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Minimum steps required to reduce all the elements of the array to zero

  • Difficulty Level : Basic
  • Last Updated : 30 Apr, 2021

Given an array arr[] of positive integers, the task is to find the minimum steps to reduce all the elements to 0. In a single step, -1 can be added to all the non-zero elements of the array at the same time.
Examples: 
 

Input: arr[] = {1, 5, 6} 
Output:
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:
 

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Naive approach: A simple approach is to first sort the array then starting from the minimum element, count the number of steps required to reduce it to 0. This count will then be reduced from the next array element as all the elements will be updated at the same time.
Efficient approach: It can be observed that the minimum number of steps will always be equal to the maximum element from the array.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
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

Javascript




<script>
// Javascript implementation of the approach
 
// Function to return the minimum steps
// required to reduce all the elements to 0
function minSteps(arr, n)
{
 
    // Maximum element from the array
    let maxVal = Math.max(...arr);
    return maxVal;
}
 
// Driver code
    let arr = [ 1, 2, 4 ];
    let n = arr.length;
 
    document.write(minSteps(arr, n));
 
</script>
Output: 
4

 




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