Given an integer N, the task is to calculate the sum of first N natural numbers adding all powers of 2 twice to the sum.
Examples:
Input: N = 4
Output: 17
Explanation:
Sum = 2+4+3+8 = 17
Since 1, 2 and 4 are 2 0, 2 1 and 2 2 respectively, they are added twice to the sum.
Input: N = 5
Output: 22
Explanation:
The sum is equal to 2+4+3+8+5 = 22,
because 1, 2 and 4 are 2 0, 2 1 and 2 2 respectively.
Naive Approach:
The simplest approach to solve this problem is to iterate upto N, and keep calculating the sum by adding every number once except the powers of 2, which needs to be added twice.
Time Complexity: O(N)
Auxiliary Space: O(1)
Efficient Approach:
Follow the steps below to optimize the above approach:
- Calculate sum of first N natural numbers by the formula (N * (N + 1)) / 2.
- Now, all powers of 2 needs to be added once more. Sum of all powers of 2 up to N can be calculated as 2 log2(N) + 1 – 1.
- Hence, the required sum is:
(N * (N + 1)) / 2 + 2 log2(N) + 1 – 1
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 raise N to the // power P and return the value double power(int N, int P) { return pow(N, P); } // Function to calculate the // log base 2 of an integer int Log2(int N) { // Calculate log2(N) indirectly // using log() method int result = (int)(log(N) / log(2)); return result; } // Function to calculate and // return the required sum double specialSum(int n) { // Sum of first N natural // numbers double sum = n * (n + 1) / 2; // Sum of all powers of 2 // up to N int a = Log2(n); sum = sum + power(2, a + 1) - 1; return sum; } // Driver code int main() { int n = 4; cout << (specialSum(n)) << endl; return 0; } // This code is contributed by divyeshrabadiya07 |
Java
// Java program to implement // the above approach import java.util.*; import java.lang.Math; class GFG { // Function to raise N to the // power P and return the value static double power(int N, int P) { return Math.pow(N, P); } // Function to calculate the // log base 2 of an integer public static int log2(int N) { // Calculate log2(N) indirectly // using log() method int result = (int)(Math.log(N) / Math.log(2)); return result; } // Function to calculate and // return the required sum static double specialSum(int n) { // Sum of first N natural // numbers double sum = n * (n + 1) / 2; // Sum of all powers of 2 // up to N int a = log2(n); sum = sum + power(2, a + 1) - 1; return sum; } // Driver Code public static void main(String[] args) { int n = 4; System.out.println(specialSum(n)); } } |
Python3
# Python3 program to implement# the above approachimport math# Function to raise N to the# power P and return the valuedef power(N, P): return math.pow(N, P)# Function to calculate the# log base 2 of an integerdef Log2(N): # Calculate log2(N) indirectly # using log() method result = (math.log(N) // math.log(2)) return result# Function to calculate and# return the required sumdef specialSum(n): # Sum of first N natural # numbers sum = n * (n + 1) // 2 # Sum of all powers of 2 # up to N a = Log2(n) sum = sum + power(2, a + 1) - 1 return sum # Driver code if __name__=="__main__": n = 4 print(specialSum(n))# This code is contributed by rutvik_56 |
C#
// C# program to implement // the above approach using System; class GFG { // Function to raise N to the // power P and return the value static double power(int N, int P) { return Math.Pow(N, P); } // Function to calculate the // log base 2 of an integer public static int log2(int N) { // Calculate log2(N) indirectly // using log() method int result = (int)(Math.Log(N) / Math.Log(2)); return result; } // Function to calculate and // return the required sum static double specialSum(int n) { // Sum of first N natural // numbers double sum = (double)(n) * (n + 1) / 2; // Sum of all powers of 2 // up to N int a = log2(n); sum = (sum) + power(2, a + 1) - 1; return sum; } // Driver Code public static void Main(string[] args) { int n = 4; Console.Write(specialSum(n)); } } // This code is contributed by Ritik Bansal |
Javascript
<script>// Javascript program to implement// the above approach// Function to raise N to the// power P and return the valuefunction power(N, P){ return Math.pow(N, P);}// Function to calculate the// log base 2 of an integerfunction Log2(N){ // Calculate log2(N) indirectly // using log() method let result = (Math.floor(Math.log(N) / Math.log(2))); return result;}// Function to calculate and// return the required sumfunction specialSum(n){ // Sum of first N natural // numbers let sum = n * (n + 1) / 2; // Sum of all powers of 2 // up to N let a = Log2(n); sum = sum + power(2, a + 1) - 1; return sum;}// Driver Code let n = 4; document.write(specialSum(n) + "<br>");// This code is contributed by Mayank Tyagi</script> |
Output:
17.0
Time Complexity: O(log2(N))
Auxiliary Space: O(1), since no extra space has been taken.
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