Generate an array of size K which satisfies the given conditions

Given two integers N and K, the task is to generate an array arr[] of length K such that:

1. arr[0] + arr[1] + … + arr[K – 1] = N.
2. arr[i] > 0 for 0 ≤ i < K.
3. arr[i] < arr[i + 1] ≤ 2 * arr[i] for 0 ≤ i < K – 1.

If there are multiple answers find any one of them, otherwise, print -1.

Examples:

Input: N = 26, K = 6 
Output: 1 2 4 5 6 8 
The generated array satisfies all of the given conditions.
Input: N = 8, K = 3 
Output: -1  

Approach: Let r = n – k * (k + 1) / 2. If r < 0 then answer is -1 already. Otherwise, let’s construct the array arr[], where all arr[i] are floor(r / k) except for rightmost r % k values, they are ceil(r / k). 
It is easy to see that the sum of this array is r, it is sorted in non-decreasing order and the difference between the maximum and the minimum element is not greater than 1. 
Let’s add 1 to arr[1], 2 to arr[2], and so on (this is what we subtract from n at the beginning). 
Then, if r != k – 1 or k = 1 then arr[] is our required array. Otherwise, we got some array of kind 1, 3, ….., arr[k]. For k = 2 or k = 3, there is no answer for this case. Otherwise, we can subtract 1 from arr[2] and add it to arr[k] and this answer will be correct.

Below is the implementation of the above approach: 

C++

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

// Function to generate and print
// the required array
void generateArray(int n, int k)
{

    // Initializing the array
    vector<int> array(k, 0);

    // Finding r (from above approach)
    int remaining = n - int(k * (k + 1) / 2);

    // If r<0
    if (remaining < 0)
        cout << ("NO");

    int right_most = remaining % k;

    // Finding ceiling and floor values
    int high = ceil(remaining / (k * 1.0));
    int low = floor(remaining / (k * 1.0));

    // Fill the array with ceiling values
    for (int i = k - right_most; i < k; i++)
        array[i]= high;

    // Fill the array with floor values
    for (int i = 0; i < (k - right_most); i++)
        array[i]= low;

    // Add 1, 2, 3, ... with corresponding values
    for (int i = 0; i < k; i++)
        array[i] += i + 1;

    if (k - 1 != remaining or k == 1)
    {
        for(int u:array) cout << u << " ";
    }

    // There is no solution for below cases
    else if (k == 2 or k == 3)
        printf("-1\n");
    else
    {

        // Modify A[1] and A[k-1] to get
        // the required array
        array[1] -= 1;
        array[k - 1] += 1;
        for(int u:array) cout << u << " ";
    }
}

// Driver Code
int main()
{
    int n = 26, k = 6;
    generateArray(n, k);
    return 0;
}

// This code is contributed
// by Mohit Kumar

Java

// Java implementation of the approach
class GFG
{

// Function to generate and print
// the required array
static void generateArray(int n, int k)
{

    // Initializing the array
    int []array = new int[k];

    // Finding r (from above approach)
    int remaining = n - (k * (k + 1) / 2);

    // If r < 0
    if (remaining < 0)
        System.out.print("NO");

    int right_most = remaining % k;

    // Finding ceiling and floor values
    int high = (int) Math.ceil(remaining / (k * 1.0));
    int low = (int) Math.floor(remaining / (k * 1.0));

    // Fill the array with ceiling values
    for (int i = k - right_most; i < k; i++)
        array[i] = high;

    // Fill the array with floor values
    for (int i = 0; i < (k - right_most); i++)
        array[i] = low;

    // Add 1, 2, 3, ... with corresponding values
    for (int i = 0; i < k; i++)
        array[i] += i + 1;

    if (k - 1 != remaining || k == 1)
    {
        for(int u:array)
            System.out.print(u + " ");
    }

    // There is no solution for below cases
    else if (k == 2 || k == 3)
        System.out.printf("-1\n");
    else
    {

        // Modify A[1] and A[k-1] to get
        // the required array
        array[1] -= 1;
        array[k - 1] += 1;
        for(int u:array)
            System.out.print(u + " ");
    }
}

// Driver Code
public static void main(String[] args)
{
    int n = 26, k = 6;
    generateArray(n, k);
}
}

// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of the approach
import sys
from math import floor, ceil

# Function to generate and print
# the required array
def generateArray(n, k):

    # Initializing the array
    array = [0] * k

    # Finding r (from above approach)
    remaining = n-int(k*(k + 1)/2)

    # If r<0
    if remaining<0:
        print("NO")
        sys.exit()

    right_most = remaining % k

    # Finding ceiling and floor values
    high = ceil(remaining / k)
    low = floor(remaining / k)

    # Fill the array with ceiling values
    for i in range(k-right_most, k):
        array[i]= high

    # Fill the array with floor values
    for i in range(k-right_most):
        array[i]= low

    # Add 1, 2, 3, ... with corresponding values
    for i in range(k):
        array[i]+= i + 1

    if k-1 != remaining or k == 1:
        print(*array)
        sys.exit()

    # There is no solution for below cases
    elif k == 2 or k == 3:
        print("-1")
        sys.exit()
    else:

        # Modify A[1] and A[k-1] to get
        # the required array
        array[1]-= 1
        array[k-1]+= 1
        print(*array)
        sys.exit()

# Driver Code
if __name__=="__main__":
    n, k = 26, 6
    generateArray(n, k)

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to generate and print
// the required array
static void generateArray(int n, int k)
{

    // Initializing the array
    int []array = new int[k];

    // Finding r (from above approach)
    int remaining = n - (k * (k + 1) / 2);

    // If r < 0
    if (remaining < 0)
        Console.Write("NO");

    int right_most = remaining % k;

    // Finding ceiling and floor values
    int high = (int) Math.Ceiling(remaining /
                                 (k * 1.0));
    int low = (int) Math.Floor(remaining /
                              (k * 1.0));

    // Fill the array with ceiling values
    for (int i = k - right_most; i < k; i++)
        array[i] = high;

    // Fill the array with floor values
    for (int i = 0;
             i < (k - right_most); i++)
        array[i] = low;

    // Add 1, 2, 3, ... with
    // corresponding values
    for (int i = 0; i < k; i++)
        array[i] += i + 1;

    if (k - 1 != remaining || k == 1)
    {
        foreach(int u in array)
            Console.Write(u + " ");
    }

    // There is no solution for below cases
    else if (k == 2 || k == 3)
        Console.Write("-1\n");
    else
    {

        // Modify A[1] and A[k-1] to get
        // the required array
        array[1] -= 1;
        array[k - 1] += 1;
        foreach(int u in array)
            Console.Write(u + " ");
    }
}

// Driver Code
public static void Main(String[] args)
{
    int n = 26, k = 6;
    generateArray(n, k);
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>
// javascript implementation of the approach
// Function to generate and print
    // the required array
    function generateArray(n , k) {

        // Initializing the array
        var array = Array(k).fill(0);

        // Finding r (from above approach)
        var remaining = n - parseInt(k * (k + 1) / 2);

        // If r < 0
        if (remaining < 0)
            document.write("NO");

        var right_most = remaining % k;

        // Finding ceiling and floor values
        var high = parseInt( Math.ceil(remaining / (k * 1.0)));
        var low = parseInt( Math.floor(remaining / (k * 1.0)));

        // Fill the array with ceiling values
        for (i = k - right_most; i < k; i++)
            array[i] = high;

        // Fill the array with floor values
        for (i = 0; i < (k - right_most); i++)
            array[i] = low;

        // Add 1, 2, 3, ... with corresponding values
        for (i = 0; i < k; i++)
            array[i] += i + 1;

        if (k - 1 != remaining || k == 1) {
            for (var u = 0;u< array.length;u++)
                document.write(array[u] + " ");
        }

        // There is no solution for below cases
        else if (k == 2 || k == 3)
            document.write("-1");
        else {

            // Modify A[1] and A[k-1] to get
            // the required array
            array[1] -= 1;
            array[k - 1] += 1;
            for (var f = 0;f< array.length;f++)
                document.write(array[f] + " ");
        }
    }

    // Driver Code

        var n = 26, k = 6;
        generateArray(n, k);

// This code is contributed by todaysgaurav
</script>

Output:

1 2 4 5 6 8

Time Complexity: O(K)

 Auxiliary Space: O(K)