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Count of subsets having sum of min and max element less than K

  • Difficulty Level : Medium
  • Last Updated : 13 Jul, 2021

Given an integer array arr[] and an integer K, the task is to find the number of non-empty subsets S such that min(S) + max(S) < K.
Examples: 
 

Input: arr[] = {2, 4, 5, 7} K = 8 
Output: 4 
Explanation: 
The possible subsets are {2}, {2, 4}, {2, 4, 5} and {2, 5}
Input:: arr[] = {2, 4, 2, 5, 7} K = 10 
Output: 26 
 

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Approach 
 

  1. Sort the input array first.
  2. Now use Two Pointer Technique to count the number of subsets.
  3. Let take two pointers left and right and set left = 0 and right = N-1.
  4.  

if (arr[left] + arr[right] < K ) 
Increment the left pointer by 1 and add 2 j – i into answer, because the left and right values make up a potential end values of a subset. All the values from [i, j – 1] also make up end of subsets which will have the sum < K. So, we need to calculate all the possible subsets for left = i and right ∊ [i, j]. So, after suming up values 2 j – i + 1 + 2 j – i – 2 + … + 2 0 of the GP, we get 2 j – i
if( arr[left] + arr[right] >= K ) 
Decrement the right pointer by 1. 
 

  1.  
  2. Repeat the below process until left <= right.

Below is the implementation of the above approach:
 

C++




// C++ program to print count
// of subsets S such that
// min(S) + max(S) < K
 
#include <bits/stdc++.h>
using namespace std;
 
// Function that return the
// count of subset such that
// min(S) + max(S) < K
int get_subset_count(int arr[], int K,
                     int N)
{
    // Sorting the array
    sort(arr, arr + N);
 
    int left, right;
    left = 0;
    right = N - 1;
 
    // ans stores total number of subsets
    int ans = 0;
 
    while (left <= right) {
        if (arr[left] + arr[right] < K) {
 
            // add all possible subsets
            // between i and j
            ans += 1 << (right - left);
            left++;
        }
        else {
            // Decrease the sum
            right--;
        }
    }
    return ans;
}
 
// Driver code
int main()
{
    int arr[] = { 2, 4, 5, 7 };
    int K = 8;
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << get_subset_count(arr, K, N);
    return 0;
}

Java




// Java program to print count
// of subsets S such that
// Math.min(S) + Math.max(S) < K
import java.util.*;
 
class GFG{
 
// Function that return the
// count of subset such that
// Math.min(S) + Math.max(S) < K
static int get_subset_count(int arr[], int K,
                                       int N)
{
     
    // Sorting the array
    Arrays.sort(arr);
 
    int left, right;
    left = 0;
    right = N - 1;
 
    // ans stores total number
    // of subsets
    int ans = 0;
 
    while (left <= right)
    {
        if (arr[left] + arr[right] < K)
        {
 
            // Add all possible subsets
            // between i and j
            ans += 1 << (right - left);
            left++;
        }
        else
        {
             
            // Decrease the sum
            right--;
        }
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 2, 4, 5, 7 };
    int K = 8;
    int N = arr.length;
     
    System.out.print(get_subset_count(arr, K, N));
}
}
 
// This code is contributed by Rajput-Ji

Python3




# Python3 program to print
# count of subsets S such
# that min(S) + max(S) < K
 
# Function that return the
# count of subset such that
# min(S) + max(S) < K
def get_subset_count(arr, K, N):
 
    # Sorting the array
    arr.sort()
 
    left = 0;
    right = N - 1;
 
    # ans stores total number of subsets
    ans = 0;
 
    while (left <= right):
        if (arr[left] + arr[right] < K):
             
            # Add all possible subsets
            # between i and j
            ans += 1 << (right - left);
            left += 1;
        else:
             
            # Decrease the sum
            right -= 1;
     
    return ans;
 
# Driver code
arr = [ 2, 4, 5, 7 ];
K = 8;
 
print(get_subset_count(arr, K, 4))
 
# This code is contributed by grand_master

C#




// C# program to print count
// of subsets S such that
// Math.Min(S) + Math.Max(S) < K
using System;
 
class GFG{
 
// Function that return the
// count of subset such that
// Math.Min(S) + Math.Max(S) < K
static int get_subset_count(int []arr, int K,
                                       int N)
{
     
    // Sorting the array
    Array.Sort(arr);
 
    int left, right;
    left = 0;
    right = N - 1;
 
    // ans stores total number
    // of subsets
    int ans = 0;
 
    while (left <= right)
    {
        if (arr[left] + arr[right] < K)
        {
             
            // Add all possible subsets
            // between i and j
            ans += 1 << (right - left);
            left++;
        }
        else
        {
             
            // Decrease the sum
            right--;
        }
    }
    return ans;
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 2, 4, 5, 7 };
    int K = 8;
    int N = arr.Length;
     
    Console.Write(get_subset_count(arr, K, N));
}
}
 
// This code is contributed by gauravrajput1

Javascript




<script>
 
// JavaScript program to print count
// of subsets S such that
// Math.min(S) + Math.max(S) < K
 
// Function that return the
// count of subset such that
// Math.min(S) + Math.max(S) < K
function get_subset_count(arr,K,N)
{
    // Sorting the array
    (arr).sort(function(a,b){return a-b;});
  
    let left, right;
    left = 0;
    right = N - 1;
  
    // ans stores total number
    // of subsets
    let ans = 0;
  
    while (left <= right)
    {
        if (arr[left] + arr[right] < K)
        {
  
            // Add all possible subsets
            // between i and j
            ans += 1 << (right - left);
            left++;
        }
        else
        {
              
            // Decrease the sum
            right--;
        }
    }
    return ans;
}
 
// Driver code
let arr=[ 2, 4, 5, 7];
let K = 8;
let N = arr.length;
document.write(get_subset_count(arr, K, N));
 
 
// This code is contributed by patel2127
 
</script>
Output: 
4

 

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




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