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
Approach
- Sort the input array first.
- Now use Two Pointer Technique to count the number of subsets.
- Let take two pointers left and right and set left = 0 and right = N-1.
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 summing 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.
- 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) < Kint 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 codeint 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) < Kimport java.util.*;class GFG{// Function that return the// count of subset such that// Math.min(S) + Math.max(S) < Kstatic 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 codepublic 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) < Kusing System;class GFG{// Function that return the// count of subset such that// Math.Min(S) + Math.Max(S) < Kstatic 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 codepublic 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) < Kfunction 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 codelet 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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