one at a time:
- Add one to the element.
- Subtract one from the element.
- Leave the element unchanged.
The task is to find the minimum cost required to convert it into a Geometric Progression and also find the common ratio.
Note: Each addition and subtraction operation costs 1 unit.
Examples:
Input: N = 6, arr[] = {1, 11, 4, 27, 15, 33}
Output: 28 2
Explanation:
For r = 1, arr[] = {1, 4, 11, 15, 27, 33}
expected[] = {1, 1, 1, 1, 1, 1}
cost[] = {0, 3, 10, 14, 26, 32}
Total cost: ∑ cost = 85For r = 2, arr[] = {1, 4, 11, 15, 27, 33}
expected[] = {1, 2, 4, 8, 16, 32}
cost[] = {0, 2, 7, 7, 11, 1}
Total cost: ∑ cost = 28
For r = 3, arr[] = {1, 4, 11, 15, 27, 33}
expected[] = {1, 3, 9, 27, 81, 243}
cost[] = {0, 1, 2, 12, 54, 210}
Total cost: ∑ cost = 279Minimum cost = 28
Common ratio = 2
Input: N = 7, arr[] = {1, 2, 4, 8, 9, 6, 7}
Output: 30 1
Approach: The idea is to iterate over range of possible common ratios and check for the minimum operations required. Follow the steps below to solve the problem:
- Sort the array.
- Find the range of possible common ratios using the formula Ceil( arr[maximum_element] / (N – 1)).
- Calculate the number of operations required for all possible common ratios.
- Find the minimum operations required for any of the common ratios.
Below is the implementation of the above approach:
C++
// C++ program for above approach#include <bits/stdc++.h>using namespace std;// Function to find minimum costvoid minCost(int arr[], int n){ if (n == 1) { cout << 0 << endl; return; } // Sort the array sort(arr, arr + n); // Maximum possible common ratios float raised = 1 / float(n - 1); float temp = pow(arr[n - 1], raised); int r = round(temp) + 1; int i, j, min_cost = INT_MAX; int common_ratio = 1; // Iterate over all possible common ratios for (j = 1; j <= r; j++) { int curr_cost = 0, prod = 1; // Calculate operations required // for the current common ratio for (i = 0; i < n; i++) { curr_cost += abs(arr[i] - prod); prod *= j; if (curr_cost >= min_cost) break; } // Calculate minimum cost if (i == n) { min_cost = min(min_cost, curr_cost); common_ratio = j; } } cout << min_cost << ' '; cout << common_ratio << ' ';}// Driver Codeint main(){ // Given N int N = 6; // Given arr[] int arr[] = { 1, 11, 4, 27, 15, 33 }; // Function Calling minCost(arr, N); return 0;} |
Java
// Java program for // the above approachimport java.util.*;class GFG{// Function to find minimum coststatic void minCost(int arr[], int n){ if (n == 1) { System.out.print(0 + "\n"); return; } // Sort the array Arrays.sort(arr); // Maximum possible common ratios float raised = 1 / (float)(n - 1); float temp = (float)Math.pow(arr[n - 1], raised); int r = (int)(temp) + 1; int i, j, min_cost = Integer.MAX_VALUE; int common_ratio = 1; // Iterate over all possible // common ratios for (j = 1; j <= r; j++) { int curr_cost = 0, prod = 1; // Calculate operations required // for the current common ratio for (i = 0; i < n; i++) { curr_cost += Math.abs(arr[i] - prod); prod *= j; if (curr_cost >= min_cost) break; } // Calculate minimum cost if (i == n) { min_cost = Math.min(min_cost, curr_cost); common_ratio = j; } } System.out.print(min_cost + " "); System.out.print(common_ratio + " ");}// Driver Codepublic static void main(String[] args){ // Given N int N = 6; // Given arr[] int arr[] = {1, 11, 4, 27, 15, 33}; // Function Calling minCost(arr, N);}}// This code is contributed by shikhasingrajput |
Python3
# Python3 program for above approachimport sys# Function to find minimum costdef minCost(arr, n): if (n == 1): print(0) return # Sort the array arr = sorted(arr) # Maximum possible common ratios raised = 1 / (n - 1) temp = pow(arr[n - 1], raised) r = round(temp) + 1 min_cost = sys.maxsize common_ratio = 1 # Iterate over all possible # common ratios for j in range(1, r + 1): curr_cost = 0 prod = 1 # Calculate operations required # for the current common ratio i = 0 while i < n: curr_cost += abs(arr[i] - prod) prod *= j if (curr_cost >= min_cost): break i += 1 # Calculate minimum cost if (i == n): min_cost = min(min_cost, curr_cost) common_ratio = j print(min_cost, common_ratio)# Driver Codeif __name__ == '__main__': # Given N N = 6 # Given arr[] arr = [ 1, 11, 4, 27, 15, 33 ] # Function calling minCost(arr, N)# This code is contributed by mohit kumar 29 |
C#
// C# program for // the above approachusing System;class GFG{// Function to find minimum coststatic void minCost(int []arr, int n){ if (n == 1) { Console.Write(0 + "\n"); return; } // Sort the array Array.Sort(arr); // Maximum possible common ratios float raised = 1 / (float)(n - 1); float temp = (float)Math.Pow(arr[n - 1], raised); int r = (int)(temp) + 1; int i, j, min_cost = int.MaxValue; int common_ratio = 1; // Iterate over all possible // common ratios for (j = 1; j <= r; j++) { int curr_cost = 0, prod = 1; // Calculate operations required // for the current common ratio for (i = 0; i < n; i++) { curr_cost += Math.Abs(arr[i] - prod); prod *= j; if (curr_cost >= min_cost) break; } // Calculate minimum cost if (i == n) { min_cost = Math.Min(min_cost, curr_cost); common_ratio = j; } } Console.Write(min_cost + " "); Console.Write(common_ratio + " ");}// Driver Codepublic static void Main(String[] args){ // Given N int N = 6; // Given []arr int []arr = { 1, 11, 4, 27, 15, 33 }; // Function Calling minCost(arr, N);}}// This code is contributed by gauravrajput1 |
28 2
Time Complexity: O(N * K), where K is the maximum possible common ratio.
Auxiliary Space: O(1)
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