Given an array arr[] of size N, the task is to count the minimum number of steps required to make all the array elements the same by adding 1, 2, or 5 to exactly (N – 1) elements of the array at each step.
Examples:
Input: N = 4, arr[] = {2, 2, 3, 7}
Output: 2
Explanation:
Step 1: {2, 2, 3, 7} -> {3, 3, 3, 8}
Step 2: {3, 3, 3, 8} -> {8, 8, 8, 8}Input: N = 3, arr[] = {10, 7, 12}
Output: 3
Naive Approach: The simplest approach is to try all possible combinations recursively of adding numbers 1, 2, and 5 such that all the elements become the same and calculate the number of steps required for all such combinations. Finally, print the minimum of them as the required answer.
Time Complexity: O(MM), where M is the maximum element present in the array.
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized by the following observations:
- Adding a number K to all indices except one (say index X) is the same as removing K from the value at index X.
- This reduces the bound to search for the final array element to lower than equal to the minimum value present in the given array.
- Let the minimum value be A, then the final value after optimal operations can either be A, A – 1, or A – 2.
- The reason A – 3 and so on are not considered in the calculation is because A + 1 takes 2 steps to reach there (-1, -2 ), A + 2 requires one step in reaching A – 3 (- 5) but can easily reach A requires in a single step(A+1 requires 1 step(-1) to reach A and A + 2 requires 1 step(-2) to reach A).
- Also, A + 3 requires 2 steps (-5, -1) to reach A – 3 and 2 steps to reach A again (-1, -2).
- Therefore, A – 3 or any lower bases are not needed to be considered.
Therefore, the idea is to find the count of operations required to reduce all the array elements to their minimum element(say minE), minE – 1, and minE – 2 by subtracting 1, 2, and 5. Print the minimum among the above three operations.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate the minimum// number of stepsint calculate_steps(int arr[], int n, int minimum){ // count stores number of operations // required to make all elements // equal to minimum value int count = 0; // Remark, the array should remain // unchanged for further calculations // with different minimum for (int i = 0; i < n; i++) { // Storing the current value of // arr[i] in val int val = arr[i]; if (arr[i] > minimum) { // Finds how much extra amount // is to be removed arr[i] = arr[i] - minimum; // Subtract the maximum number // of 5 and stores remaining count += arr[i] / 5; arr[i] = arr[i] % 5; // Subtract the maximum number // of 2 and stores remaining count += arr[i] / 2; arr[i] = arr[i] % 2; if (arr[i]) { count++; } } // Restores the actual value // of arr[i] arr[i] = val; } // Return the count return count;}// Function to find the minimum number// of steps to make array elements sameint solve(int arr[], int n){ // Sort the array in descending order sort(arr, arr + n, greater<int>()); // Stores the minimum array element int minimum = arr[n - 1]; int count1 = 0, count2 = 0, count3 = 0; // Stores the operations required // to make array elements equal to minimum count1 = calculate_steps(arr, n, minimum); // Stores the operations required // to make array elements equal to minimum - 1 count2 = calculate_steps(arr, n, minimum - 1); // Stores the operations required // to make array elements equal to minimum - 2 count3 = calculate_steps(arr, n, minimum - 2); // Return minimum of the three counts return min(count1, min(count2, count3));}// Driver Codeint main(){ int arr[] = { 3, 6, 6 }; int N = sizeof(arr) / sizeof(arr[0]); cout << solve(arr, N);} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to calculate the minimum// number of stepsstatic int calculate_steps(Integer arr[], int n, int minimum){ // count stores number of operations // required to make all elements // equal to minimum value int count = 0; // Remark, the array should remain // unchanged for further calculations // with different minimum for (int i = 0; i < n; i++) { // Storing the current value of // arr[i] in val int val = arr[i]; if (arr[i] > minimum) { // Finds how much extra amount // is to be removed arr[i] = arr[i] - minimum; // Subtract the maximum number // of 5 and stores remaining count += arr[i] / 5; arr[i] = arr[i] % 5; // Subtract the maximum number // of 2 and stores remaining count += arr[i] / 2; arr[i] = arr[i] % 2; if (arr[i] > 0) { count++; } } // Restores the actual value // of arr[i] arr[i] = val; } // Return the count return count;}// Function to find the minimum number// of steps to make array elements samestatic int solve(Integer arr[], int n){ // Sort the array in descending order Arrays.sort(arr, Collections.reverseOrder()); // Stores the minimum array element int minimum = arr[n - 1]; int count1 = 0, count2 = 0, count3 = 0; // Stores the operations required // to make array elements equal // to minimum count1 = calculate_steps(arr, n, minimum); // Stores the operations required // to make array elements equal to // minimum - 1 count2 = calculate_steps(arr, n, minimum - 1); // Stores the operations required // to make array elements equal to // minimum - 2 count3 = calculate_steps(arr, n, minimum - 2); // Return minimum of the three counts return Math.min(count1, Math.min(count2, count3));}// Driver Codepublic static void main(String[] args){ Integer arr[] = {3, 6, 6}; int N = arr.length; System.out.print(solve(arr, N));}}// This code is contributed by Rajput-Ji |
Python3
# Python3 program for the above approach# Function to calculate the minimum# number of stepsdef calculate_steps(arr, n, minimum): # count stores number of operations # required to make all elements # equal to minimum value count = 0 # Remark, the array should remain # unchanged for further calculations # with different minimum for i in range(n): # Storing the current value of # arr[i] in val val = arr[i] if (arr[i] > minimum): # Finds how much extra amount # is to be removed arr[i] = arr[i] - minimum # Subtract the maximum number # of 5 and stores remaining count += arr[i] // 5 arr[i] = arr[i] % 5 # Subtract the maximum number # of 2 and stores remaining count += arr[i] // 2 arr[i] = arr[i] % 2 if (arr[i]): count += 1 # Restores the actual value # of arr[i] arr[i] = val # Return the count return count# Function to find the minimum number# of steps to make array elements samedef solve(arr, n): # Sort the array in descending order arr = sorted(arr) arr = arr[::-1] # Stores the minimum array element minimum = arr[n - 1] count1 = 0 count2 = 0 count3 = 0 # Stores the operations required # to make array elements equal to minimum count1 = calculate_steps(arr, n, minimum) # Stores the operations required # to make array elements equal to minimum - 1 count2 = calculate_steps(arr, n, minimum - 1) # Stores the operations required # to make array elements equal to minimum - 2 count3 = calculate_steps(arr, n, minimum - 2) # Return minimum of the three counts return min(count1, min(count2, count3))# Driver Codeif __name__ == '__main__': arr = [ 3, 6, 6 ] N = len(arr) print(solve(arr, N))# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;public class GFG{// Function to calculate the minimum// number of stepsstatic int calculate_steps(int []arr, int n, int minimum){ // count stores number of operations // required to make all elements // equal to minimum value int count = 0; // Remark, the array should remain // unchanged for further calculations // with different minimum for (int i = 0; i < n; i++) { // Storing the current value of // arr[i] in val int val = arr[i]; if (arr[i] > minimum) { // Finds how much extra amount // is to be removed arr[i] = arr[i] - minimum; // Subtract the maximum number // of 5 and stores remaining count += arr[i] / 5; arr[i] = arr[i] % 5; // Subtract the maximum number // of 2 and stores remaining count += arr[i] / 2; arr[i] = arr[i] % 2; if (arr[i]>0) { count++; } } // Restores the actual value // of arr[i] arr[i] = val; } // Return the count return count;}// Function to find the minimum number// of steps to make array elements samestatic int solve(int []arr, int n){ // Sort the array in descending order Array.Sort(arr); Array.Reverse(arr); // Stores the minimum array element int minimum = arr[n - 1]; int count1 = 0, count2 = 0, count3 = 0; // Stores the operations required // to make array elements equal to minimum count1 = calculate_steps(arr, n, minimum); // Stores the operations required // to make array elements equal to minimum - 1 count2 = calculate_steps(arr, n, minimum - 1); // Stores the operations required // to make array elements equal to minimum - 2 count3 = calculate_steps(arr, n, minimum - 2); // Return minimum of the three counts return Math.Min(count1, Math.Min(count2, count3));}// Driver Codepublic static void Main(String[] args){ int []arr = { 3, 6, 6 }; int N = arr.Length; Console.Write(solve(arr, N));}} // This code contributed by Rajput-Ji |
Javascript
<script> // JavaScript program for the above approach // Function to calculate the minimum // number of steps function calculate_steps(arr, n, minimum) { // count stores number of operations // required to make all elements // equal to minimum value var count = 0; // Remark, the array should remain // unchanged for further calculations // with different minimum for (var i = 0; i < n; i++) { // Storing the current value of // arr[i] in val var val = arr[i]; if (arr[i] > minimum) { // Finds how much extra amount // is to be removed arr[i] = arr[i] - minimum; // Subtract the maximum number // of 5 and stores remaining count += parseInt(arr[i] / 5); arr[i] = arr[i] % 5; // Subtract the maximum number // of 2 and stores remaining count += parseInt(arr[i] / 2); arr[i] = arr[i] % 2; if (arr[i]) { count++; } } // Restores the actual value // of arr[i] arr[i] = val; } // Return the count return count; } // Function to find the minimum number // of steps to make array elements same function solve(arr, n) { // Sort the array in descending order arr.sort((a, b) => b - a); // Stores the minimum array element var minimum = arr[n - 1]; var count1 = 0, count2 = 0, count3 = 0; // Stores the operations required // to make array elements equal to minimum count1 = calculate_steps(arr, n, minimum); // Stores the operations required // to make array elements equal to minimum - 1 count2 = calculate_steps(arr, n, minimum - 1); // Stores the operations required // to make array elements equal to minimum - 2 count3 = calculate_steps(arr, n, minimum - 2); // Return minimum of the three counts return Math.min(count1, Math.min(count2, count3)); } // Driver Code var arr = [3, 6, 6]; var N = arr.length; document.write(solve(arr, N)); </script> |
Output:
3
Time Complexity: O(N), where N is the size of the given array.
Auxiliary Space: O(1)
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