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

Given a positive integer **X ***(1 ≤ X ≤ 10 ^{6})*, the task is to find the minimum value

**N**, such that the sum of first N natural numbers is ≥

**X**.

**Examples:**

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Input:X = 14Output:5Explanation: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 = 91Output: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 :

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