# Minimum value of K such that sum of cubes of first K natural number is greater than equal to N

Given a number **N**, the task is to find the minimum value **K** such that the sum of cubes of the first **K** natural number is greater than or equal to **N**. **Examples:**

Input:N = 100Output:4Explanation:

The sum of cubes of first 4 natural number is 100 which is equal to N = 100Input:N = 15Output:3Explanation:

The sum of cubes of first 2 natural number is 9 which is lesser than K = 15 and sum of first

3 natural number is 36 which is just greater than K. So the answer is 3.

**Naive Approach:** The naive approach for this problem is to run a loop from and find the sum of cubes. Whenever the sum exceeds the value of N, break from the loop.

Below is the implementation of the above approach:

## C++

`// C++ program to determine the` `// minimum value of K such that the` `// sum of cubes of first K` `// natural number is greater than` `// or equal to N` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to determine the` `// minimum value of K such that the` `// sum of cubes of first K` `// natural number is greater than` `// or equal to N` `int` `naive_find_x(` `int` `N)` `{` ` ` `// Variable to store the` ` ` `// sum of cubes` ` ` `int` `c = 0, i;` ` ` `// Loop to find the number` ` ` `// K` ` ` `for` `(i = 1; i < N; i++)` ` ` `{` ` ` `c += i * i * i;` ` ` ` ` `// If C is just greater then` ` ` `// N, then break the loop` ` ` `if` `(c >= N)` ` ` `break` `;` ` ` `}` ` ` `return` `i;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `N = 100;` ` ` `cout << naive_find_x(N);` ` ` `return` `0;` `}` `// This code is contributed by sapnasingh4991` |

## Java

`// Java program to determine the` `// minimum value of K such that the` `// sum of cubes of first K` `// natural number is greater than` `// or equal to N` `class` `GFG {` ` ` `// Function to determine the` `// minimum value of K such that the` `// sum of cubes of first K` `// natural number is greater than` `// or equal to N` `static` `int` `naive_find_x(` `int` `N)` `{` ` ` `// Variable to store the` ` ` `// sum of cubes` ` ` `int` `c = ` `0` `, i;` ` ` `// Loop to find the number` ` ` `// K` ` ` `for` `(i = ` `1` `; i < N; i++)` ` ` `{` ` ` `c += i * i * i;` ` ` ` ` `// If C is just greater then` ` ` `// N, then break the loop` ` ` `if` `(c >= N)` ` ` `break` `;` ` ` `}` ` ` `return` `i;` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `N = ` `100` `;` ` ` ` ` `System.out.println(naive_find_x(N));` `}` `}` `// This code is contributed by sapnasingh4991` |

## Python3

`# Python3 program to determine the` `# minimum value of K such that the` `# sum of cubes of first K` `# natural number is greater than` `# or equal to N` `# Function to determine the` `# minimum value of K such that the` `# sum of cubes of first K` `# natural number is greater than` `# or equal to N` `def` `naive_find_x(N):` ` ` `# Variable to store the` ` ` `# sum of cubes` ` ` `c ` `=` `0` ` ` `# Loop to find the number` ` ` `# K` ` ` `for` `i ` `in` `range` `(` `1` `, N):` ` ` `c ` `+` `=` `i` `*` `i` `*` `i` ` ` `# If C is just greater then` ` ` `# N, then break the loop` ` ` `if` `c>` `=` `N:` ` ` `break` ` ` `return` `i` `# Driver code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` ` ` `N ` `=` `100` ` ` `print` `(naive_find_x(N))` |

## C#

`// C# program to determine the` `// minimum value of K such that the` `// sum of cubes of first K` `// natural number is greater than` `// or equal to N` `using` `System;` `class` `GFG {` ` ` `// Function to determine the` `// minimum value of K such that the` `// sum of cubes of first K` `// natural number is greater than` `// or equal to N` `static` `int` `naive_find_x(` `int` `N)` `{` ` ` `// Variable to store the` ` ` `// sum of cubes` ` ` `int` `c = 0, i;` ` ` `// Loop to find the number` ` ` `// K` ` ` `for` `(i = 1; i < N; i++)` ` ` `{` ` ` `c += i * i * i;` ` ` ` ` `// If C is just greater then` ` ` `// N, then break the loop` ` ` `if` `(c >= N)` ` ` `break` `;` ` ` `}` ` ` `return` `i;` `}` `// Driver code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `N = 100;` ` ` ` ` `Console.WriteLine(naive_find_x(N));` `}` `}` `// This code is contributed by 29AjayKumar` |

## Javascript

`<script>` `// javascript program to determine the` `// minimum value of K such that the` `// sum of cubes of first K` `// natural number is greater than` `// or equal to N` `// Function to determine the` `// minimum value of K such that the` `// sum of cubes of first K` `// natural number is greater than` `// or equal to N` `function` `naive_find_x(N)` `{` ` ` `// Variable to store the` ` ` `// sum of cubes` ` ` `var` `c = 0, i;` ` ` `// Loop to find the number` ` ` `// K` ` ` `for` `(i = 1; i < N; i++)` ` ` `{` ` ` `c += i * i * i;` ` ` ` ` `// If C is just greater then` ` ` `// N, then break the loop` ` ` `if` `(c >= N)` ` ` `break` `;` ` ` `}` ` ` `return` `i;` `}` `// Driver code` `var` `N = 100;` `document.write(naive_find_x(N));` `// This code is contributed by Amit Katiyar` `</script>` |

**Output:**

4

**Time Complexity:** O(K), where K is the number which needs to be found.**Efficient Approach:** One observation which needs to be made is that the sum of cubes first N natural numbers is given by the formula:

sum = ((N * (N + 1))/2)^{2}

And, this is a monotonically increasing function. Therefore, the idea is to apply binary search to find the value of K. If the sum is greater than N for some number ‘i’, then we know that the answer is less than or equal to ‘i’. So, iterate to the left half. Else, iterate through the right half.

Below is the implementation of the above approach:

## C++

`// C++ program to determine the` `// minimum value of K such that` `// the sum of cubes of first K` `// natural number is greater than` `// or equal to N` `#include <bits/stdc++.h>` `using` `namespace` `std;` ` ` `// Function to determine the` `// minimum value of K such that` `// the sum of cubes of first K` `// natural number is greater than` `// or equal to N` `int` `binary_searched_find_x(` `int` `k)` `{` ` ` ` ` `// Left bound` ` ` `int` `l = 0;` ` ` `// Right bound` ` ` `int` `r = k;` ` ` `// Variable to store the` ` ` `// answer` ` ` `int` `ans = 0;` ` ` `// Applying binary search` ` ` `while` `(l <= r)` ` ` `{` ` ` ` ` `// Calculating mid value` ` ` `// of the range` ` ` `int` `mid = l + (r - l) / 2;` ` ` `if` `(` `pow` `(((mid * (mid + 1)) / 2), 2) >= k)` ` ` `{` ` ` ` ` `// If the sum of cubes of` ` ` `// first mid natural numbers` ` ` `// is greater than equal to N` ` ` `// iterate the left half` ` ` `ans = mid;` ` ` `r = mid - 1;` ` ` `}` ` ` `else` ` ` `{` ` ` `// Sum of cubes of first` ` ` `// mid natural numbers is` ` ` `// less than N, then move` ` ` `// to the right segment` ` ` `l = mid + 1;` ` ` `}` ` ` `}` ` ` `return` `ans;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `N = 100;` ` ` ` ` `cout << binary_searched_find_x(N);` ` ` `return` `0;` `}` `// This code is contributed by shubhamsingh10` |

## Java

`// Java program to determine the` `// minimum value of K such that the` `// sum of cubes of first K` `// natural number is greater than` `// or equal to N` `class` `GFG{` ` ` `// Function to determine the` `// minimum value of K such that the` `// sum of cubes of first K` `// natural number is greater than` `// or equal to N` `static` `int` `binary_searched_find_x(` `int` `k)` `{` ` ` `// Left bound` ` ` `int` `l = ` `0` `;` ` ` `// Right bound` ` ` `int` `r = k;` ` ` `// Variable to store the` ` ` `// answer` ` ` `int` `ans = ` `0` `;` ` ` `// Applying binary search` ` ` `while` `(l <= r)` ` ` `{` ` ` `// Calculating mid value` ` ` `// of the range` ` ` `int` `mid = l + (r - l) / ` `2` `;` ` ` `if` `(Math.pow(((mid * (mid + ` `1` `)) / ` `2` `), ` `2` `) >= k)` ` ` `{` ` ` `// If the sum of cubes of` ` ` `// first mid natural numbers` ` ` `// is greater than equal to N` ` ` `// iterate the left half` ` ` `ans = mid;` ` ` `r = mid - ` `1` `;` ` ` `}` ` ` `else` ` ` `{` ` ` `// Sum of cubes of first` ` ` `// mid natural numbers is` ` ` `// less than N, then move` ` ` `// to the right segment` ` ` `l = mid + ` `1` `;` ` ` `}` ` ` `}` ` ` `return` `ans;` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `N = ` `100` `;` ` ` `System.out.println(binary_searched_find_x(N));` `}` `}` `// This code is contributed by 29AjayKumar` |

## Python3

`# Python program to determine the` `# minimum value of K such that the` `# sum of cubes of first K` `# natural number is greater than` `# or equal to N` `# Function to determine the` `# minimum value of K such that the` `# sum of cubes of first K` `# natural number is greater than` `# or equal to N` `def` `binary_searched_find_x(k):` ` ` `# Left bound` ` ` `l ` `=` `0` ` ` `# Right bound` ` ` `r ` `=` `k` ` ` `# Variable to store the` ` ` `# answer` ` ` `ans ` `=` `0` ` ` `# Applying binary search` ` ` `while` `l<` `=` `r:` ` ` `# Calculating mid value` ` ` `# of the range` ` ` `mid ` `=` `l` `+` `(r` `-` `l)` `/` `/` `2` ` ` `if` `((mid` `*` `(mid ` `+` `1` `))` `/` `/` `2` `)` `*` `*` `2` `>` `=` `k:` ` ` `# If the sum of cubes of` ` ` `# first mid natural numbers` ` ` `# is greater than equal to N` ` ` `# iterate the left half` ` ` `ans ` `=` `mid` ` ` `r ` `=` `mid` `-` `1` ` ` `else` `:` ` ` `# Sum of cubes of first` ` ` `# mid natural numbers is` ` ` `# less than N, then move` ` ` `# to the right segment` ` ` `l ` `=` `mid ` `+` `1` ` ` `return` `ans ` `# Driver code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `N ` `=` `100` ` ` `print` `(binary_searched_find_x(N))` |

## C#

`// C# program to determine the` `// minimum value of K such that the` `// sum of cubes of first K` `// natural number is greater than` `// or equal to N` `using` `System;` `class` `GFG{` ` ` `// Function to determine the` `// minimum value of K such that the` `// sum of cubes of first K` `// natural number is greater than` `// or equal to N` `static` `int` `binary_searched_find_x(` `int` `k)` `{` ` ` `// Left bound` ` ` `int` `l = 0;` ` ` `// Right bound` ` ` `int` `r = k;` ` ` `// Variable to store the` ` ` `// answer` ` ` `int` `ans = 0;` ` ` `// Applying binary search` ` ` `while` `(l <= r)` ` ` `{` ` ` `// Calculating mid value` ` ` `// of the range` ` ` `int` `mid = l + (r - l) / 2;` ` ` `if` `(Math.Pow(((mid * (mid + 1)) / 2), 2) >= k)` ` ` `{` ` ` `// If the sum of cubes of` ` ` `// first mid natural numbers` ` ` `// is greater than equal to N` ` ` `// iterate the left half` ` ` `ans = mid;` ` ` `r = mid - 1;` ` ` `}` ` ` `else` ` ` `{` ` ` `// Sum of cubes of first` ` ` `// mid natural numbers is` ` ` `// less than N, then move` ` ` `// to the right segment` ` ` `l = mid + 1;` ` ` `}` ` ` `}` ` ` `return` `ans;` `}` `// Driver code` `public` `static` `void` `Main()` `{` ` ` `int` `N = 100;` ` ` `Console.Write(binary_searched_find_x(N));` `}` `}` `// This code is contributed by Nidhi_biet` |

## Javascript

`<script>` `// javascript program to determine the` `// minimum value of K such that the` `// sum of cubes of first K` `// natural number is greater than` `// or equal to N` ` ` `// Function to determine the` `// minimum value of K such that the` `// sum of cubes of first K` `// natural number is greater than` `// or equal to N` `function` `binary_searched_find_x(k)` `{` ` ` `// Left bound` ` ` `var` `l = 0;` ` ` `// Right bound` ` ` `var` `r = k;` ` ` `// Variable to store the` ` ` `// answer` ` ` `var` `ans = 0;` ` ` `// Applying binary search` ` ` `while` `(l <= r)` ` ` `{` ` ` `// Calculating mid value` ` ` `// of the range` ` ` `var` `mid = parseInt(l + (r - l) / 2);` ` ` `if` `(Math.pow(((mid * (mid + 1)) / 2), 2) >= k)` ` ` `{` ` ` `// If the sum of cubes of` ` ` `// first mid natural numbers` ` ` `// is greater than equal to N` ` ` `// iterate the left half` ` ` `ans = mid;` ` ` `r = mid - 1;` ` ` `}` ` ` `else` ` ` `{` ` ` `// Sum of cubes of first` ` ` `// mid natural numbers is` ` ` `// less than N, then move` ` ` `// to the right segment` ` ` `l = mid + 1;` ` ` `}` ` ` `}` ` ` `return` `ans;` `}` `// Driver code` `var` `N = 100;` `document.write(binary_searched_find_x(N));` `// This code contributed by shikhasingrajput` `</script>` |

**Output:**

4

**Time Complexity:** O(log(K)).