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 >= Xbool 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 >= Xint 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 Codeint 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 approachimport 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 >= Xdef isGreaterEqual(N, X): return (N * (N + 1) // 2) >= X# Finds minimum value of# N such that sum of first# N natural number >= Xdef 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 Codeif __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 approachusing 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 >= Xfunction isGreaterEqual(N, X){ return parseInt((N * (N + 1)) / 2) >= X;}// Finds minimum value of// N such that sum of first// N natural number >= Xfunction 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// Inputlet X = 14;// Finds minimum value of// N such that sum of first// N natural number >= Xdocument.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 >= Xbool 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 >= Xint 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 Codeint 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 approachimport java.util.*;class GFG{// Function to check if sum of first// N natural numbers is >= Xstatic 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 >= Xstatic 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 Codepublic 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 >= Xdef isGreaterEqual(N, X): return (N * (N + 1) // 2) >= X;# Finds minimum value of# N such that sum of first# N natural number >= Xdef 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 Codeif __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 approachusing 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 >= Xfunction isGreaterEqual(N, X){ return parseInt((N * (N + 1)) / 2) >= X;}// Finds minimum value of// N such that sum of first// N natural number >= Xfunction 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// Inputlet X = 14;// Finds minimum value of// N such that sum of first// N natural number >= Xdocument.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 >= Xint 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 moduleimport 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 moduleimport math# Finds minimum value of# N such that sum of first# N natural number >= Xdef minimumPossible(X): return math.ceil(((2*X+1/4)**(0.5))-1/2)# Driver Codeif __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 moduleusing 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 >= Xfunction minimumPossible(X){ return Math.ceil(((2*X+1/4)**(0.5))-1/2)}// Driver Code// Inputlet X = 14 // Finds minimum value of// N such that sum of first// N natural number >= Xconsole.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.
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