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 <bits/stdc++.h>using namespace std;// Function to split the first N// natural numbers into two sets// having minimum absolute// difference of their sumsint 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 Codeint main(){ int N = 6; cout << minAbsDiff(N);} |
Java
// Java program to implement// the above approachimport java.io.*; class GFG{ // Function to split the first N// natural numbers into two sets// having minimum absolute// difference of their sumsstatic 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 codepublic 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 sumsdef 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 CodeN = 6print(minAbsDiff(N))# This code is contributed by sallagondaavinashreddy7 |
C#
// C# program to implement// the above approachusing System;class GFG{// Function to split the first N// natural numbers into two sets// having minimum absolute// difference of their sumsstatic 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 codestatic void Main() { int N = 6; Console.Write(minAbsDiff(N));}}// This code is contributed by divyeshrabadiya07 |
Javascript
<script>// Javascript program to implement// the above approach// Function to split the first N// natural numbers into two sets// having minimum absolute// difference of their sumsfunction minAbsDiff(N){ // Stores the sum of // elements of set1 var sumSet1 = 0; // Stores the sum of // elements of set2 var sumSet2 = 0; // Traverse first N // natural numbers for(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 codevar N = 6;document.write(minAbsDiff(N));// This code is contributed by umadevi9616</script> |
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 <bits/stdc++.h>using namespace std;// Function to split the first N// natural numbers into two sets// having minimum absolute// difference of their sumsint minAbsDiff(int N){ if (N % 4 == 0 || N % 4 == 3) { return 0; } return 1;}// Driver Codeint main(){ int N = 6; cout << minAbsDiff(N);} |
Java
// Java program to implement// the above approachimport java.io.*;import java.util.*;class GFG{ // Function to split the first N// natural numbers into two sets// having minimum absolute// difference of their sumsstatic int minAbsDiff(int N){ if (N % 4 == 0 || N % 4 == 3) { return 0; } return 1;}// Driver Codepublic 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 sumsdef minAbsDiff(N): if (N % 4 == 0 or N % 4 == 3): return 0 return 1# Driver CodeN = 6print(minAbsDiff(N))# This code is contributed by sallagondaavinashreddy7 |
C#
// C# program to implement// the above approachusing System;class GFG{ // Function to split the first N// natural numbers into two sets// having minimum absolute// difference of their sumsstatic int minAbsDiff(int N){ if (N % 4 == 0 || N % 4 == 3) { return 0; } return 1;}// Driver Codepublic static void Main(String[] args){ int N = 6; Console.WriteLine(minAbsDiff(N));}}// This code is contributed by 29AjayKumar |
Javascript
<script>// javascript program to implement// the above approach // Function to split the first N // natural numbers into two sets // having minimum absolute // difference of their sums function minAbsDiff(N) { if (N % 4 == 0 || N % 4 == 3) { return 0; } return 1; } // Driver Code var N = 6; document.write(minAbsDiff(N));// This code contributed by gauravrajput1 </script> |
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 <climits>#include <cmath>#include <iostream>#include <numeric>#include <set>// Function to calculate the minimum absolute difference of// two subsetsint 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<Integer> 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<Integer> subset1 = new HashSet<>(); Set<Integer> 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<Integer> 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<<j): subset1.add(j+1) else: subset2.add(j+1) diff = abs(sum(subset1)-sum(subset2)) if diff < min_diff: min_diff = diff return min_diff# Example usageprint(min_absolute_diff(5)) # Output: 1print(min_absolute_diff(6)) # Output: 1 |
Output
1 1
Time complexity: O(2^N)
Space complexity: O(2^N)
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