Skip to content
Related Articles

Related Articles

Improve Article
Find n positive integers that satisfy the given equations
  • Difficulty Level : Easy
  • Last Updated : 19 May, 2021

Given three integers N, X, and Y. The task is to find N positive integers that satisfy the given equations. 
 

  1. a12 + a22 + …. + an2 ≥ X
  2. a1 + a2 + …. + an ≤ Y

If no such sequence of integers is possible then print -1.
Examples: 
 

Input: N = 3, X = 254, Y = 18 
Output: 1 1 16 
12 + 12 + 162 = 1 + 1 + 256 = 258 which is ≥ X 
1 + 1 + 16 = 18 which is ≤ Y
Input: N = 2, X = 3, Y = 2 
Output: -1 
No such sequence exists. 
 

 

Approach: It is easy to see that in order to maximize the sum of squares, one should make all numbers except the first one equal to 1 and maximize the first number. Keeping this in mind we only need to check whether the given value of y is large enough to satisfy a restriction that all n numbers are positive. If y is not too small, then all we need is to ensure that X ≤ 1 + 1 + … + (y – (n – 1))2.
Below is the implementation of the above approach: 
 



C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find n positive integers
// that satisfy the given conditions
void findIntegers(int n, int x, int y)
{
 
    // To store n positive integers
    vector<int> ans;
 
    // Place N - 1 one's
    for (int i = 0; i < n - 1; i++)
        ans.push_back(1);
 
    // If can not place (y - (n - 1))
    // as the Nth integer
    if (y - (n - 1) <= 0) {
        cout << "-1";
        return;
    }
 
    // Place Nth integer
    ans.push_back(y - (n - 1));
 
    // To store the sum of
    // squares of N integers
    int store = 0;
    for (int i = 0; i < n; i++)
        store += ans[i] * ans[i];
 
    // If it is less than x
    if (store < x) {
        cout << "-1";
        return;
    }
 
    // Print the required integers
    for (int i = 0; i < n; i++)
        cout << ans[i] << " ";
}
 
// Driver code
int main()
{
    int n = 3, x = 254, y = 18;
    findIntegers(n, x, y);
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
     
// Function to find n positive integers
// that satisfy the given conditions
static void findIntegers(int n, int x, int y)
{
 
    // To store n positive integers
    ArrayList<Integer> ans = new ArrayList<Integer>();
 
    // Place N - 1 one's
    for (int i = 0; i < n - 1; i++)
        ans.add(1);
 
    // If can not place (y - (n - 1))
    // as the Nth integer
    if (y - (n - 1) <= 0)
    {
        System.out.print("-1");
        return;
    }
 
    // Place Nth integer
    ans.add(y - (n - 1));
 
    // To store the sum of
    // squares of N integers
    int store = 0;
    for (int i = 0; i < n; i++)
        store += ans.get(i) * ans.get(i);
 
    // If it is less than x
    if (store < x)
    {
        System.out.print("-1");
        return;
    }
 
    // Print the required integers
    for (int i = 0; i < n; i++)
        System.out.print(ans.get(i)+" ");
}
 
// Driver code
public static void main (String[] args)
{
    int n = 3, x = 254, y = 18;
    findIntegers(n, x, y);
}
}
 
// This code is contributed by mits

Python3




# Python3 implementation of the approach
 
# Function to find n positive integers
# that satisfy the given conditions
def findIntegers(n, x, y):
 
    # To store n positive integers
    ans = []
 
    # Place N - 1 one's
    for i in range(n - 1):
        ans.append(1)
 
    # If can not place (y - (n - 1))
    # as the Nth integer
    if (y - (n - 1) <= 0):
        print("-1", end = "")
        return
 
    # Place Nth integer
    ans.append(y - (n - 1))
 
    # To store the sum of
    # squares of N integers
    store = 0
 
    for i in range(n):
        store += ans[i] * ans[i]
 
    # If it is less than x
    if (store < x):
        print("-1", end = "")
        return;
 
    # Print the required integers
    for i in range(n):
        print(ans[i], end = " ")
 
# Driver code
n, x, y = 3, 254, 18
findIntegers(n, x, y)
 
# This code is contributed by mohit kumar

C#




// C# implementation of the approach
using System;
using System.Collections;
 
class GFG
{
     
// Function to find n positive integers
// that satisfy the given conditions
static void findIntegers(int n, int x, int y)
{
 
    // To store n positive integers
    ArrayList ans = new ArrayList();
 
    // Place N - 1 one's
    for (int i = 0; i < n - 1; i++)
        ans.Add(1);
 
    // If can not place (y - (n - 1))
    // as the Nth integer
    if (y - (n - 1) <= 0)
    {
        Console.Write("-1");
        return;
    }
 
    // Place Nth integer
    ans.Add(y - (n - 1));
 
    // To store the sum of
    // squares of N integers
    int store = 0;
    for (int i = 0; i < n; i++)
        store += (int)ans[i] *(int)ans[i];
 
    // If it is less than x
    if (store < x)
    {
        Console.Write("-1");
        return;
    }
 
    // Print the required integers
    for (int i = 0; i < n; i++)
        Console.Write((int)ans[i]+" ");
}
 
// Driver code
static void Main()
{
    int n = 3, x = 254, y = 18;
    findIntegers(n, x, y);
}
}
 
// This code is contributed by mits

PHP




<?php
// Php implementation of the approach
 
// Function to find n positive integers
// that satisfy the given conditions
function findIntegers($n, $x, $y)
{
 
    // To store n positive integers
    $ans = array();
 
    // Place N - 1 one's
    for ($i = 0; $i < $n - 1; $i++)
        array_push($ans,1) ;
 
    // If can not place (y - (n - 1))
    // as the Nth integer
    if ($y - ($n - 1) <= 0)
    {
        echo "-1";
        return;
    }
 
    // Place Nth integer
    array_push($ans,$y - ($n - 1));
 
    // To store the sum of
    // squares of N integers
    $store = 0;
    for ($i = 0; $i < $n; $i++)
        $store += $ans[$i] * $ans[$i];
 
    // If it is less than x
    if ($store < $x)
    {
        echo "-1";
        return;
    }
 
    // Print the required integers
    for ($i = 0; $i < $n; $i++)
        echo $ans[$i]," ";
}
 
    // Driver code
    $n = 3; $x = 254; $y = 18;
    findIntegers($n, $x, $y);
     
    // This code is contributed by Ryuga
?>

Javascript




<script>
 
// JavaScript implementation of the approach
 
// Function to find n positive letegers
// that satisfy the given conditions
function findIntegers(n, x, y)
{
   
    // To store n positive letegers
    let ans = [];
   
    // Place N - 1 one's
    for (let i = 0; i < n - 1; i++)
        ans.push(1);
   
    // If can not place (y - (n - 1))
    // as the Nth leteger
    if (y - (n - 1) <= 0)
    {
        document.write("-1");
        return;
    }
   
    // Place Nth leteger
    ans.push(y - (n - 1));
   
    // To store the sum of
    // squares of N letegers
    let store = 0;
    for (let i = 0; i < n; i++)
        store += ans[i] * ans[i];
   
    // If it is less than x
    if (store < x)
    {
        document.write("-1");
        return;
    }
   
    // Print the required letegers
    for (let i = 0; i < n; i++)
        document.write(ans[(i)]+" ");
}     
     
// Driver Code
 
     let n = 3, x = 254, y = 18;
    findIntegers(n, x, y);
           
</script>
Output: 
1 1 16

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live




My Personal Notes arrow_drop_up
Recommended Articles
Page :