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

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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.

5. Repeat the below process until left <= right.

Below is the implementation of the above approach:

 `// C++ program to print count ` `// of subsets S such that ` `// min(S) + max(S) < K ` ` `  `#include ` `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 posible 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 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 posible 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 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 posible 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# 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 posible 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 `

Output:
```4
```

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

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