Divide 1 to n into two groups with minimum sum difference

Last Updated : 20 Mar, 2023

Given a positive integer n such that n > 2. Divide numbers from 1 to n in two groups such that absolute difference of sum of each group is minimum. Print any two groups with their size in first line and in next line print elements of that group.
Examples:

Input : 5
Output : 2
         5 2
         3
         4 3 1
Here sum of group 1 is 7 and sum of group 2 is 8.
Their absolute difference is 1 which is minimum.
We can have multiple correct answers. (1, 2, 5) and 
(3, 4) is another such group.

Input : 6
Output : 2
         6 4
         4
         5 3 2 1

We can always divide sum of n integers in two groups such that their absolute difference of their sum is 0 or 1. So sum of group at most differ by 1. We define sum of group1 as half of n elements sum.
Now run a loop from n to 1 and insert i into group1 if inserting an element doesn’t exceed group1 sum otherwise insert that i into group2.

C++

// CPP program to divide n integers
// in two groups such that absolute
// difference of their sum is minimum
#include <bits/stdc++.h>
using namespace std;
 
// To print vector along size
void printVector(vector<int> v)
{
    // Print vector size
    cout << v.size() << endl;
 
    // Print vector elements
    for (int i = 0; i < v.size(); i++)
        cout << v[i] << " ";
 
    cout << endl;
}
 
// To divide n in two groups such that
// absolute difference of their sum is 
// minimum
void findTwoGroup(int n)
{
    // Find sum of all elements upto n
    int sum = n * (n + 1) / 2;
 
    // Sum of elements of group1
    int group1Sum = sum / 2;
 
    vector<int> group1, group2;
 
    for (int i = n; i > 0; i--) {
 
        // If sum is greater than or equal
        // to 0 include i in group 1
        // otherwise include in group2
        if (group1Sum - i >= 0) {
 
            group1.push_back(i);
 
            // Decrease sum of group1
            group1Sum -= i;
        }
        else {
            group2.push_back(i);
        }
    }
 
    // Print both the groups
    printVector(group1);
    printVector(group2);
}
 
// Driver program to test above functions
int main()
{
    int n = 5;
    findTwoGroup(n);
    return 0;
}

Java

// Java program to divide n integers
// in two groups such that absolute
// difference of their sum is minimum
import java.io.*;
import java.util.*;
 
class GFG 
{
    // To print vector along size
    static void printVector(Vector<Integer> v)
    {
        // Print vector size
        System.out.println(v.size());
     
        // Print vector elements
        for (int i = 0; i < v.size(); i++)
            System.out.print(v.get(i) + " ");
     
        System.out.println();
    }
     
    // To divide n in two groups such that
    // absolute difference of their sum is 
    // minimum
    static void findTwoGroup(int n)
    {
        // Find sum of all elements upto n
        int sum = n * (n + 1) / 2;
     
        // Sum of elements of group1
        int group1Sum = sum / 2;
     
        Vector<Integer> group1 = new Vector<Integer>();
        Vector<Integer> group2 = new Vector<Integer>();
     
        for (int i = n; i > 0; i--) {
     
            // If sum is greater than or equal
            // to 0 include i in group1
            // otherwise include in group2
            if (group1Sum - i >= 0) {
     
                group1.add(i);
     
                // Decrease sum of group1
                group1Sum -= i;
            }
            else {
                group2.add(i);
            }
        }
     
        // Print both the groups
        printVector(group1);
        printVector(group2);
    }
     
    // Driver code
    public static void main (String[] args) 
    {
        int n = 5;
        findTwoGroup(n);
    }
}
 
// This code is contributed by Gitanjali.

Python3

# Python program to divide n integers
# in two groups such that absolute
# difference of their sum is minimum
import math
 
# To print vector along size
def printVector( v):
 
    # Print vector size
    print(len(v))
 
    # Print vector elements
    for i in range( 0, len(v)):
        print(v[i] , end =  " ")
 
    print()
 
 
# To divide n in two groups such that
# absolute difference of their sum is 
# minimum
def findTwoGroup(n):
 
    # Find sum of all elements upto n
    sum = n * (n + 1) / 2
 
    # Sum of elements of group1
    group1Sum = sum / 2
 
    group1=[]
    group2=[]
    for i in range(n, 0, -1):
 
        # If sum is greater than or equal
        # to 0 include i in group 1
        # otherwise include in group2
        if (group1Sum - i >= 0) :
            group1.append(i)
 
            # Decrease sum of group1
            group1Sum -= i
         
        else :
            group2.append(i)
 
    # Print both the groups
    printVector(group1)
    printVector(group2)
 
# driver code
n = 5
findTwoGroup(n)
 
# This code is contributed by Gitanjali.

C#

// C# program to divide n integers
// in two groups such that absolute
// difference of their sum is minimum
using System;
using System.Collections;
 
class GFG 
{
// To print vector along size
static void printVector(ArrayList v)
{
    // Print vector size
    Console.WriteLine(v.Count);
 
    // Print vector elements
    for (int i = 0; i < v.Count; i++)
        Console.Write(v[i] + " ");
 
    Console.WriteLine();
}
 
// To divide n in two groups 
// such that absolute difference 
// of their sum is minimum
static void findTwoGroup(int n)
{
    // Find sum of all elements upto n
    int sum = n * (n + 1) / 2;
 
    // Sum of elements of group1
    int group1Sum = sum / 2;
     
    ArrayList group1 = new ArrayList();
    ArrayList group2 = new ArrayList();
 
    for (int i = n; i > 0; i--) 
    {
 
        // If sum is greater than 
        // or equal to 0 include i 
        // in group1 otherwise 
        // include in group2
        if (group1Sum - i >= 0) 
        {
            group1.Add(i);
 
            // Decrease sum of group1
            group1Sum -= i;
        }
        else
        {
            group2.Add(i);
        }
    }
 
    // Print both the groups
    printVector(group1);
    printVector(group2);
}
 
// Driver code
public static void Main() 
{
    int n = 5;
    findTwoGroup(n);
}
}
 
// This code is contributed by mits

PHP

<?php
// PHP program to divide n 
// integers in two groups 
// such that absolute 
// difference of their
// sum is minimum
 
// To print vector
// along size
function printVector($v)
{
    // Print vector size
    echo count($v) . "\n";
 
    // Print vector elements
    for ($i = 0; 
         $i < count($v); $i++)
        echo $v[$i] . " ";
 
    echo "\n";
}
 
// To divide n in two groups 
// such that absolute difference 
// of their sum is minimum
function findTwoGroup($n)
{
    // Find sum of all 
    // elements upto n
    $sum = $n * ($n + 1) / 2;
 
    // Sum of elements 
    // of group1
    $group1Sum = (int)($sum / 2);
 
    $group1;
    $group2;
    $x = 0;
    $y = 0;
 
    for ($i = $n; $i > 0; $i--) 
    {
 
        // If sum is greater than 
        // or equal to 0 include 
        // i in group 1 otherwise 
        // include in group2
        if ($group1Sum - $i >= 0) 
        {
 
            $group1[$x++] = $i;
 
            // Decrease sum
            // of group1
            $group1Sum -= $i;
        }
        else
        {
            $group2[$y++] = $i;
        }
    }
 
    // Print both the groups
    printVector($group1);
    printVector($group2);
}
 
// Driver Code
$n = 5;
findTwoGroup($n);
 
// This code is contributed by mits. 
?>

Javascript

<script>
 
 
// Javascript program to divide n integers
// in two groups such that absolute
// difference of their sum is minimum
 
// To print vector along size
function printVector(v)
{
    // Print vector size
    document.write( v.length + "<br>");
 
    // Print vector elements
    for (var i = 0; i < v.length; i++)
        document.write( v[i] + " ");
 
    document.write("<br>");
}
 
// To divide n in two groups such that
// absolute difference of their sum is 
// minimum
function findTwoGroup(n)
{
    // Find sum of all elements upto n
    var sum = n * (n + 1) / 2;
 
    // Sum of elements of group1
    var group1Sum = parseInt(sum / 2);
 
    var group1 = [], group2 = [];
 
    for (var i = n; i > 0; i--) {
 
        // If sum is greater than or equal
        // to 0 include i in group 1
        // otherwise include in group2
        if (group1Sum - i >= 0) {
 
            group1.push(i);
 
            // Decrease sum of group1
            group1Sum -= i;
        }
        else {
            group2.push(i);
        }
    }
 
    // Print both the groups
    printVector(group1);
    printVector(group2);
}
 
// Driver program to test above functions
var n = 5;
findTwoGroup(n);
 
</script>

Output:

Time Complexity: O(n)

Auxiliary Space: O(n)

Suggest improvement

Minimum Fibonacci terms with sum equal to K

Paper Cut into Minimum Number of Squares

Share your thoughts in the comments

Standard Greedy algorithms

Greedy algorithm on Array

Greedy algorithm on Operating System

Greedy algorithm on Graph

Approximate Greedy algorithm for NP complete problems

Greedy algorithm for special cases of DP

Some easy problems on Greedy algorithm

Some medium level problems on Greedy algorithm

Some hard problems on Greedy algorithm