Generate an array of K elements such that sum of elements is N and the condition a[i] < a[i+1] <= 2*a[i] is met

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

arr[0] + arr[1] + … + arr[K – 1] = N.
arr[i] > 0 for 0 ≤ i < K.
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 3 4 6 10
The above array satisfies all the conditions.

Input: N = 8, k = 3
Output: -1

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Initially we form the array with the lowest possible configuration which is filling up the array with 1, 2, 3, 4.. which satisfies the given conditions. If the summation of 1..K is greater than N, then the array cannot be formed. In order to form the array, fill up the array initially with 1, 2, 3, .. K. Again add (n-sum)/k to every element in the array, because in adding so, no conditions are void, because we are adding equal elements to every number.

The remaining number rem is greedily added from the back, to make every number twice of its previous number. After filling up the array, if any of the given conditions are not met, print -1, else the formed array will be our desired answer.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function that print the 
// desired array which 
// satisfies the given conditions 
void solve(int n, int k) 
{ 
  
    int mini = 0; 
    int x1 = 1; 
    int a[k]; 
    for (int i = 1; i <= k; i++) { 
        mini += x1; 
        a[i - 1] = x1; 
        x1 += 1; 
    } 
  
    // If the lowest filling condition 
    // is void, then it is not possible to 
    // generate the required array 
    if (n < mini) { 
        cout << "-1"; 
        return; 
    } 
  
    int rem = n - mini; 
    int cnt = rem / k; 
    rem = rem % k; 
  
    // Increase all the elements by cnt 
    for (int i = 0; i < k; i++) 
        a[i] += cnt; 
  
    // Start filling from the back 
    // till the number is a[i+1] <= 2*a[i] 
    for (int i = k - 1; i > 0 && rem > 0; i--) { 
  
        // Get the number to be filled 
        int xx = a[i - 1] * 2; 
        int left = xx - a[i]; 
  
        // If it is less than the 
        // remaining numbers to be filled 
        if (rem >= left) { 
            a[i] = xx; 
            rem -= left; 
        } 
  
        // less than remaining numbers 
        // to be filled 
        else { 
            a[i] += rem; 
            rem = 0; 
        } 
    } 
  
    // Get the sum of the array 
    int sum = a[0]; 
    for (int i = 1; i < k; i++) { 
  
        // If this condition is void at any stage 
        // during filling up, then print -1 
        if (a[i] > 2 * a[i - 1]) { 
            cout << "-1"; 
            return; 
        } 
  
        // Else add it to the sum 
        sum += a[i]; 
    } 
  
    // If the sum conditon is not 
    // satisified, then print -1 
    if (sum != n) { 
        cout << "-1"; 
        return; 
    } 
  
    // Print the generated array 
    for (int i = 0; i < k; i++) 
        cout << a[i] << " "; 
} 
  
// Driver code 
int main() 
{ 
    int n = 26, k = 6; 
    solve(n, k); 
  
    return 0; 
} 

Java

// Java implementation of the approach 
import java.util.*; 
  
class GFG { 
  
    // Function that print the 
    // desired array which 
    // satisfies the given conditions 
    static void solve(int n, int k) 
    { 
  
        int mini = 0; 
        int x1 = 1; 
        int[] a = new int[k]; 
        for (int i = 1; i <= k; i++) { 
            mini += x1; 
            a[i - 1] = x1; 
            x1 += 1; 
        } 
  
        // If the lowest filling condition 
        // is void, then it is not possible to 
        // generate the required array 
        if (n < mini) { 
            System.out.print("-1"); 
            return; 
        } 
  
        int rem = n - mini; 
        int cnt = rem / k; 
        rem = rem % k; 
  
        // Increase all the elements by cnt 
        for (int i = 0; i < k; i++) 
            a[i] += cnt; 
  
        // Start filling from the back 
        // till the number is a[i+1] <= 2*a[i] 
        for (int i = k - 1; i > 0 && rem > 0; i--) { 
  
            // Get the number to be filled 
            int xx = a[i - 1] * 2; 
            int left = xx - a[i]; 
  
            // If it is less than the 
            // remaining numbers to be filled 
            if (rem >= left) { 
                a[i] = xx; 
                rem -= left; 
            } 
  
            // less than remaining numbers 
            // to be filled 
            else { 
                a[i] += rem; 
                rem = 0; 
            } 
        } 
  
        // Get the sum of the array 
        int sum = a[0]; 
        for (int i = 1; i < k; i++) { 
  
            // If this condition is void at any stage 
            // during filling up, then print -1 
            if (a[i] > 2 * a[i - 1]) { 
                System.out.print("-1"); 
                return; 
            } 
  
            // Else add it to the sum 
            sum += a[i]; 
        } 
  
        // If the sum conditon is not 
        // satisified, then print -1 
        if (sum != n) { 
            System.out.print("-1"); 
            return; 
        } 
  
        // Print the generated array 
        for (int i = 0; i < k; i++) 
            System.out.print(a[i] + " "); 
    } 
  
    // Driver code 
    public static void main(String[] args) 
    { 
        int n = 26, k = 6; 
        solve(n, k); 
    } 
} 
  
// This code contributed by Rajput-Ji 

C#

// C# implementation of the approach 
using System; 
  
class GFG  
{ 
  
    // Function that print the 
    // desired array which 
    // satisfies the given conditions 
    static void solve(int n, int k) 
    { 
  
        int mini = 0; 
        int x1 = 1; 
        int[] a = new int[k]; 
        for (int i = 1; i <= k; i++)  
        { 
            mini += x1; 
            a[i - 1] = x1; 
            x1 += 1; 
        } 
  
        // If the lowest filling condition 
        // is void, then it is not possible to 
        // generate the required array 
        if (n < mini)  
        { 
            Console.Write("-1"); 
            return; 
        } 
  
        int rem = n - mini; 
        int cnt = rem / k; 
        rem = rem % k; 
  
        // Increase all the elements by cnt 
        for (int i = 0; i < k; i++) 
            a[i] += cnt; 
  
        // Start filling from the back 
        // till the number is a[i+1] <= 2*a[i] 
        for (int i = k - 1; i > 0 && rem > 0; i--)  
        { 
  
            // Get the number to be filled 
            int xx = a[i - 1] * 2; 
            int left = xx - a[i]; 
  
            // If it is less than the 
            // remaining numbers to be filled 
            if (rem >= left)  
            { 
                a[i] = xx; 
                rem -= left; 
            } 
  
            // less than remaining numbers 
            // to be filled 
            else 
            { 
                a[i] += rem; 
                rem = 0; 
            } 
        } 
  
        // Get the sum of the array 
        int sum = a[0]; 
        for (int i = 1; i < k; i++) 
        { 
  
            // If this condition is void at any stage 
            // during filling up, then print -1 
            if (a[i] > 2 * a[i - 1]) 
            { 
                Console.Write("-1"); 
                return; 
            } 
  
            // Else add it to the sum 
            sum += a[i]; 
        } 
  
        // If the sum conditon is not 
        // satisified, then print -1 
        if (sum != n) { 
            Console.Write("-1"); 
            return; 
        } 
  
        // Print the generated array 
        for (int i = 0; i < k; i++) 
            Console.Write(a[i] + " "); 
    } 
  
    // Driver code 
    public static void Main() 
    { 
        int n = 26, k = 6; 
        solve(n, k); 
    } 
} 
  
// This code contributed by anuj_67.. 

PHP

<?php 
// PHP implementation of the approach  
  
// Function that print the  
// desired array which  
// satisfies the given conditions  
function solve($n, $k)  
{  
  
    $mini = 0;  
    $x1 = 1;  
    $a = array();  
    for ($i = 1; $i <= $k; $i++)  
    {  
        $mini += $x1;  
        $a[$i - 1] = $x1;  
        $x1 += 1;  
    }  
  
    // If the lowest filling condition  
    // is void, then it is not possible to  
    // generate the required array  
    if ($n < $mini) 
    {  
        echo "-1";  
        return;  
    }  
  
    $rem = $n - $mini;  
    $cnt = floor($rem / $k);  
    $rem = $rem % $k;  
  
    // Increase all the elements by cnt  
    for ($i = 0; $i < $k; $i++)  
        $a[$i] += $cnt;  
  
    // Start filling from the back  
    // till the number is a[i+1] <= 2*a[i]  
    for ($i = $k - 1; $i > 0 && $rem > 0; $i--)  
    {  
          
        // Get the number to be filled  
        $xx = $a[$i - 1] * 2;  
        $left = $xx - $a[$i];  
  
        // If it is less than the  
        // remaining numbers to be filled  
        if ($rem >= $left)  
        {  
            $a[$i] = $xx;  
            $rem -= $left;  
        }  
  
        // less than remaining numbers  
        // to be filled  
        else
        {  
            $a[$i] += $rem;  
            $rem = 0;  
        }  
    }  
  
    // Get the sum of the array  
    $sum = $a[0];  
    for ($i = 1; $i < $k; $i++)  
    {  
  
        // If this condition is void at any stage  
        // during filling up, then print -1  
        if ($a[$i] > 2 * $a[$i - 1])  
        {  
            echo "-1";  
            return;  
        }  
  
        // Else add it to the sum  
        $sum += $a[$i];  
    }  
  
    // If the sum conditon is not  
    // satisified, then print -1  
    if ($sum != $n) 
    {  
        echo "-1";  
        return;  
    }  
  
    // Print the generated array  
    for ($i = 0; $i < $k; $i++)  
        echo $a[$i], " ";  
}  
  
// Driver code  
$n = 26; $k = 6;  
solve($n, $k);  
  
// This code is contributed by AnkitRai01 
?> 