将数组的所有元素减少为零所需的最少步骤

给定一个正整数数组arr [] ，任务是找到使所有元素均减为0的最小步骤。在单个步骤中，可以将-1同时添加到数组的所有非零元素中。

例子：

Input: arr[] = {1, 5, 6}
Output: 6
Operation 1: arr[] = {0, 4, 5}
Operation 2: arr[] = {0, 3, 4}
Operation 3: arr[] = {0, 2, 3}
Operation 4: arr[] = {0, 1, 2}
Operation 5: arr[] = {0, 0, 1}
Operation 6: arr[] = {0, 0, 0}

Input: arr[] = {1, 1}
Output: 1

天真的方法：一种简单的方法是先对数组进行排序，然后从最小元素开始，计算将其减少到0所需的步骤数。然后，从下一个数组元素开始减少此计数，因为所有元素都将在更新同时。

高效的方法：可以观察到，最小步数将始终等于数组中的最大元素。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the minimum steps
// required to reduce all the elements to 0
int minSteps(int arr[], int n)
{
  
    // Maximum element from the array
    int maxVal = *max_element(arr, arr + n);
    return maxVal;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 4 };
    int n = sizeof(arr) / sizeof(int);
  
    cout << minSteps(arr, n);
  
    return 0;
}

Java

// Java implementation of the approach
class GFG 
{
      
    // method to get maximum number from array elements
    static int getMax(int inputArray [])
    {
        int maxValue = inputArray[0];
  
        for(int i = 1; i < inputArray.length; i++)
        {
            if(inputArray[i] > maxValue)
            {
                maxValue = inputArray[i];
            }
        }
        return maxValue;
    }
      
    // Function to return the minimum steps 
    // required to reduce all the elements to 0 
    static int minSteps(int arr[], int n) 
    { 
      
        // Maximum element from the array 
        int maxVal = getMax(arr); 
        return maxVal; 
    } 
      
    // Driver code 
    public static void main (String[] args) 
    { 
        int arr[] = { 1, 2, 4 }; 
        int n = arr.length; 
      
        System.out.println(minSteps(arr, n)); 
    } 
}
  
// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach
  
# Function to return the minimum steps
# required to reduce all the elements to 0
def minSteps(arr, n):
  
    # Maximum element from the array
    maxVal = max(arr)
    return maxVal
  
# Driver code
arr = [1, 2, 4]
n = len(arr)
  
print(minSteps(arr, n))
  
# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;
  
class GFG 
{
      
    // method to get maximum number from array elements
    static int getMax(int []inputArray)
    {
        int maxValue = inputArray[0];
  
        for(int i = 1; i < inputArray.Length; i++)
        {
            if(inputArray[i] > maxValue)
            {
                maxValue = inputArray[i];
            }
        }
        return maxValue;
    }
      
    // Function to return the minimum steps 
    // required to reduce all the elements to 0 
    static int minSteps(int []arr, int n) 
    { 
      
        // Maximum element from the array 
        int maxVal = getMax(arr); 
        return maxVal; 
    } 
      
    // Driver code 
    public static void Main(String []args) 
    { 
        int []arr = { 1, 2, 4 }; 
        int n = arr.Length; 
      
        Console.WriteLine(minSteps(arr, n)); 
    } 
}
  
// This code is contributed by Arnab Kundu

输出：

4