Find the smallest value of N such that sum of first N natural numbers is ≥ X

Given a positive integer X (1 ? X ? 10⁶), the task is to find the minimum value N, such that the sum of first N natural numbers is ? X.

Examples:

Input: X = 14
Output: 5
Explanation: Sum of first 5 natural numbers is 15 which is greater than X( = 14).

• 1 + 2 = 3( < 14)
• 1 + 2 + 3 = 6( < 14)
• 1 + 2 + 3 + 4 = 10( < 15)
• 1 + 2 + 3 + 4 + 5 = 15( > 14)

Input: X = 91
Output: 13

Naive Approach: The simplest approach to solve this problem is to check every value in the range [1, X] and return the first value from this range for which the sum of the first N natural numbers is found to be ? X.

Below is the implementation of the above approach:

C++

// C++ Program to implement
// the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to check if sum of first
// N natural numbers is >= X
bool isGreaterEqual(int N, int X)
{
    return (N * 1LL * (N + 1) / 2) >= X;
}

// Finds minimum value of
// N such that sum of first
// N natural number >= X
int minimumPossible(int X)
{
    for (int i = 1; i <= X; i++) {

        // Check if sum of first i
        // natural number >= X
        if (isGreaterEqual(i, X))
            return i;
    }
}

// Driver Code
int main()
{
    // Input
    int X = 14;

    // Finds minimum value of
    // N such that sum of first
    // N natural number >= X
    cout << minimumPossible(X);
    return 0;
}

Java

// Java Program to implement
// the above approach
import java.io.*;
class GFG
{

  // Function to check if sum of first
  // N natural numbers is >= X
  static boolean isGreaterEqual(int N, int X)
  {
    return (N * (N + 1) / 2) >= X;
  }

  // Finds minimum value of
  // N such that sum of first
  // N natural number >= X
  static int minimumPossible(int X)
  {
    for (int i = 1; i <= X; i++)
    {

      // Check if sum of first i
      // natural number >= X
      if (isGreaterEqual(i, X))
        return i;
    }
    return 0;
  }

  // Driver Code
  public static void main (String[] args)
  {

    // Input
    int X = 14;

    // Finds minimum value of
    // N such that sum of first
    // N natural number >= X
    System.out.print(minimumPossible(X));
  }
}

// This code is contributed by Dharanendra L V.

Python3

# Python3 Program to implement
# the above approach

# Function to check if sum of first
# N natural numbers is >= X
def isGreaterEqual(N, X):
    return (N * (N + 1) // 2) >= X

# Finds minimum value of
# N such that sum of first
# N natural number >= X
def minimumPossible(X):

    for i in range(1, X + 1):

        # Check if sum of first i
        # natural number >= X
        if (isGreaterEqual(i, X)):
            return i

# Driver Code
if __name__ == '__main__':

    # Input
    X = 14

    # Finds minimum value of
    # N such that sum of first
    # N natural number >= X
    print (minimumPossible(X))

    # This code is contributed by mohit kumar 29.

C#

// C# Program to implement
// the above approach
using System;
public class GFG
{

  // Function to check if sum of first
  // N natural numbers is >= X
  static bool isGreaterEqual(int N, int X)
  {
    return (N * (N + 1) / 2) >= X;
  }

  // Finds minimum value of
  // N such that sum of first
  // N natural number >= X
  static int minimumPossible(int X)
  {
    for (int i = 1; i <= X; i++)
    {

      // Check if sum of first i
      // natural number >= X
      if (isGreaterEqual(i, X))
        return i;
    }
    return 0;
  }

  // Driver Code
  static public void Main ()
  {

    // Input
    int X = 14;

    // Finds minimum value of
    // N such that sum of first
    // N natural number >= X
    Console.Write(minimumPossible(X));
  }
}

// This code is contributed by Dharanendra L V.

Javascript

<script>

// Javascript program to implement
// the above approach

// Function to check if sum of first
// N natural numbers is >= X
function isGreaterEqual(N, X)
{
    return parseInt((N * (N + 1)) / 2) >= X;
}

// Finds minimum value of
// N such that sum of first
// N natural number >= X
function minimumPossible(X)
{
    for(let i = 1; i <= X; i++)
    {

        // Check if sum of first i
        // natural number >= X
        if (isGreaterEqual(i, X))
            return i;
    }
}

// Driver Code

// Input
let X = 14;

// Finds minimum value of
// N such that sum of first
// N natural number >= X
document.write(minimumPossible(X));

// This code is contributed by rishavmahato348

</script>

Output:

5

Time Complexity : O(N)
Auxiliary Space : O(1)

Efficient Method: Below is the implementation of above approach :

1. The idea is to use binary search to solve this problem.
2. Initialize variables low = 1, high = X and perform binary search on this range.
3. Calculate mid = low + (high – low) / 2 and check if the sum of first mid numbers is greater than or equal to x or not.
4. If sum ? X, store it in a variable res and set high = mid-1
5. Otherwise, set low = mid + 1
6. Print res, which is the required answer.

Below is the implementation of the above approach:

C++

#include <bits/stdc++.h>
using namespace std;

// Function to check if sum of first
// N natural numbers is >= X
bool isGreaterEqual(int N, int X)
{
    return (N * 1LL * (N + 1) / 2) >= X;
}

// Finds minimum value of
// N such that sum of first
// N natural number >= X
int minimumPossible(int X)
{

    int low = 1, high = X, res = -1;

    // Binary Search
    while (low <= high) {
        int mid = low + (high - low) / 2;

        // Checks if sum of first 'mid' natural
        // numbers is greater than equal to X
        if (isGreaterEqual(mid, X)) {
            // Update res
            res = mid;
            // Update high
            high = mid - 1;
        }
        else
            // Update low
            low = mid + 1;
    }
    return res;
}

// Driver Code
int main()
{
    // Input
    int X = 14;

    // Finds minimum value of
    // N such that sum of first
    // N natural number >= X
    cout << minimumPossible(X);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{

// Function to check if sum of first
// N natural numbers is >= X
static boolean isGreaterEqual(int N, int X)
{
    return (N  * (N + 1) / 2) >= X;
}

// Finds minimum value of
// N such that sum of first
// N natural number >= X
static int minimumPossible(int X)
{
    int low = 1, high = X, res = -1;

    // Binary Search
    while (low <= high)
    {
        int mid = low + (high - low) / 2;

        // Checks if sum of first 'mid' natural
        // numbers is greater than equal to X
        if (isGreaterEqual(mid, X))
        {

            // Update res
            res = mid;

            // Update high
            high = mid - 1;
        }
        else
            // Update low
            low = mid + 1;
    }
    return res;
}

// Driver Code
public static void main(String[] args)
{

    // Input
    int X = 14;

    // Finds minimum value of
    // N such that sum of first
    // N natural number >= X
    System.out.print( minimumPossible(X));
}
}

// This code is contributed by code_hunt.

Python3

# Function to check if sum of first
# N natural numbers is >= X
def isGreaterEqual(N, X):
    return (N * (N + 1) // 2) >= X;

# Finds minimum value of
# N such that sum of first
# N natural number >= X
def minimumPossible(X):
  low = 1
  high = X
  res = -1;

  # Binary Search
  while (low <= high):
        mid = low + (high - low) // 2;

        # Checks if sum of first 'mid' natural
        # numbers is greater than equal to X
        if (isGreaterEqual(mid, X)):

            # Update res
            res = mid;

            # Update high
            high = mid - 1;

        else:
            # Update low
            low = mid + 1;

  return res

# Driver Code
if __name__ == "__main__":

    # Input
    X = 14;

    # Finds minimum value of
    # N such that sum of first
    # N natural number >= X
    print(minimumPossible(X));

    # This code is contributed by chitranayal.

C#

// C# program for the above approach
using System;
class GFG{

  // Function to check if sum of first
  // N natural numbers is >= X
  static bool isGreaterEqual(int N, int X)
  {
    return (N  * (N + 1) / 2) >= X;
  }

  // Finds minimum value of
  // N such that sum of first
  // N natural number >= X
  static int minimumPossible(int X)
  {
    int low = 1, high = X, res = -1;

    // Binary Search
    while (low <= high)
    {
      int mid = low + (high - low) / 2;

      // Checks if sum of first 'mid' natural
      // numbers is greater than equal to X
      if (isGreaterEqual(mid, X))
      {

        // Update res
        res = mid;

        // Update high
        high = mid - 1;
      }
      else
        // Update low
        low = mid + 1;
    }
    return res;
  }


  // Driver Code
  static public void Main()
  {
    // Input
    int X = 14;

    // Finds minimum value of
    // N such that sum of first
    // N natural number >= X
    Console.Write( minimumPossible(X));
  }
}

// This code is contributed by susmitakundugoaldanga.

Javascript

<script>

// Function to check if sum of first
// N natural numbers is >= X
function isGreaterEqual(N, X)
{
    return parseInt((N * (N + 1)) / 2) >= X;
}

// Finds minimum value of
// N such that sum of first
// N natural number >= X
function minimumPossible(X)
{

    let low = 1, high = X, res = -1;

    // Binary Search
    while (low <= high) {
        let mid = low + parseInt((high - low) / 2);

        // Checks if sum of first 'mid' natural
        // numbers is greater than equal to X
        if (isGreaterEqual(mid, X)) {
            // Update res
            res = mid;
            // Update high
            high = mid - 1;
        }
        else
            // Update low
            low = mid + 1;
    }
    return res;
}

// Driver Code
// Input
let X = 14;

// Finds minimum value of
// N such that sum of first
// N natural number >= X
document.write(minimumPossible(X));

// This code is contributed by rishavmahato348.
</script>

Output:

5

Time Complexity: O(log(X))
Auxiliary Space: O(1)

Another Efficient Approach:

As we know that expression for the sum of the first n natural number is  (n*(n+1))/2.so this expression can be very useful in finding out the value of n from the given X.
From the problem statement, it is quite clear that n * (n + 1) / 2 <= X, so first solve this equation and then use the simplified version of the approach to calculate the value of n for given X.

         n * (n + 1) / 2 >=X
=>    n² + n >= 2X
=>    n² + n + (1/4 – 1/4) >= 2X
=>    (n + 1/2)² – 1/4 >= 2X
=>    (n + 1/2)²  >= (2X+1/4)
=>    (n + 1/2) >= sqrt(2X + 1/4)
=>    n >= sqrt(2X + 1/4) – 1/2
The required value of n will be.
=>   n = ceil(sqrt(2X + 1/4) – 1/2)                                                                           

Below is the implementation of the above approach:

C++

// C++ program to implement the approach
#include <cmath>
#include <iostream>

using namespace std;

// Finds minimum value of
// N such that sum of first
// N natural number >= X
int minimumPossible(int X)
{
    return ceil((pow(2 * X + 1 / 4, 0.5) - 0.5));
}

int main()
{
    // Input
    int X = 14;

    // Finds minimum value of
    // N such that sum of first
    // N natural number >= X
    cout << minimumPossible(X) << endl;

    return 0;
}

// This code is contributed by phasing17.

Java

// Java Program to implement
// the above approach
// importing math module
import java.util.*;

class GFG
{

  // Finds minimum value of
  // N such that sum of first
  // N natural number >= X
  static int minimumPossible(int X)
  {
    return (int)Math.ceil((int)Math.pow((2*X+1/4), 0.5)-0.5);
  }

  // Driver Code
  public static void main(String[] args)
  {
    // Input
    int X = 14;

    // Finds minimum value of
    // N such that sum of first
    // N natural number >= X
    System.out.print(minimumPossible(X));
  }
}

// This code is contributed by phasing17

Python3

# Python3 Program to implement
# the above approach
# importing math module
import math

# Finds minimum value of
# N such that sum of first
# N natural number >= X


def minimumPossible(X):
    return math.ceil(((2*X+1/4)**(0.5))-1/2)


# Driver Code
if __name__ == "__main__":
    # Input
    X = 14
    # Finds minimum value of
    # N such that sum of first
    # N natural number >= X
    print(minimumPossible(X))
"""This code is contributed by rajatkumargla19 (RAJAT KUMAR)."""

C#

// C# Program to implement
// the above approach
// importing math module
using System;

class GFG
{
  // Finds minimum value of
  // N such that sum of first
  // N natural number >= X
  static int minimumPossible(int X)
  {
    return (int)Math.Ceiling((int)Math.Pow((2*X+1/4), 0.5)-0.5);
  }

  // Driver Code
  public static void Main(string[] args)
  {
    // Input
    int X = 14;

    // Finds minimum value of
    // N such that sum of first
    // N natural number >= X
    Console.Write(minimumPossible(X));
  }
}

// This code is contributed by phasing17

Javascript

// JS Program to implement
// the above approach
// importing math module

// Finds minimum value of
// N such that sum of first
// N natural number >= X
function minimumPossible(X)
{
    return Math.ceil(((2*X+1/4)**(0.5))-1/2)
}

// Driver Code

// Input
let X = 14

// Finds minimum value of
// N such that sum of first
// N natural number >= X
console.log(minimumPossible(X))

// This code is contributed by phasing17

Output: 5

Time Complexity: O(log(X)), for calculating the square root of the given X value.
Auxiliary Space: O(1) as no extra space is required here.