# Find K consecutive integers such that their sum is N

• Difficulty Level : Medium
• Last Updated : 09 Aug, 2022

Given two integers N and K, the task is to find K consecutive integers such that their sum if N.
Note: If there is no such K integers print -1.
Examples:

Input: N = 15, K = 5
Output: 1 2 3 4 5
Explanation:
N can be represented as sum of 5 consecutive integers as follows –
=> N => 1 + 2 + 3 + 4 + 5 = 15
Input: N = 33, K = 6
Output: 3 4 5 6 7 8
Explanation:
N can be represented as sum of 6 consecutive integers as follows –
=> N => 3 + 4 + 5 + 6 + 7 + 8 = 33

Naive Approach: A simple solution is to run a loop from i = 0 to N – (K – 1) to check if K consecutive integers starting from i is having sum as N.
Efficient Approach: The idea is to use Arithmetic Progression to solve this problem, where sum of K terms of arithmetic progression with common difference is 1 can be defined as follows –

1. Sum of K Terms – => 2. Solving the equation further to get the first term possible
=> 3. Here aK is the Kth term which can be written as a1 + K – 1
=> => 4. Finally, check the first term computed is an integer, If yes then K consecutive number exists whose sum if N.

Below is the implementation of the above approach:

## C++

 // C++ implementation to check if// a number can be expressed as// sum of K consecutive integer #include using namespace std; // Function to check if a number can be// expressed as the sum of k consecutivevoid checksum(int n, int k){    // Finding the first    // term of AP    float first_term = ((2 * n) / k                        + (1 - k))                       / 2.0;     // Checking if first    // term is an integer    if (first_term - int(first_term) == 0) {         // Loop to print the K        // consecutive integers        for (int i = first_term;             i <= first_term + k - 1; i++) {            cout << i << " ";        }    }    else        cout << "-1";} // Driver Codeint main(){    int n = 33, k = 6;    checksum(n, k);    return 0;}

## Java

 // Java implementation to check if// a number can be expressed as// sum of K consecutive integerclass GFG{ // Function to check if a number can be// expressed as the sum of k consecutivestatic void checksum(int n, int k){         // Finding the first    // term of AP    float first_term = (float) (((2 * n) / k +                                 (1 - k)) / 2.0);     // Checking if first    // term is an integer    if (first_term - (int)(first_term) == 0)    {         // Loop to print the K        // consecutive integers        for(int i = (int)first_term;            i <= first_term + k - 1; i++)        {           System.out.print(i + " ");        }    }    else        System.out.print("-1");} // Driver Codepublic static void main(String[] args){    int n = 33, k = 6;         checksum(n, k);}} // This code is contributed by 29AjayKumar

## Python3

 # Python3 implementation to check # if a number can be expressed as# sum of K consecutive integer # Function to check if a number can be# expressed as the sum of k consecutivedef checksum(n, k):         # Finding the first    # term of AP    first_term = ((2 * n) / k + (1 - k)) / 2.0         # Checking if first    # term is an integer    if (first_term - int(first_term) == 0):                 # Loop to print the K        # consecutive integers        for i in range(int(first_term),                       int(first_term) + k):            print(i, end = ' ')    else:        print('-1') # Driver Codeif __name__=='__main__':         (n, k) = (33, 6)    checksum(n, k) # This code is contributed by rutvik_56

## C#

 // C# implementation to check if// a number can be expressed as// sum of K consecutive integerusing System;class GFG{ // Function to check if a number can be// expressed as the sum of k consecutivestatic void checksum(int n, int k){         // Finding the first    // term of AP    float first_term = (float)(((2 * n) / k +                                (1 - k)) / 2.0);     // Checking if first    // term is an integer    if (first_term - (int)(first_term) == 0)    {         // Loop to print the K        // consecutive integers        for(int i = (int)first_term;                i <= first_term + k - 1; i++)        {            Console.Write(i + " ");        }    }    else        Console.Write("-1");} // Driver Codepublic static void Main(String[] args){    int n = 33, k = 6;         checksum(n, k);}} // This code is contributed by sapnasingh4991

## Javascript

 

Output:

3 4 5 6 7 8

Time Complexity: O(n)

Auxiliary Space: O(1)

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