# Split first N natural numbers into two sets with minimum absolute difference of their sums

Given an integer N, split the first N natural numbers into two sets such that the absolute difference between their sum is minimum. The task is to print the minimum absolute difference that can be obtained.

Examples:

Input: N = 5
Output: 1
Explanation:
Split the first N (= 5) natural numbers into sets {1, 2, 5} (sum = 8) and {3, 4} (sum = 7).
Therefore, the required output is 1.

Input: N = 6
Output: 1

Naive Approach: This problem can be solved using the Greedy technique. Follow the steps below to solve the problem:

• Initialize two variables, say sumSet1 and sumSet2 to store the sum of the elements from the two sets.
• Traverse first N natural numbers from N to 1. For every number, check if the current sum of elements in set1 is less than or equal to the sum of elements in set2. If found to be true, add the currently traversed number into set1 and update sumSet1.
• Otherwise, add the value of the current natural number to set2 and update sumSet2.
• Finally, print abs(sumSet1 – sumSet2) as the required answer.

Below is the implementation of the above approach:

## C++14

 `// C++ program to implement` `// the above approach`   `#include ` `using` `namespace` `std;`   `// Function to split the first N` `// natural numbers into two sets` `// having minimum absolute` `// difference of their sums` `int` `minAbsDiff(``int` `N)` `{` `    ``// Stores the sum of` `    ``// elements of set1` `    ``int` `sumSet1 = 0;`   `    ``// Stores the sum of` `    ``// elements of set2` `    ``int` `sumSet2 = 0;`   `    ``// Traverse first N` `    ``// natural numbers` `    ``for` `(``int` `i = N; i > 0; i--) {`   `        ``// Check if sum of elements of` `        ``// set1 is less than or equal` `        ``// to sum of elements of set2` `        ``if` `(sumSet1 <= sumSet2) {` `            ``sumSet1 += i;` `        ``}` `        ``else` `{` `            ``sumSet2 += i;` `        ``}` `    ``}` `    ``return` `abs``(sumSet1 - sumSet2);` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `N = 6;` `    ``cout << minAbsDiff(N);` `}`

## Java

 `// Java program to implement` `// the above approach` `import` `java.io.*;` ` `  `class` `GFG{` `    `  `// Function to split the first N` `// natural numbers into two sets` `// having minimum absolute` `// difference of their sums` `static` `int` `minAbsDiff(``int` `N)` `{` `    `  `    ``// Stores the sum of` `    ``// elements of set1` `    ``int` `sumSet1 = ``0``;` ` `  `    ``// Stores the sum of` `    ``// elements of set2` `    ``int` `sumSet2 = ``0``;` ` `  `    ``// Traverse first N` `    ``// natural numbers` `    ``for``(``int` `i = N; i > ``0``; i--) ` `    ``{` `        `  `        ``// Check if sum of elements of` `        ``// set1 is less than or equal` `        ``// to sum of elements of set2` `        ``if` `(sumSet1 <= sumSet2) ` `        ``{` `            ``sumSet1 += i;` `        ``}` `        ``else` `        ``{` `            ``sumSet2 += i;` `        ``}` `    ``}` `    ``return` `Math.abs(sumSet1 - sumSet2);` `}`   `// Driver code` `public` `static` `void` `main (String[] args)` `{` `    ``int` `N = ``6``;` `    `  `    ``System.out.println(minAbsDiff(N));` `}` `}`   `// This code is contributed by offbeat`

## Python3

 `# Python3 program to implement` `# the above approach`   `# Function to split the first N` `# natural numbers into two sets` `# having minimum absolute` `# difference of their sums` `def` `minAbsDiff(N):` `    `  `    ``# Stores the sum of` `    ``# elements of set1` `    ``sumSet1 ``=` `0`   `    ``# Stores the sum of` `    ``# elements of set2` `    ``sumSet2 ``=` `0`   `    ``# Traverse first N` `    ``# natural numbers` `    ``for` `i ``in` `reversed``(``range``(N ``+` `1``)):` `        `  `        ``# Check if sum of elements of` `        ``# set1 is less than or equal` `        ``# to sum of elements of set2` `        ``if` `sumSet1 <``=` `sumSet2:` `           ``sumSet1 ``=` `sumSet1 ``+` `i` `        ``else``:` `           ``sumSet2 ``=` `sumSet2 ``+` `i` `      `  `    ``return` `abs``(sumSet1 ``-` `sumSet2)`   `# Driver Code` `N ``=` `6`   `print``(minAbsDiff(N))`   `# This code is contributed by sallagondaavinashreddy7`

## C#

 `// C# program to implement` `// the above approach` `using` `System;`   `class` `GFG{`   `// Function to split the first N` `// natural numbers into two sets` `// having minimum absolute` `// difference of their sums` `static` `int` `minAbsDiff(``int` `N)` `{` `    `  `    ``// Stores the sum of` `    ``// elements of set1` `    ``int` `sumSet1 = 0;` ` `  `    ``// Stores the sum of` `    ``// elements of set2` `    ``int` `sumSet2 = 0;` ` `  `    ``// Traverse first N` `    ``// natural numbers` `    ``for``(``int` `i = N; i > 0; i--)` `    ``{` `        `  `        ``// Check if sum of elements of` `        ``// set1 is less than or equal` `        ``// to sum of elements of set2` `        ``if` `(sumSet1 <= sumSet2) ` `        ``{` `            ``sumSet1 += i;` `        ``}` `        ``else` `        ``{` `            ``sumSet2 += i;` `        ``}` `    ``}` `    ``return` `Math.Abs(sumSet1 - sumSet2);` `}`   `// Driver code` `static` `void` `Main() ` `{` `    ``int` `N = 6;` `    `  `    ``Console.Write(minAbsDiff(N));` `}` `}`   `// This code is contributed by divyeshrabadiya07`

## Javascript

 ``

Output

```1

```

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

Efficient Approach: To optimize the above approach the idea is based on the following observations:

Splitting any 4 consecutive integers into 2 sets gives the minimum absolute difference of their sum equal to 0.

Mathematical proof:
Considering 4 consecutive integers {a1, a2, a3, a4}
a4 = a3 + 1
a1=a2  – 1
=> a4 + a1 = a3 + 1 + a2  – 1
=> a4 + a1 = a2 + a3

Follow the steps below to solve the problem:

• If N % 4 == 0 or N % 4 == 3, then print 0.
• Otherwise, print 1.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement` `// the above approach` `#include ` `using` `namespace` `std;`   `// Function to split the first N` `// natural numbers into two sets` `// having minimum absolute` `// difference of their sums` `int` `minAbsDiff(``int` `N)` `{` `    ``if` `(N % 4 == 0 || N % 4 == 3) {` `        ``return` `0;` `    ``}` `    ``return` `1;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `N = 6;` `    ``cout << minAbsDiff(N);` `}`

## Java

 `// Java program to implement` `// the above approach` `import` `java.io.*;` `import` `java.util.*;`   `class` `GFG{` `    `  `// Function to split the first N` `// natural numbers into two sets` `// having minimum absolute` `// difference of their sums` `static` `int` `minAbsDiff(``int` `N)` `{` `    ``if` `(N % ``4` `== ``0` `|| N % ``4` `== ``3``)` `    ``{` `        ``return` `0``;` `    ``}` `    ``return` `1``;` `}`   `// Driver Code` `public` `static` `void` `main (String[] args)` `{` `    ``int` `N = ``6``;` `    `  `    ``System.out.println(minAbsDiff(N));` `}` `}`   `// This code is contributed by sallagondaavinashreddy7`

## Python3

 `# Python3 program to implement` `# the above approach`   `# Function to split the first N` `# natural numbers into two sets` `# having minimum absolute` `# difference of their sums` `def` `minAbsDiff(N):` `    `  `    ``if` `(N ``%` `4` `=``=` `0` `or` `N ``%` `4` `=``=` `3``):` `        ``return` `0` `        `  `    ``return` `1`   `# Driver Code` `N ``=` `6`   `print``(minAbsDiff(N))`   `# This code is contributed by sallagondaavinashreddy7`

## C#

 `// C# program to implement` `// the above approach` `using` `System;` `class` `GFG{` `    `  `// Function to split the first N` `// natural numbers into two sets` `// having minimum absolute` `// difference of their sums` `static` `int` `minAbsDiff(``int` `N)` `{` `  ``if` `(N % 4 == 0 || ` `      ``N % 4 == 3)` `  ``{` `    ``return` `0;` `  ``}` `  ``return` `1;` `}`   `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `  ``int` `N = 6;` `  ``Console.WriteLine(minAbsDiff(N));` `}` `}`   `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output

```1

```

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

### Approach 3:

Approach:

• Create a set of the first N natural numbers.
• Initialize a variable min_diff to infinity to keep track of the minimum absolute difference found so far.
• Loop through all possible subsets of the set of numbers. Since we are only interested in splitting the set into two subsets, we can loop through all
• binary numbers from 1 to 2^(N-1), where each bit in the binary number represents whether the corresponding number in the set belongs to subset1 or subset2.
• For each subset, calculate the absolute difference between the sum of the numbers in subset1 and subset2.
• If the absolute difference is less than the current minimum difference, update the minimum difference.
• After all subsets have been checked, return the minimum absolute difference found.

## C++

 `#include ` `#include ` `#include ` `#include ` `#include `   `// Function to calculate the minimum absolute difference of` `// two subsets` `int` `minAbsoluteDiff(``int` `N)` `{` `    ``// Create a set containing numbers from 1 to N` `    ``std::set<``int``> numSet;` `    ``for` `(``int` `i = 1; i <= N; i++) {` `        ``numSet.insert(i);` `    ``}`   `    ``int` `minDiff = INT_MAX;`   `    ``// Generate all possible subsets using bitwise` `    ``// operations` `    ``for` `(``int` `i = 1; i < std::``pow``(2, N - 1); i++) {` `        ``std::set<``int``> subset1, subset2;`   `        ``// Divide the numbers into two subsets based on the` `        ``// binary representation of i` `        ``for` `(``int` `j = 0; j < N; j++) {` `            ``if` `(i & (1 << j)) {` `                ``subset1.insert(j + 1);` `            ``}` `            ``else` `{` `                ``subset2.insert(j + 1);` `            ``}` `        ``}`   `        ``// Calculate the absolute difference between the` `        ``// sums of the two subsets` `        ``int` `sum1 = std::accumulate(subset1.begin(),` `                                   ``subset1.end(), 0);` `        ``int` `sum2 = std::accumulate(subset2.begin(),` `                                   ``subset2.end(), 0);` `        ``int` `diff = std::``abs``(sum1 - sum2);`   `        ``// Update the minimum difference if a smaller one is` `        ``// found` `        ``if` `(diff < minDiff) {` `            ``minDiff = diff;` `        ``}` `    ``}`   `    ``return` `minDiff;` `}`   `int` `main()` `{` `    ``// Example usage and output` `    ``std::cout << minAbsoluteDiff(5)` `              ``<< std::endl; ``// Output: 1` `    ``std::cout << minAbsoluteDiff(6)` `              ``<< std::endl; ``// Output: 1`   `    ``return` `0;` `}`

## Java

 `import` `java.util.HashSet;` `import` `java.util.Set;`   `public` `class` `MinAbsoluteDiff {`   `    ``// Function to find the minimum absolute difference between two subsets of {1, 2, ..., N}` `    ``public` `static` `int` `minAbsoluteDiff(``int` `N) {` `        ``Set numSet = ``new` `HashSet<>();` `        ``for` `(``int` `i = ``1``; i <= N; i++) {` `            ``numSet.add(i);` `        ``}`   `        ``int` `minDiff = Integer.MAX_VALUE;`   `        ``// Iterate through all possible non-empty subsets of {1, 2, ..., N-1}` `        ``for` `(``int` `i = ``1``; i < (``1` `<< (N - ``1``)); i++) {` `            ``Set subset1 = ``new` `HashSet<>();` `            ``Set subset2 = ``new` `HashSet<>();`   `            ``// Partition the numbers into two subsets based on the binary representation of 'i'` `            ``for` `(``int` `j = ``0``; j < N; j++) {` `                ``if` `((i & (``1` `<< j)) != ``0``) {` `                    ``subset1.add(j + ``1``);` `                ``} ``else` `{` `                    ``subset2.add(j + ``1``);` `                ``}` `            ``}`   `            ``// Calculate the absolute difference between the sums of the two subsets` `            ``int` `diff = Math.abs(sum(subset1) - sum(subset2));`   `            ``// Update the minimum difference if 'diff' is smaller` `            ``if` `(diff < minDiff) {` `                ``minDiff = diff;` `            ``}` `        ``}`   `        ``return` `minDiff;` `    ``}`   `    ``// Function to calculate the sum of elements in a set` `    ``public` `static` `int` `sum(Set set) {` `        ``int` `sum = ``0``;` `        ``for` `(``int` `num : set) {` `            ``sum += num;` `        ``}` `        ``return` `sum;` `    ``}`   `    ``public` `static` `void` `main(String[] args) {` `        ``// Example usage` `        ``System.out.println(minAbsoluteDiff(``5``)); ``// Output: 1` `        ``System.out.println(minAbsoluteDiff(``6``)); ``// Output: 1` `    ``}` `}`

## Python3

 `def` `min_absolute_diff(N):` `    ``num_set ``=` `set``(``range``(``1``,N``+``1``))` `    ``min_diff ``=` `float``(``'inf'``)` `    ``for` `i ``in` `range``(``1``, ``2``*``*``(N``-``1``)):` `        ``subset1 ``=` `set``()` `        ``subset2 ``=` `set``()` `        ``for` `j ``in` `range``(N):` `            ``if` `i & (``1``<

## C#

 `using` `System;` `using` `System.Collections.Generic;`   `class` `Program {` `    ``// Function to calculate the minimum absolute difference` `    ``// of two subsets` `    ``static` `int` `MinAbsoluteDiff(``int` `N)` `    ``{` `        ``// Create a set containing numbers from 1 to N` `        ``HashSet<``int``> numSet = ``new` `HashSet<``int``>();` `        ``for` `(``int` `i = 1; i <= N; i++) {` `            ``numSet.Add(i);` `        ``}`   `        ``int` `minDiff = ``int``.MaxValue;`   `        ``// Generate all possible subsets using bitwise` `        ``// operations` `        ``for` `(``int` `i = 1; i < Math.Pow(2, N - 1); i++) {` `            ``HashSet<``int``> subset1 = ``new` `HashSet<``int``>();` `            ``HashSet<``int``> subset2 = ``new` `HashSet<``int``>();`   `            ``// Divide the numbers into two subsets based on` `            ``// the binary representation of i` `            ``for` `(``int` `j = 0; j < N; j++) {` `                ``if` `((i & (1 << j)) != 0) {` `                    ``subset1.Add(j + 1);` `                ``}` `                ``else` `{` `                    ``subset2.Add(j + 1);` `                ``}` `            ``}`   `            ``// Calculate the absolute difference between the` `            ``// sums of the two subsets` `            ``int` `sum1 = 0;` `            ``foreach``(``int` `num ``in` `subset1) { sum1 += num; }`   `            ``int` `sum2 = 0;` `            ``foreach``(``int` `num ``in` `subset2) { sum2 += num; }`   `            ``int` `diff = Math.Abs(sum1 - sum2);`   `            ``// Update the minimum difference if a smaller` `            ``// one is found` `            ``if` `(diff < minDiff) {` `                ``minDiff = diff;` `            ``}` `        ``}`   `        ``return` `minDiff;` `    ``}`   `    ``static` `void` `Main()` `    ``{` `        ``// Example usage and output` `        ``Console.WriteLine(MinAbsoluteDiff(5)); ``// Output: 1` `        ``Console.WriteLine(MinAbsoluteDiff(6)); ``// Output: 1` `    ``}` `}`

## Javascript

 ``

Output

```1
1

```

Time complexity: O(2^N)
Space complexity: O(2^N)

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