Check if a number can be expressed as a sum of consecutive numbers

Try it on GfG Practice

Given a number n, the task is to check whether it can be expressed as a sum of two or more consecutive numbers or not. 
Example

Input  : n = 10 
Output : true
It can be expressed as sum of two consecutive
numbers 1 + 2 + 3 + 4.
Input  : n = 16  
Output : false
It cannot be expressed as sum of two consecutive
numbers.
Input  : n = 5  
Output : true
2 + 3 = 5

There is a direct and quick method to solve this. If a number is a power of two, then it cannot be expressed as a sum of consecutive numbers otherwise Yes.
The idea is based on below two facts. 
1) Sum of any two consecutive numbers is odd as one of them has to be even and the other odd. 
2) 2ⁿ = 2^n-1 + 2^n-1
If we take a closer look at 1) and 2), we can get the intuition behind the fact.
Below is the implementation of the above idea. 

// C++ program to check if a number can
// be expressed as sum of consecutive numbers
#include<bits/stdc++.h>
using namespace std;

// This function returns true if n can be
// expressed sum of consecutive.
bool canBeSumofConsec(unsigned int n)
{
    // We basically return true if n is a
    // power of two
    return ((n&(n-1)) && n);
}

// Driver code
int main()
{
    unsigned int n = 15;
    canBeSumofConsec(n)? cout << "true" :
                         cout << "false";
    return 0;
}

// Java program to check if a number can
// be expressed as sum of consecutive numbers

class Test
{
    // This function returns true if n can be
    // expressed sum of consecutive.
    static boolean canBeSumofConsec(int n)
    {
        // We basically return true if n is a
        // power of two
        return (((n&(n-1))!=0) && n!=0);
    }
    
    // Driver method
    public static void main(String[] args) 
    {
        int n = 15;
        System.out.println(canBeSumofConsec(n) ? "true" : "false");
    }
}

# Python 3 program to check if a number can
# be expressed as sum of consecutive numbers


# This function returns true if n 
# can be expressed sum of consecutive.
def canBeSumofConsec(n) :

    # We basically return true if n is a
    # power of two
    return ((n&(n-1)) and n)


# Driver code
n = 15
if(canBeSumofConsec(n)) :
    print("true")
else :
    print("false")
    
# This code is contributed by Nikita Tiwari.

// C# program to check if a number can be
// expressed as sum of consecutive numbers
using System;

class Test
{
    // This function returns true if n
    // can be expressed sum of consecutive.
    static bool canBeSumofConsec(int n)
    {
        // We basically return true if n is a
        // power of two
        return (((n & (n - 1)) != 0) && n != 0);
    }
    
    // Driver Code
    public static void Main() 
    {
        int n = 15;
        Console.Write(canBeSumofConsec(n) ? "True" : "False");
    }
}

// This code is contributed by Nitin Mittal.

<script>

// Javascript program to check if a number can
// be expressed as sum of consecutive numbers


    // This function returns true if n can be
    // expressed sum of consecutive.
    function canBeSumofConsec(n)
    {
        // We basically return true if n is a
        // power of two
        return (((n&(n-1))!=0) && n!=0);
    }
      
      
// function call
    
    let n = 15;
    document.write(canBeSumofConsec(n) ? "true" : "false");
    
</script>

<?php
// php program to check if a number
// can be expressed as sum of 
// consecutive numbers

// This function returns true if n
// can be expressed sum of consecutive.
function canBeSumofConsec($n)
{
    
    // We basically return true if n is a
    // power of two
    return (($n & ($n - 1)) && $n);
}

// Driver code
    $n = 15;
    if(canBeSumofConsec($n)) 
        echo "true" ;
    else
        echo "false";

// This code is contributed by
// nitin mittal. 
?>

true

Time Complexity: O(1)
Auxiliary Space: O(1)

Another Approach :

Let number chosen to represent N as a sum of consecutive numbers be X + 1, X + 2, X + 3 .... Y 

Sum of these chosen numbers = Sum of first Y natural numbers - Sum of first X natural number 

Sum of first Y natural number = \frac {Y \cdot (Y + 1)} {2}   Sum of first X natural number = \frac {X \cdot (X + 1)} {2}   We know that, N = Sum of first Y natural number - Sum of first X natural number N = \frac {Y \cdot (Y + 1)} {2} - \frac {X \cdot (X + 1)} {2}   2N = Y \cdot (Y + 1) - X \cdot (X + 1)   2N = Y ^ 2 - X ^ 2 + Y - X   2N = (Y - X) \cdot (Y + X + 1)   Let Y - X = a, Y + X + 1 = b Y + X + 1 > Y - X, b > a X = \frac {a - b - 1} {2}   , Y = \frac {a + b + 1} {2}   2N = a * b It means that a and b are factor of 2N, we know that X and Y are integers so, 1. b - a - 1 => multiple of 2 (Even number) 2. b + a + 1 => multiple of 2 (Even number)

Both conditions must be satisfied

From 1 and 2 we can say that either one of them (a, b) should be Odd and another one Even

So if the number (2N) has only odd factors (can't be possible as it is an even number (2N not N) ) or only even factors we can't represent it as a sum of any consecutive natural numbers

So now, we have to now only check whether it has an odd factor or not

1. If the number (2N not N) does not have any odd factor (contains only even factor means can be represented as 2 ^ k                  ) then we can't represent it as a sum of consecutive number

2. If the number (2N not N) has an odd factor then we can represent it as a sum of a consecutive number

After this we have to only check whether we can represent (2N as 2^k   ) or not

• if Yes then answer is false or 0
• if No then answer is true or 1

Below is the implementation of the above idea : 

#include <bits/stdc++.h>
using namespace std;

long long int canBeSumofConsec(long long int n) 
{ 
    // Updating n with 2n
    n = 2 * n;
    // (n & (n - 1)) => Checking whether we can write 2n as 2^k
    // if yes (can't represent 2n as 2^k) then answer 1
    // if no (can represent 2n as 2^k) then answer 0
    return ((n & (n - 1)) != 0);
}

int main()
{
    long long int n = 10;
    cout<<canBeSumofConsec(n)<<"\n";
}

#include <stdio.h>

long long int canBeSumofConsec(long long int n) 
{ 
    // Updating n with 2n
    n = 2 * n;
    // (n & (n - 1)) => Checking whether we can write 2n as 2^k
    // if yes (can't represent 2n as 2^k) then answer 1
    // if no (can represent 2n as 2^k) then answer 0
    return ((n & (n - 1)) != 0);
}

int main()
{
    long long int n = 10;
    printf("%lld", canBeSumofConsec(n));
}

import java.util.*;
class GFG{

 static int canBeSumofConsec( int n) 
{ 
    // Updating n with 2n
    n = 2 * n;
   
    // (n & (n - 1)) => Checking whether we can write 2n as 2^k
    // if yes (can't represent 2n as 2^k) then answer 1
    // if no (can represent 2n as 2^k) then answer 0
    return ((n & (n - 1)) != 0)?1:0;
}

public static void main(String[] args)
{
    int n = 10;
    System.out.print(canBeSumofConsec(n)+"\n");
}
}

// This code is contributed by umadevi9616 

def canBeSumofConsec(n):
  
    # Updating n with 2n
    n = 2 * n;

    # (n & (n - 1)) => Checking whether we can write 2n as 2^k
    # if yes (can't represent 2n as 2^k) then answer 1
    # if no (can represent 2n as 2^k) then answer 0
    if((n & (n - 1)) != 0):
        return 1;
    else:
        return 0;

if __name__ == '__main__':
    n = 10;
    print(canBeSumofConsec(n));

# This code is contributed by umadevi9616 

using System;

public class GFG {

    static int canBeSumofConsec(int n)
    {
      
        // Updating n with 2n
        n = 2 * n;

        // (n & (n - 1)) => Checking whether we can write 2n as 2^k
        // if yes (can't represent 2n as 2^k) then answer 1
        // if no (can represent 2n as 2^k) then answer 0
        return ((n & (n - 1)) != 0) ? 1 : 0;
    }

    public static void Main(String[] args) {
        int n = 10;
        Console.Write(canBeSumofConsec(n) + "\n");
    }
}

// This code is contributed by umadevi9616

<script>

    function canBeSumofConsec(n) {
        // Updating n with 2n
        n = 2 * n;

        // (n & (n - 1)) => Checking whether we can write 2n as 2^k
        // if yes (can't represent 2n as 2^k) then answer 1
        // if no (can represent 2n as 2^k) then answer 0
        return ((n & (n - 1)) != 0) ? 1 : 0;
    }

        var n = 10;
        document.write(canBeSumofConsec(n) + "\n");

// This code is contributed by umadevi9616
</script>

1

Time Complexity: O(1)
Auxiliary Space: O(1)

Approach#3:Using Brute Force

Start a loop from 1 to n/2 (inclusive) For each value i in the loop, start another loop from i+1 to n/2+1 (inclusive) Calculate the sum of consecutive numbers from i to j If the sum is equal to n, return True If the sum is greater than n, break the inner loop and start the next iteration of the outer loop If no consecutive sum is found, return False

Algorithm

1. Define a function is_sum_of_consecutive_numbers(n)
2. Start a loop from 1 to n/2 (inclusive)
3. For each value i in the loop, start another loop from i+1 to n/2+1 (inclusive)
4. Calculate the sum of consecutive numbers from i to j
5. If the sum is equal to n, return True
6. If the sum is greater than n, break the inner loop and start the next iteration of the outer loop
7. If no consecutive sum is found, return False

#include <iostream>

bool is_sum_of_consecutive_numbers(int n) {
    for (int i = 1; i < (n / 2 + 2); i++) {
        for (int j = i + 1; j < (n / 2 + 2); j++) {
            int sum_consecutive = (j - i + 1) * (i + j) / 2;
            if (sum_consecutive == n) {
                return true;
            }
            else if (sum_consecutive > n) {
                break;
            }
        }
    }
    return false;
}

int main() {
    int n = 10;
    std::cout << std::boolalpha << is_sum_of_consecutive_numbers(n) << std::endl;
    // This Code Is Contributed By Shubham Tiwari.
    return 0;
}

public class Main {
    // Function to check if a given number 'n' can be expressed as the sum of consecutive numbers
    public static boolean isSumOfConsecutiveNumbers(int n) {
        // Iterate through possible consecutive number combinations
        for (int i = 1; i < (n / 2 + 2); i++) {
            for (int j = i + 1; j < (n / 2 + 2); j++) {
                // Calculate the sum of consecutive numbers between 'i' and 'j'
                int sumConsecutive = (j - i + 1) * (i + j) / 2;
                
                // If the calculated sum matches 'n', it can be expressed as consecutive numbers
                if (sumConsecutive == n) {
                    return true;
                } else if (sumConsecutive > n) {
                    // If the sum exceeds 'n', there's no need to continue with larger 'i' values
                    break;
                }
            }
        }
        // If no consecutive number sum matches 'n', return false
        return false;
    }

    public static void main(String[] args) {
        int n = 10;
        // Check if 'n' can be expressed as the sum of consecutive numbers and print the result
        System.out.println(isSumOfConsecutiveNumbers(n));
        // This code is contributed by Shivam Tiwari
    }
}

def is_sum_of_consecutive_numbers(n):
    for i in range(1, n//2+2):
        for j in range(i+1, n//2+2):
            sum_consecutive = (j-i+1)*(i+j)//2
            if sum_consecutive == n:
                return True
            elif sum_consecutive > n:
                break
    return False
n=10
print(is_sum_of_consecutive_numbers(n))

using System;

class GFG
{
    // Function to check if a number can be represented 
   // as the sum of consecutive numbers
    static bool IsSumOfConsecutiveNumbers(int n)
    {
        for (int i = 1; i < (n / 2 + 2); i++)
        {
            for (int j = i + 1; j < (n / 2 + 2); j++)
            {
                int sumConsecutive = (j - i + 1) * (i + j) / 2;
                if (sumConsecutive == n)
                {
                    return true;
                }
                else if (sumConsecutive > n)
                {
                    break;
                }
            }
        }
        return false;
    }

    static void Main(string[] args)
    {
        int n = 10;
        Console.WriteLine($"{IsSumOfConsecutiveNumbers(n)}");

        // This Code Is Contributed By Shubham Tiwari.
    }
}

// Function to check if a number can be represented 
// as the sum of consecutive numbers
function is_sum_of_consecutive_numbers(n) {
    //loop from 1 to n/2+1
    for (let i = 1; i < (n / 2 + 2); i++) {
        //loop from i+1 to n/2+1 
        for (let j = i + 1; j < (n / 2 + 2); j++) {
            //sum of consecutive numbers from i to j
            let sum_consecutive = (j - i + 1) * (i + j) / 2;
            //When sum is equal to n
            if (sum_consecutive == n) {
                return true;
            }
            else if (sum_consecutive > n) {
                break;
            }
        }
    }
    //When no consecutive sum is found
    return false;
}

let n = 10;
console.log(is_sum_of_consecutive_numbers(n));

True

Time complexity: O(n^2)
Space complexity: O(1)

Reference: 
http://www.cut-the-knot.org/arithmetic/UnpropertyOfPowersOf2.shtml