Given a positive integer N, the task is to count the number of integers upto N which can be represented as the sum of two or more consecutive numbers.
Examples:
Input: N = 7
Output: 4
Explanation: In the range [1, 7]: {3, 5, 6, 7} can be represented as sum of consecutive numbers.
- 3 = 1 + 2
- 5 = 2 + 3
- 6 = 1 + 2 + 3
- 7 = 3 + 4
Input: N = 15
Output: 11
Naive Approach: Follow the steps below to solve the problem:
- Iterate over all integers in the range [1, N] and for each integer, check if it can be represented as the sum of two or more consecutive integers or not.
- To check whether a number can be expressed as the sum of two or more consecutive numbers or not, refer to this article.
Below is the implementation of the above approach:
// C++ Program to implement // the above approach #include <bits/stdc++.h> using namespace std;
// Function to check if the number N // can be expressed as sum of 2 or // more consecutive numbers or not bool isPossible( int N)
{ return ((N & (N - 1)) && N);
} // Function to count integers in the range // [1, N] that can be expressed as sum of // 2 or more consecutive numbers void countElements( int N)
{ // Stores the required count
int count = 0;
for ( int i = 1; i <= N; i++) {
if (isPossible(i))
count++;
}
cout << count;
} // Driver Code int main()
{ int N = 15;
countElements(N);
return 0;
} |
// Java program of the above approach import java.io.*;
class GFG
{ // Function to check if the number N // can be expressed as sum of 2 or // more consecutive numbers or not static int isPossible( int N)
{ return (((N & (N - 1 )) & N));
} // Function to count integers in the range // [1, N] that can be expressed as sum of // 2 or more consecutive numbers static void countElements( int N)
{ // Stores the required count
int count = 0 ;
for ( int i = 1 ; i <= N; i++)
{
if (isPossible(i) != 0 )
count++;
}
System.out.println(count);
} // Driver Code public static void main(String[] args)
{ int N = 15 ;
countElements(N);
} } // This code is contributed by jana_sayantan. |
# Python3 Program to implement # the above approach # Function to check if the number N # can be expressed as sum of 2 or # more consecutive numbers or not def isPossible(N):
return ((N & (N - 1 )) and N)
# Function to count integers in the range # [1, N] that can be expressed as sum of # 2 or more consecutive numbers def countElements(N):
# Stores the required count
count = 0
for i in range ( 1 , N + 1 ):
if (isPossible(i)):
count + = 1
print (count)
# Driver Code if __name__ = = '__main__' :
N = 15
countElements(N)
# This code is contributed by mohit kumar 29
|
// C# program for the above approach using System;
class GFG
{ // Function to check if the number N // can be expressed as sum of 2 or // more consecutive numbers or not static int isPossible( int N)
{ return (((N & (N - 1)) & N));
} // Function to count integers in the range // [1, N] that can be expressed as sum of // 2 or more consecutive numbers static void countElements( int N)
{ // Stores the required count
int count = 0;
for ( int i = 1; i <= N; i++)
{
if (isPossible(i) != 0)
count++;
}
Console.Write(count);
} // Driver Code static public void Main()
{ int N = 15;
countElements(N);
} } // This code is contributed by code_hunt. |
<script> // JavaScript program of the above approach // Function to check if the number N
// can be expressed as sum of 2 or
// more consecutive numbers or not
function isPossible(N) {
return (((N & (N - 1)) & N));
}
// Function to count integers in the range
// [1, N] that can be expressed as sum of
// 2 or more consecutive numbers
function countElements(N) {
// Stores the required count
var count = 0;
for (i = 1; i <= N; i++) {
if (isPossible(i) != 0)
count++;
}
document.write(count);
}
// Driver Code
var N = 15;
countElements(N);
// This code contributed by Rajput-Ji </script> |
11
Time Complexity: O(N)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, follow the steps below to solve the problem
- All numbers except the ones which are powers of 2 can be expressed as a sum of consecutive numbers.
- Count the number of powers of 2 ? N and store it in a variable, k say Count
- Print (N – Count) as the required answer.
Below is the implementation of the above approach:
// C++ implementation // of the above approach #include <bits/stdc++.h> using namespace std;
// Function to count integers in the range // [1, N] that can be expressed as sum of // 2 or more consecutive numbers void countElements( int N)
{ int Cur_Ele = 1;
int Count = 0;
// Count powers of 2 up to N
while (Cur_Ele <= N) {
// Increment count
Count++;
// Update current power of 2
Cur_Ele = Cur_Ele * 2;
}
cout << N - Count;
} // Driver Code int main()
{ int N = 15;
countElements(N);
return 0;
} |
// Java implementation // of the above approach import java.util.*;
class GFG
{ // Function to count integers in the range
// [1, N] that can be expressed as sum of
// 2 or more consecutive numbers
static void countElements( int N)
{
int Cur_Ele = 1 ;
int Count = 0 ;
// Count powers of 2 up to N
while (Cur_Ele <= N)
{
// Increment count
Count++;
// Update current power of 2
Cur_Ele = Cur_Ele * 2 ;
}
System.out.print(N - Count);
}
// Driver Code
public static void main(String[] args)
{
int N = 15 ;
countElements(N);
}
} // This code is contributed by shikhasingrajput |
# python 3 implementation # of the above approach # Function to count integers in the range # [1, N] that can be expressed as sum of # 2 or more consecutive numbers def countElements(N):
Cur_Ele = 1
Count = 0
# Count powers of 2 up to N
while (Cur_Ele < = N):
# Increment count
Count + = 1
# Update current power of 2
Cur_Ele = Cur_Ele * 2
print (N - Count)
# Driver Code if __name__ = = '__main__' :
N = 15
countElements(N)
|
// C# implementation // of the above approach using System;
public class GFG
{ // Function to count integers in the range
// [1, N] that can be expressed as sum of
// 2 or more consecutive numbers
static void countElements( int N)
{
int Cur_Ele = 1;
int Count = 0;
// Count powers of 2 up to N
while (Cur_Ele <= N)
{
// Increment count
Count++;
// Update current power of 2
Cur_Ele = Cur_Ele * 2;
}
Console.Write(N - Count);
}
// Driver Code
public static void Main(String[] args)
{
int N = 15;
countElements(N);
}
} // This code is contributed by 29AjayKumar |
<script> // Javascript implementation // of the above approach // Function to count integers in the range // [1, N] that can be expressed as sum of // 2 or more consecutive numbers function countElements(N)
{ var Cur_Ele = 1;
var Count = 0;
// Count powers of 2 up to N
while (Cur_Ele <= N)
{
// Increment count
Count++;
// Update current power of 2
Cur_Ele = Cur_Ele * 2;
}
document.write(N - Count);
} // Driver Code var N = 15;
countElements(N); // This code is contributed by todaysgaurav </script> |
11
Time Complexity: O(log(N))
Auxiliary Space: O(1)