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 | Set 2

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 3 4 6 10
The above array satisfies all the conditions.

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

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++

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// 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;
}

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Java














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// 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 



















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Python3














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# Python 3 implementation of the approach 


  


  


# Function that print the 


# desired array which 


# satisfies the given conditions 


def solve(n, k): 


    mini = 0


    x1 = 1


    a = [0 for i in range(k)] 


    for i in range(1, k + 1): 


        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): 


        print("-1",end = "") 


        return


  


    rem = n - mini 


    cnt = int(rem / k) 


    rem = rem % k 


  


    # Increase all the elements by cnt 


    for i in range(k): 


        a[i] += cnt 


  


    # Start filling from the back 


    # till the number is a[i+1] <= 2*a[i] 


    i = k - 1


    while (i > 0 and rem > 0): 


        # 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


              


        i -= 1


  


    # Get the sum of the array 


    sum = a[0] 


    for i in range(1, k): 


        # If this condition is void at any stage 


        # during filling up, then print -1 


        if (a[i] > 2 * a[i - 1]): 


            print("-1", end = "") 


            return


  


        # Else add it to the sum 


        sum += a[i] 


  


    # If the sum conditon is not 


    # satisified, then print -1 


    if (sum != n): 


        print("-1", end = "") 


        return


  


    # Print the generated array 


    for i in range(k): 


        print(a[i], end = " ") 


  


# Driver code 


if __name__ == '__main__': 


    n = 26


    k = 6


    solve(n, k) 


  


# This code is contributed by 


# Surendra_Gangwar 



















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C#














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// 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.. 



















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PHP














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<?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 


?> 



















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Output:


1 2 3 4 6 10






          
          
          
          

            

    

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