Given an array arr[] of N integers, the task is to split the array into two subsets such that the absolute difference between the maximum of first subset and minimum of second subset is minimum.
Examples:
Input: arr[] = {3, 1, 2, 6, 4}
Output: 1
Explanation:
Splitting the given array in two subsets, A = [1, 2, 4], B = [3, 6]. Difference of maximum of first set is 4 and minimum of second set is 3 and their difference is 1.
Input: arr[] = {2, 1, 3, 2, 4, 3}
Output: 0
Explanation:
Splitting the given array in two subsets, A = [1, 2, 2, 3], B = [3, 4]. Difference of maximum of first set is 3 and minimum of second set is 3 and their difference is 0.
Approach: To solve the above problem we have to find the two integers such that m and n such that max of first set is m and the min of second set is n. The idea is to sort the given array ascending order and after sorting the array, the minimum difference between the consecutive element is the required minimum difference after partitioning the array elements into subsets.
Below is the implementation of above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to split the array int splitArray( int arr[], int N)
{ // Sort the array in increasing order
sort(arr, arr + N);
int result = INT_MAX;
// Calculating the max difference
// between consecutive elements
for ( int i = 1; i < N; i++) {
result = min(result,
arr[i] - arr[i - 1]);
}
// Return the final minimum difference
return result;
} // Driver Code int main()
{ // Given array arr[]
int arr[] = { 3, 1, 2, 6, 4 };
// Size of array
int N = sizeof (arr) / sizeof (arr[0]);
// Function Call
cout << splitArray(arr, N);
return 0;
} |
// java program for the above approach import java.util.*;
class GFG{
// Function to split the array static int splitArray( int arr[], int N)
{ // Sort the array in increasing order
Arrays.sort(arr);
int result = Integer.MAX_VALUE;
// Calculating the max difference
// between consecutive elements
for ( int i = 1 ; i < N; i++)
{
result = Math.min(result,
arr[i] - arr[i - 1 ]);
}
// Return the final minimum difference
return result;
} // Driver Code public static void main(String[] args)
{ // Given array arr[]
int arr[] = { 3 , 1 , 2 , 6 , 4 };
// Size of array
int N = arr.length;
// Function Call
System.out.print(splitArray(arr, N));
} } // This code is contributed by shivanisinghss2110 |
# Python3 program for the above approach # Function to split the array def splitArray(arr, N):
# Sort the array in increasing
# order
arr = sorted (arr)
result = 10 * * 9
# Calculating the max difference
# between consecutive elements
for i in range ( 1 , N):
result = min (result, arr[i] - arr[i - 1 ])
# Return the final minimum difference
return result
# Driver Code if __name__ = = '__main__' :
# Given array arr[]
arr = [ 3 , 1 , 2 , 6 , 4 ]
# Size of array
N = len (arr)
# Function Call
print (splitArray(arr, N))
# This code is contributed by mohit kumar 29 |
// C# program for the above approach using System;
class GFG{
// Function to split the array static int splitArray( int []arr, int N)
{ // Sort the array in increasing order
Array.Sort(arr);
int result = Int32.MaxValue;
// Calculating the max difference
// between consecutive elements
for ( int i = 1; i < N; i++)
{
result = Math.Min(result,
arr[i] - arr[i - 1]);
}
// Return the final minimum difference
return result;
} // Driver Code public static void Main()
{ // Given array arr[]
int []arr = { 3, 1, 2, 6, 4 };
// Size of array
int N = arr.Length;
// Function Call
Console.Write(splitArray(arr, N));
} } // This code is contributed by Code_Mech |
<script> // Javascript program for the above approach
// Function to split the array
function splitArray(arr, N)
{
// Sort the array in increasing order
arr.sort();
let result = Number.MAX_VALUE;
// Calculating the max difference
// between consecutive elements
for (let i = 1; i < N; i++) {
result = Math.min(result,
arr[i] - arr[i - 1]);
}
// Return the final minimum difference
return result;
}
// Given array arr[]
let arr = [ 3, 1, 2, 6, 4 ];
// Size of array
let N = arr.length;
// Function Call
document.write(splitArray(arr, N));
</script> |
1
Time Complexity: O(N*log N)
Space Complexity : O(1)