Subtraction in the Array

Given an integer k and an array arr[], the task is to repeat the following operation exactly k times: 
Find the minimum non-zero element in the array, print it and then subtract this number from all the non-zero elements of the array. If all the elements of the array are < 0, just print 0.

Examples:

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

k = 1; Pick 2 and update arr[] = {1, 4, 2, 0} 
k = 2; Pick 1, arr[] = {0, 3, 1, 0} 
k = 3; Pick 1, arr[] = {0, 2, 0, 0} 
k = 4; Pick 2, arr[] = {0, 0, 0, 0} 
k = 5; Nothing to pick so print 0 

Input: arr[] = {1, 2}, k = 3 
Output: 1 1 0 

Approach: Sort the array and take an extra variable named sum which will store previous element which became 0. 

Taking arr[] = {3, 6, 4, 2} and initially sum = 0 after sorting the array, it becomes arr[] = {2, 3, 4, 6}. 
Now sum = 0, and we print first nonzero element i.e. 2 and assign sum = 2. 
In the next iteration, pick second element i.e. 3 and print 3 – sum i.e. 1 as 2 has already been subtracted from all the other non-zero elements. Repeat these steps exactly k times.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long int

// Function to perform the given operation on arr[]
void operations(int arr[], int n, int k)
{
    sort(arr, arr + n);
    ll i = 0, sum = 0;
    while (k--) {

        // Skip elements which are 0
        while (i < n && arr[i] - sum == 0)
            i++;

        // Pick smallest non-zero element
        if (i < n && arr[i] - sum > 0) {
            cout << arr[i] - sum << " ";
            sum = arr[i];
        }

        // If all the elements of arr[] are 0
        else
            cout << 0 << endl;
    }
}

// Driver code
int main()
{
    int k = 5;
    int arr[] = { 3, 6, 4, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    operations(arr, n, k);
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG
{

// Function to perform the given operation on arr[]
static void operations(int arr[], int n, int k)
{
    Arrays.sort(arr);
    int i = 0, sum = 0;
    while (k-- > 0)
    {

        // Skip elements which are 0
        while (i < n && arr[i] - sum == 0)
            i++;

        // Pick smallest non-zero element
        if (i < n && arr[i] - sum > 0)
        {
            System.out.print(arr[i] - sum + " ");
            sum = arr[i];
        }

        // If all the elements of arr[] are 0
        else
            System.out.println("0");
    }
}

// Driver code
public static void main(String args[])
{
    int k = 5;
    int arr[] = { 3, 6, 4, 2 };
    int n = arr.length;
    operations(arr, n, k);
}
}

// This code is contributed by Princi Singh

Python3

# Python implementation of the approach

# Function to perform the given operation on arr[]
def operations(arr, n, k):
    arr.sort();
    i = 0; sum = 0;
    while (k > 0):

        # Skip elements which are 0
        while (i < n and arr[i] - sum == 0):
            i+=1;

        # Pick smallest non-zero element
        if (i < n and arr[i] - sum > 0):
            print(arr[i] - sum, end= " ");
            sum = arr[i];

        # If all the elements of arr[] are 0
        else:
            print(0);
        k-=1;

# Driver code
k = 5;
arr = [ 3, 6, 4, 2 ];
n = len(arr);
operations(arr, n, k);

# This code is contributed by PrinciRaj1992

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to perform the given operation on arr[]
static void operations(int []arr, int n, int k)
{
    Array.Sort(arr);
    int i = 0, sum = 0;
    while (k-- > 0)
    {

        // Skip elements which are 0
        while (i < n && arr[i] - sum == 0)
            i++;

        // Pick smallest non-zero element
        if (i < n && arr[i] - sum > 0)
        {
            Console.Write(arr[i] - sum + " ");
            sum = arr[i];
        }

        // If all the elements of arr[] are 0
        else
            Console.WriteLine("0");
    }
}

// Driver code
public static void Main(String []args)
{
    int k = 5;
    int []arr = { 3, 6, 4, 2 };
    int n = arr.Length;
    operations(arr, n, k);
}
}

// This code has been contributed by 29AjayKumar

Javascript

<script>

// JavaScript program implementation of the approach

// Function to perform the given operation on arr[]
function operations(arr, n, k)
{
    arr.sort();
    let i = 0, sum = 0;
    while (k-- > 0)
    {

        // Skip elements which are 0
        while (i < n && arr[i] - sum == 0)
            i++;

        // Pick smallest non-zero element
        if (i < n && arr[i] - sum > 0)
        {
            document.write(arr[i] - sum + " ");
            sum = arr[i];
        }

        // If all the elements of arr[] are 0
        else
            document.write("0");
    }
}

// Driver code

    let k = 5;
    let arr = [ 3, 6, 4, 2 ];
    let n = arr.length;
    operations(arr, n, k);

</script>

Output

2 1 1 2 0

Time Complexity: O(nlog(n)+n*k)
Auxiliary Space: O(1)