Minimize count of Subsets with difference between maximum and minimum element not exceeding K
Last Updated :
19 May, 2021
Given an array arr[ ] and an integer K, the task is to split the given array into minimum number of subsets having the difference between the maximum and the minimum element ≤ K.
Examples:
Input: arr[ ] = {1, 3, 7, 9, 10}, K = 3
Output: 2
Explanation:
One of the possible subsets of arr[] are {1, 3} and {7, 9, 10} where the difference between maximum and minimum element does not greater than K i.e, 3.
Input: arr[ ] = {1, 10, 8, 3, 9}, K =3
Output: 2.
Approach: Follow the steps below to solve the problem:
- Sort the array in ascending order.
- Iterate over the array, setting currMin as the first element of the array and keep updating currMax with the elements traversed.
- If at any index, the difference between currMax and currMin exceeds K, increment answer by 1 and set currMax and currMin to arr[i].
- Finally, return answer.
Below is the implementation of the above approach:
C++
// C++ Program to implement
// above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the minimum count
// of subsets of required type
int findCount(int arr[], int N, int K)
{
sort(arr, arr + N);
// Stores the result
int result = 1;
// Store the maximum and minimum
// element of the current subset
int cur_max = arr[0];
int cur_min = arr[0];
for (int i = 1; i < N; i++) {
// Update current maximum
cur_max = arr[i];
// If difference exceeds K
if (cur_max - cur_min > K) {
// Update count
result++;
// Update maximum and minimum
// to the current subset
cur_max = arr[i];
cur_min = arr[i];
}
}
return result;
}
// Driver Code
int main()
{
int arr[] = { 1,10, 8, 3, 9 };
int K = 3;
int N = sizeof(arr) / sizeof(arr[0]);
cout << findCount(arr, N, K);
return 0;
}
// Java program to implement
// above approach
import java.util.*;
class GFG{
// Function to find the minimum count
// of subsets of required type
static int findCount(int arr[], int N, int K)
{
Arrays.sort(arr);
// Stores the result
int result = 1;
// Store the maximum and minimum
// element of the current subset
int cur_max = arr[0];
int cur_min = arr[0];
for(int i = 1; i < N; i++)
{
// Update current maximum
cur_max = arr[i];
// If difference exceeds K
if (cur_max - cur_min > K)
{
// Update count
result++;
// Update maximum and minimum
// to the current subset
cur_max = arr[i];
cur_min = arr[i];
}
}
return result;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 1, 10, 8, 3, 9 };
int K = 3;
int N = arr.length;
System.out.print(findCount(arr, N, K));
}
}
// This code is contributed by amal kumar choubey
# Python3 program to implement
# the above approach
# Function to find the minimum count
# of subsets of required type
def findCount(arr, N, K):
arr.sort()
# Stores the result
result = 1
# Store the maximum and minimum
# element of the current subset
cur_max = arr[0]
cur_min = arr[0]
for i in range(1, N):
# Update current maximum
cur_max = arr[i]
# If difference exceeds K
if(cur_max - cur_min > K):
# Update count
result += 1
# Update maximum and minimum
# to the current subset
cur_max = arr[i]
cur_min = arr[i]
return result
# Driver Code
arr = [ 1, 10, 8, 3, 9 ]
K = 3
N = len(arr)
# Function call
print(findCount(arr, N, K))
# This code is contributed by Shivam Singh
// C# program to implement
// above approach
using System;
class GFG{
// Function to find the minimum count
// of subsets of required type
static int findCount(int []arr,
int N, int K)
{
Array.Sort(arr);
// Stores the result
int result = 1;
// Store the maximum and minimum
// element of the current subset
int cur_max = arr[0];
int cur_min = arr[0];
for(int i = 1; i < N; i++)
{
// Update current maximum
cur_max = arr[i];
// If difference exceeds K
if (cur_max - cur_min > K)
{
// Update count
result++;
// Update maximum and minimum
// to the current subset
cur_max = arr[i];
cur_min = arr[i];
}
}
return result;
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 1, 10, 8, 3, 9 };
int K = 3;
int N = arr.Length;
Console.Write(findCount(arr, N, K));
}
}
// This code is contributed by gauravrajput1
<script>
// javascript program for the
// above approach
// Function to find the minimum count
// of subsets of required type
function findCount(arr, N, K)
{
arr.sort();
// Stores the result
let result = 1;
// Store the maximum and minimum
// element of the current subset
let cur_max = arr[0];
let cur_min = arr[0];
for(let i = 1; i < N; i++)
{
// Update current maximum
cur_max = arr[i];
// If difference exceeds K
if (cur_max - cur_min > K)
{
// Update count
result++;
// Update maximum and minimum
// to the current subset
cur_max = arr[i];
cur_min = arr[i];
}
}
return result;
}
// Driver Code
let arr = [ 1, 10, 8, 3, 9 ];
let K = 3;
let N = arr.length;
document.write(findCount(arr, N, K));
// This code is contributed by target_2.
</script>
Time Complexity: O(NLog(N))
Auxiliary Space: O(1)
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