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: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

**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` |

**Output:**

4

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