Minimum number of steps required to obtain the given Array by the given operations

Given an array arr[] of N positive integers, the task is to find the minimum number of operations required of the following types to obtain the array arr[] from an array of zeroes only.

• Select any index i and increment all the elements at the indices [i, N - 1] by 1.
• Select any index i and decrease all the elements at the indices [i, N - 1] by 1.

Examples: 
 

Input: arr[]={1,1,2,2,1} 
Output: 3 
Explanation: 
Initially arr[] = {0,0,0,0,0} 
Step 1: arr[]={1,1,1,1,1}(Operation 1 on index 0) 
Step 2: arr[]={1,1,2,2,2}(Operation 1 on index 2) 
Step 3: arr[]={1,1,2,2,1}(Operation 2 on index 4)
Input: arr[]={1,2,3,4} 
Output: 4 

Naive Approach: The simplest approach is to convert each element of the array result array, brr[] to arr[] by performing one of the above operations over the indices [i, N - 1] and increase the count of each operation performed. 

Algorithm

1. Start by naming function min_operations that takes a list arr as input.
2. Then Create a new list brr with the same length as arr, filled with zeros.
3. Initialize a variable count to 0.
4. Loop over each element of arr using an index variable i:
    a. Loop until the corresponding element of brr is equal to the corresponding element of arr:
        i. If the current element of brr is less than the corresponding element of arr:
            1. Loop over each element of brr starting from the current index i:
                a. Increment each element of brr until it matches the corresponding element of arr.
            2. Increment the count variable for each increment operation.
        ii. If the current element of brr is greater than the corresponding element of arr:
            1. Loop over each element of brr starting from the current index i:
                a. Decrement each element of brr until it matches the corresponding element of arr.
            2. Increment the count variable for each decrement operation.
5. Return the final count value.

#include <iostream>
#include <vector>

using namespace std;

int min_operations(vector<int>& arr)
{
    // Create a new vector brr with the same
    // length as arr, filled with zeros
    vector<int> brr(arr.size(), 0);
    // Initialize a variable count to 0
    int count = 0;
    // Loop over each element of arr
    // using an index variable i
    for (int i = 0; i < arr.size(); i++) {
        // Loop until the corresponding element of brr
        // is equal to the corresponding element of arr
        while (brr[i] != arr[i]) {
            // If the current element of brr is less
            // than the corresponding element of arr
            if (brr[i] < arr[i]) {
                // Loop over each element of brr starting
                // from the current index i
                for (int j = i; j < arr.size(); j++) {
                    // Increment each element of brr until
                    // it
                    // matches the corresponding element of
                    // arr
                    brr[j]++;
                }
                // Increment the count variable for each
                // increment operation
                count++;
            }
            // If the current element of brr is greater
            // than the corresponding element of arr
            else {
                // Loop over each element of brr starting
                // from the current index i
                for (int j = i; j < arr.size(); j++) {
                    // Decrement each element of brr until
                    // it matches the corresponding element
                    // of arr
                    brr[j]--;
                }
                // Increment the count variable for each
                // decrement operation
                count++;
            }
        }
    }
    // Return the final count value
    return count;
}

int main()
{
    vector<int> arr{ 1, 2, 3, 4 };
    cout << min_operations(arr) << endl; // Output: 4
    return 0;
}

import java.io.*;

class GFG {
  static int min_operations(int arr[])
{
    int length=arr.length;
    /* Create a new array brr with the same
      length as arr.*/
    int []brr=new int[length];
    // Initialize a variable count to 0
    int count = 0;
    // Loop over each element of arr using an index variable i
    for (int i = 0; i < length; i++) {
        /* Loop until the corresponding element of brr
           is equal to the corresponding element of arr.*/
        while (brr[i] != arr[i]) {
            // If the current element of brr is less
            // than the corresponding element of arr
            if (brr[i] < arr[i]) {
                // Loop over each element of brr starting
                // from the current index i
                for (int j = i; j < length; j++) {
                    /* Increment each element of brr until it
                     matches the corresponding element of arr*/
                    brr[j]++;
                }
                // Increment the count variable for each increment operation
                count++;
            }
            /* If the current element of brr is greater
             than the corresponding element of arr*/
            else {
                /* Loop over each element of brr starting
                 from the current index i*/
                for (int j = i; j < length; j++) {
                    /* Decrement each element of brr until
                       it matches the corresponding element
                       of arr*/
                    brr[j]--;
                }
                // Increment the count variable for each decrement operation
                count++;
            }
        }
    }
    // Return the final count value
    return count;
}
  //Driver Code
    public static void main (String[] args) {
        int []arr=new int[]{1, 2, 3, 4};
        //Function call
        int minOperations= min_operations(arr); // Output: 4
        System.out.println(minOperations);
    }
}

def min_operations(arr):
    # Create a new list brr with the
    # same length as arr, filled with zeros
    brr = [0] * len(arr)
    # Initialize a variable count to 0
    count = 0
    # Loop over each element of arr using an index variable i
    for i in range(len(arr)):
        # Loop until the corresponding element
        # of brr is equal to the corresponding
        # element of arr
        while brr[i] != arr[i]:
            # If the current element of brr is
            # less than the
            # corresponding element of arr
            if brr[i] < arr[i]:
                # Loop over each element of brr starting
                # from the current index i
                for j in range(i, len(arr)):
                    # Increment each element of brr until it
                    # matches the corresponding element of arr
                    brr[j] += 1
                # Increment the count variable for
                # each increment operation
                count += 1
            # If the current element of brr is
            # greater than the corresponding element of arr
            else:
                # Loop over each element of brr starting
                # from the current index i
                for j in range(i, len(arr)):
                    # Decrement each element of brr until
                    # it matches the corresponding element of arr
                    brr[j] -= 1
                # Increment the count variable for
                # each decrement operation
                count += 1
    # Return the final count value
    return count


# Example usage:
arr = [1, 2, 3, 4]
print(min_operations(arr))  # Output: 4

using System;
using System.Collections.Generic;

class MainClass {
  static int MinOperations(List<int> arr) {
    // Create a new list brr with the same length as arr, filled with zeros
    List<int> brr = new List<int>(new int[arr.Count]);
    // Initialize a variable count to 0
    int count = 0;
    // Loop over each element of arr using an index variable i
    for (int i = 0; i < arr.Count; i++) {
      // Loop until the corresponding element of brr 
      // is equal to the corresponding element of arr
      while (brr[i] != arr[i]) {
        // If the current element of brr is less than the 
        // corresponding element of arr
        if (brr[i] < arr[i]) {
          // Loop over each element of brr starting from the current index i
          for (int j = i; j < arr.Count; j++) {
            // Increment each element of brr until it matches 
            // the corresponding element of arr
            brr[j]++;
          }
          // Increment the count variable for each increment operation
          count++;
        }
        // If the current element of brr is greater than the corresponding 
        // element of arr
        else {
          // Loop over each element of brr starting from the current index i
          for (int j = i; j < arr.Count; j++) {
            // Decrement each element of brr until it matches the 
            // corresponding element of arr
            brr[j]--;
          }
          // Increment the count variable for each decrement operation
          count++;
        }
      }
    }
    // Return the final count value
    return count;
  }

  public static void Main(string[] args) {
    List<int> arr = new List<int>{ 1, 2, 3, 4 };
    Console.WriteLine(MinOperations(arr)); // Output: 4
  }
}

function min_operations(arr)
{
    // Create a new vector brr with the same
    // length as arr, filled with zeros
    let brr = new Array(arr.length).fill(0);
    // Initialize a variable count to 0
    let count = 0;
    // Loop over each element of arr
    // using an index variable i
    for (let i = 0; i < arr.length; i++) {
        // Loop until the corresponding element of brr
        // is equal to the corresponding element of arr
        while (brr[i] != arr[i]) {
            // If the current element of brr is less
            // than the corresponding element of arr
            if (brr[i] < arr[i]) {
                // Loop over each element of brr starting
                // from the current index i
                for (let j = i; j < arr.length; j++) {
                    // Increment each element of brr until
                    // it
                    // matches the corresponding element of
                    // arr
                    brr[j]++;
                }
                // Increment the count variable for each
                // increment operation
                count++;
            }
            // If the current element of brr is greater
            // than the corresponding element of arr
            else {
                // Loop over each element of brr starting
                // from the current index i
                for (let j = i; j < arr.length; j++) {
                    // Decrement each element of brr until
                    // it matches the corresponding element
                    // of arr
                    brr[j]--;
                }
                // Increment the count variable for each
                // decrement operation
                count++;
            }
        }
    }
    // Return the final count value
    return count;
}

let arr = [ 1, 2, 3, 4 ];
console.log(min_operations(arr)); // Output: 4

4

Time Complexity: O(N²) 
Auxiliary Space: O(1)
 

Efficient Approach: The above approach can be optimized using Greedy Approach. Follow the steps below to solve the problem:  

• For the 0^th index, convert the number 0 to arr[0]. Therefore, the number of steps required will always be a[0]. Hence, add arr[0] to the answer.
• For all the other indices, the common greedy observation is to use the increase or the decrease operation by abs(a[i]-a[i-1]) times.
• The intuition behind this approach is, if the number is less than a[i-1], then increase everything from ((i-1).. n-1) by a[i], and then decrease (a[i-1] - a[i]) to get a[i].
• If a[i] > a[i-1], the approach is to use the increase operation from ((i-1)..n-1) by a[i-1], and for the remaining value, we increase it by (a[i]-a[i-1]) operations from (i..(n-1)).
• Therefore, traverse the array and for every element after the first, add the absolute difference of consecutive pairs to the answer.
• Finally, print the answer.

Below is the implementation of the above approach:

// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to calculate the minimum
// steps to obtain the desired array
int min_operation(int a[], int n)
{
    // Initialize variable
    int ans = 0;

    // Iterate over the array arr[]
    for (int i = 0; i < n; i++) {

        // Check if i > 0
        if (i > 0)

            // Update the answer
            ans += abs(a[i] - a[i - 1]);

        else
            ans += abs(a[i]);
    }

    // Return the result
    return ans;
}

// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);

    cout << min_operation(arr, n);

    return 0;
}

// Java Program to implement
// the above approach
import java.util.*;
class GFG{

// Function to calculate the minimum
// steps to obtain the desired array
static int min_operation(int a[], int n)
{
    // Initialize variable
    int ans = 0;

    // Iterate over the array arr[]
    for (int i = 0; i < n; i++)
    {

        // Check if i > 0
        if (i > 0)

            // Update the answer
            ans += Math.abs(a[i] - a[i - 1]);

        else
            ans += Math.abs(a[i]);
    }

    // Return the result
    return ans;
}

// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4 };
    int n = arr.length;

    System.out.print(min_operation(arr, n));
}
}

// This code is contributed by gauravrajput1 

# Python3 program to implement
# the above approach

# Function to calculate the minimum
# steps to obtain the desired array
def min_operation(a, n):

    # Initialize variable
    ans = 0

    # Iterate over the array arr[]
    for i in range(n):

        # Check if i > 0
        if (i > 0):

            # Update the answer
            ans += abs(a[i] - a[i - 1])
        else:
            ans += abs(a[i])

    # Return the result
    return ans

# Driver Code
if __name__ == "__main__":

    arr = [ 1, 2, 3, 4 ]
    n = len(arr)

    print(min_operation(arr, n))

# This code is contributed by chitranayal

// C# Program to implement
// the above approach
using System;
class GFG{

// Function to calculate the minimum
// steps to obtain the desired array
static int min_operation(int []a, int n)
{
    // Initialize variable
    int ans = 0;

    // Iterate over the array []arr
    for (int i = 0; i < n; i++)
    {

        // Check if i > 0
        if (i > 0)

            // Update the answer
            ans += Math.Abs(a[i] - a[i - 1]);

        else
            ans += Math.Abs(a[i]);
    }

    // Return the result
    return ans;
}

// Driver Code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 3, 4 };
    int n = arr.Length;

    Console.Write(min_operation(arr, n));
}
}

// This code is contributed by Amit Katiyar 

<script>
// javascript Program to implement
// the above approach

    // Function to calculate the minimum
    // steps to obtain the desired array
    function min_operation(a , n) {
        // Initialize variable
        var ans = 0;

        // Iterate over the array arr
        for (i = 0; i < n; i++) {

            // Check if i > 0
            if (i > 0)

                // Update the answer
                ans += Math.abs(a[i] - a[i - 1]);

            else
                ans += Math.abs(a[i]);
        }

        // Return the result
        return ans;
    }

    // Driver Code
    
        var arr = [ 1, 2, 3, 4 ];
        var n = arr.length;

        document.write(min_operation(arr, n));

// This code contributed by Rajput-Ji
</script>

4

Time Complexity: O(N)
Auxiliary Space: O(1)