Count all possible unique sum of series K, K+1, K+2, K+3, K+4, …, K+N

Given a number N and for any number K for a series formed as K, K + 1, K + 2, K + 3, K + 4, ……., K + N. The task is to find the number of unique sums that can be obtained by adding two or more integers from the series of N + 1 integers.

Examples:

Input: N = 3 
Output: 10 
Explanation:
The possible unique combinations are: 
(K) + (K + 1) = 2*K + 1 
(K) + (K + 2) = 2*K + 2 
(K) + (K + 3) = 2*K + 3 
(K + 1) + (K + 3) = 2*K + 4 
(K + 2) + (K + 3) = 2*K + 5 
(K) + (K + 1) + (K + 2) = 3*K + 3 
(K) + (K + 1) + (K + 3) = 3*K + 4 
(K) + (K + 2) + (K + 3) = 3*K + 5 
(K + 1) + (K + 2) + (K + 3) = 3*K + 6 
(K) + (K + 1) + (K + 2) + (K + 3) = 4*K + 6 
So in total, there are 10 unique ways. 
Input: N = 4 
Output: 20 
Explanation:
The possible unique combinations are 20 in number. 

Approach: The following observations are to be made:

• Since K is number the only significance it has is that two subsets with different sizes cannot have the same sum.
• A unique sum can be obtained from the minimum possible sum of the subset to the maximum possible sum of subset having size as X where (2 ≤ X ≤ (N + 1)).
  

For Example:N = 4
The Series is = {K, K + 1, K + 2, K + 3, K + 4}
For K = 2, minimum_sum = (K, K + 1) = 2*K + 1
and maximum_sum = (K + 3, K + 4) = 2*K + 7
The maximum distinct sums possible with K = 2 are (maximum_sum – minimum_sum + 1) = (7 – 1 + 1) = 7.

Follow the steps below to solve the problem: 

1. Initialize two arrays array fsum[] and rsum[] each of size N + 1 to 0.
2. For each element of both the arrays fsum[] and rsum[], update fsum[i] with ar[i] + fsum[i – 1] and rsum[i] with ar[i] + fsum[i + 1].
3. Initialize a variable ans to 1 that stores the count of different possible sums of the given series.
4. For each possible subset size X, where (2 ≤ X ≤ (N + 1)), add the value 1 + rsum[n + 1 – k] + fsum[k] to ans.
5. The value of ans is the required answer hence return it.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

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

// Function to count the unique sum
int count_unique_sum(int n)
{

    int i, ar[n + 1], fsum[n + 1];
    int rsum[n + 1], ans = 1;

    // Initialize array fsum[] with 0
    memset(fsum, 0, sizeof fsum);

    // Initialize array rsum[] with 0
    memset(rsum, 0, sizeof rsum);

    for (i = 0; i <= n; i++) {
        ar[i] = i;
    }

    // Set fsum[0] as ar[0]
    fsum[0] = ar[0];

    // Set rsum[0] as ar[n]
    rsum[n] = ar[n];

    // For each i update fsum[i] with
    // ar[i] + fsum[i - 1]
    for (i = 1; i <= n; i++) {
        fsum[i] = ar[i] + fsum[i - 1];
    }

    // For each i from n-1, update
    // rsum[i] with ar[i] + fsum[i + 1]
    for (i = n - 1; i >= 0; i--) {
        rsum[i] = ar[i] + rsum[i + 1];
    }

    // K represent size of subset as
    // explained above
    for (int k = 2; k <= n; k++) {

        // Using above relation
        ans += 1 + rsum[n + 1 - k]
               - fsum[k - 1];
    }

    // Return the result
    return ans;
}

// Driver Code
int main()
{
    // Given a number N
    int N = 4;

    // Function Call
    cout << count_unique_sum(N);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG{

// Function to count the unique sum
static int count_unique_sum(int n)
{
    int i;
    int ar[] = new int[n + 1];
    int fsum[] = new int[n + 1];
    int rsum[] = new int[n + 1];
    int ans = 1;

    // Initialize array fsum[] with 0
    Arrays.fill(fsum, 0);

    // Initialize array rsum[] with 0
    Arrays.fill(rsum, 0);

    for (i = 0; i <= n; i++)
    {
        ar[i] = i;
    }

    // Set fsum[0] as ar[0]
    fsum[0] = ar[0];

    // Set rsum[0] as ar[n]
    rsum[n] = ar[n];

    // For each i update fsum[i] with
    // ar[i] + fsum[i - 1]
    for (i = 1; i <= n; i++)
    {
        fsum[i] = ar[i] + fsum[i - 1];
    }

    // For each i from n-1, update
    // rsum[i] with ar[i] + fsum[i + 1]
    for (i = n - 1; i >= 0; i--)
    {
        rsum[i] = ar[i] + rsum[i + 1];
    }

    // K represent size of subset as
    // explained above
    for (int k = 2; k <= n; k++)
    {

        // Using above relation
        ans += 1 + rsum[n + 1 - k] -
                     fsum[k - 1];
    }

    // Return the result
    return ans;
}

// Driver Code
public static void main(String[] args)
{
    // Given a number N
    int N = 4;

    // Function Call
    System.out.print(count_unique_sum(N));
}
}

// This code is contributed by rock__cool

Python3

# Python3 program for the above approach

# Function to count the unique sum
def count_unique_sum(n):

    ar = [0] * (n + 1)
    fsum = [0] * (n + 1)
    rsum = [0] * (n + 1)
    ans = 1

    for i in range(0, n + 1):
        ar[i] = i

    # Set fsum[0] as ar[0]
    fsum[0] = ar[0]

    # Set rsum[0] as ar[n]
    rsum[n] = ar[n]

    # For each i update fsum[i] with
    # ar[i] + fsum[i - 1]
    for i in range(1, n + 1):
        fsum[i] = ar[i] + fsum[i - 1]

    # For each i from n-1, update
    # rsum[i] with ar[i] + fsum[i + 1]
    for i in range(n - 1, -1, -1):
        rsum[i] = ar[i] + rsum[i + 1]

    # K represent size of subset as
    # explained above
    for k in range(2, n + 1):

        # Using above relation
        ans += (1 + rsum[n + 1 - k] -
                    fsum[k - 1])

    # Return the result
    return ans

# Driver Code

# Given a number N
N = 4

# Function call
print(count_unique_sum(N))

# This code is contributed by sanjoy_62

C#

// C# program for the above approach
using System;
class GFG{

// Function to count the unique sum
static int count_unique_sum(int n)
{
    int i;
    int []ar = new int[n + 1];
    int []fsum = new int[n + 1];
    int []rsum = new int[n + 1];
    int ans = 1;

    for (i = 0; i <= n; i++)
    {
        ar[i] = i;
    }

    // Set fsum[0] as ar[0]
    fsum[0] = ar[0];

    // Set rsum[0] as ar[n]
    rsum[n] = ar[n];

    // For each i update fsum[i] with
    // ar[i] + fsum[i - 1]
    for (i = 1; i <= n; i++)
    {
        fsum[i] = ar[i] + fsum[i - 1];
    }

    // For each i from n-1, update
    // rsum[i] with ar[i] + fsum[i + 1]
    for (i = n - 1; i >= 0; i--)
    {
        rsum[i] = ar[i] + rsum[i + 1];
    }

    // K represent size of subset as
    // explained above
    for (int k = 2; k <= n; k++)
    {

        // Using above relation
        ans += 1 + rsum[n + 1 - k] -
                   fsum[k - 1];
    }

    // Return the result
    return ans;
}

// Driver Code
public static void Main(String[] args)
{
    // Given a number N
    int N = 4;

    // Function Call
    Console.Write(count_unique_sum(N));
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>
    // Javascript program for the above approach

    // Function to count the unique sum
    function count_unique_sum(n)
    {

        let i;
        let ans = 1;
        let ar = new Array(n + 1);
        let fsum = new Array(n + 1);
        let rsum = new Array(n + 1);

        // Initialize array fsum[] with 0
        fsum.fill(0);

        // Initialize array rsum[] with 0
        rsum.fill(0);

        for (i = 0; i <= n; i++) {
            ar[i] = i;
        }

        // Set fsum[0] as ar[0]
        fsum[0] = ar[0];

        // Set rsum[0] as ar[n]
        rsum[n] = ar[n];

        // For each i update fsum[i] with
        // ar[i] + fsum[i - 1]
        for (i = 1; i <= n; i++) {
            fsum[i] = ar[i] + fsum[i - 1];
        }

        // For each i from n-1, update
        // rsum[i] with ar[i] + fsum[i + 1]
        for (i = n - 1; i >= 0; i--) {
            rsum[i] = ar[i] + rsum[i + 1];
        }

        // K represent size of subset as
        // explained above
        for (let k = 2; k <= n; k++) {

            // Using above relation
            ans += 1 + rsum[n + 1 - k]
                   - fsum[k - 1];
        }

        // Return the result
        return ans;
    }

    // Given a number N
    let N = 4;

    // Function Call
    document.write(count_unique_sum(N));

//This code is contributed by suresh07.
</script>

Output

20

Time Complexity: O(N) 
Auxiliary Space: O(N),  since N extra space has been taken.