Given an array, arr[], and an integer K, the task is to find the total number of ways in which we can divide the array into two groups such that the sum of each group is greater than equal to K and each element belongs to one of these groups.
Examples:
Input: arr[ ] = [6, 6], K = 2
Output: 2
Explanation: The partitions are ([6], [6]) and ([6], [6]) since both the 6 at index 0 and 1 are treated differently.Input: arr[ ] = [1, 2, 3, 4], K = 4
Output: 6
Explanation: The partitions are: ([1, 2, 3], [4]), ([1, 3], [2, 4]), ([1, 4], [2, 3]), ([2, 3], [1, 4]), ([2, 4], [1, 3]) and ([4], [1, 2, 3]).
Approach: The problem can be solved based on the following observation:
Try to solve this problem recursively where we will find out the number of subsets whose sum is less than K and store this in an array to avoid calculating the same thing again and again. Finally, we will subtract this from the total subsets to find the desired answer. This in turn will reduce our time complexity. Yes, you guess it right we are about to apply the dynamic programming concept.
Follow the steps mentioned below to implement the idea:
- If the sum of elements in the array is less than twice K then a way cannot be made.
- we will make dp with all pairs whose sum is less than k
- Generate pairs for all sums from 0 to K – 1
- Collect all the power sets.
- Finally, we will perform an operation where (way whose sum is greater than K = Total partitions – partitions whose (sum < K)).
Below is the Implementation of the above approach:
C++
// C++ code for the above approach:#include <bits/stdc++.h>using namespace std;// Function to count waysint cntWays(vector<int>& arr, int K){ int ans = 1; // Dp will be contain all pairs whose // sum is from 0 to K-1 vector<int> dp(K, 0); long total = 0; dp[0] = 1; int lessThanK = K - 1; for (auto& n : arr) { // Generate pairs for all sum // from 0 to K-1 for (int i = lessThanK - n; i >= 0; i--) { dp[i + n] = (dp[i + n] + dp[i]); } // Collecting all power sets // i.e 2^n ans = ans * 2; total += n; } if (total < 2 * K) return 0; // Collect groups whose sum is less // than K long sumlessthanK = 0; for (int i = 0; i <= lessThanK; ++i) { sumlessthanK = (sumlessthanK + dp[i]); } // A set less than k will be part of // either of the two groups, hence we // multiply by 2 for the same and // remove it from the power set ans = (ans - (2 * sumlessthanK)); return ans;}// Driver codeint main(){ vector<int> arr{ 1, 2, 3, 4 }; int K = 4; // Function call cout << cntWays(arr, K); return 0;} |
Java
import java.util.*;public class Gfg { // Function to count ways public static int cntWays(List<Integer> arr, int K) { int ans = 1; // Dp will be contain all pairs whose // sum is from 0 to K-1 List<Integer> dp = new ArrayList<>(Collections.nCopies(K, 0)); long total = 0; dp.set(0, 1); int lessThanK = K - 1; for (int n : arr) { // Generate pairs for all sum // from 0 to K-1 for (int i = lessThanK - n; i >= 0; i--) { dp.set(i + n, dp.get(i + n) + dp.get(i)); } // Collecting all power sets // i.e 2^n ans *= 2; total += n; } if (total < 2 * K) return 0; // Collect groups whose sum is less // than K long sumlessthanK = 0; for (int i = 0; i <= lessThanK; ++i) { sumlessthanK += dp.get(i); } // A set less than k will be part of // either of the two groups, hence we // multiply by 2 for the same and // remove it from the power set ans -= 2 * sumlessthanK; return ans; } // Driver code public static void main(String[] args) { List<Integer> arr = Arrays.asList(1, 2, 3, 4); int K = 4; // Function call System.out.println(cntWays(arr, K)); }}// This code is contributed by hkdass001. |
Python3
# Python3 code for the above approachfrom typing import List# Function to count waysdef cntWays(arr: List[int], K: int) -> int: ans = 1 # Dp will be contain all pairs whose # sum is from 0 to K-1 dp = [0] * K total = 0 dp[0] = 1 lessThanK = K - 1 for n in arr: # Generate pairs for all sum # from 0 to K-1 for i in range(lessThanK - n, -1, -1): dp[i + n] = (dp[i + n] + dp[i]) # Collecting all power sets # i.e 2^n ans = ans * 2 total += n if total < 2 * K: return 0 # Collect groups whose sum is less # than K sumlessthanK = 0 for i in range(lessThanK + 1): sumlessthanK = (sumlessthanK + dp[i]) # A set less than k will be part of # either of the two groups, hence we # multiply by 2 for the same and # remove it from the power set ans = (ans - (2 * sumlessthanK)) return ans# Driver codeif __name__ == "__main__": arr = [1, 2, 3, 4] K = 4 # Function call print(cntWays(arr, K))# This code is contributed by ik_9 |
Javascript
// Javascript code for the above approach:// Function to count waysfunction cntWays( arr, K){ let ans = 1; // Dp will be contain all pairs whose // sum is from 0 to K-1 let dp=new Array(K).fill(0); let total = 0; dp[0] = 1; let lessThanK = K - 1; for (let n of arr) { // Generate pairs for all sum // from 0 to K-1 for (let i = lessThanK - n; i >= 0; i--) { dp[i + n] = (dp[i + n] + dp[i]); } // Collecting all power sets // i.e 2^n ans = ans * 2; total += n; } if (total < 2 * K) return 0; // Collect groups whose sum is less // than K let sumlessthanK = 0; for (let i = 0; i <= lessThanK; ++i) { sumlessthanK = (sumlessthanK + dp[i]); } // A set less than k will be part of // either of the two groups, hence we // multiply by 2 for the same and // remove it from the power set ans = (ans - (2 * sumlessthanK)); return ans;}// Driver codelet arr=[ 1, 2, 3, 4 ];let K = 4;// Function calldocument.write(cntWays(arr, K)); |
C#
using System;public class GFG{ // Function to count ways public static int cntWays(int[] arr, int K) { int ans = 1; // Dp will be contain all pairs whose // sum is from 0 to K-1 int[] dp = new int[K]; int total = 0; dp[0] = 1; int lessThanK = K - 1; foreach (int n in arr) { // Generate pairs for all sum // from 0 to K-1 for (int i = lessThanK - n; i >= 0; i--) { dp[i + n] = dp[i + n] + dp[i]; } // Collecting all power sets // i.e 2^n ans *= 2; total += n; } if (total < 2 * K) return 0; // Collect groups whose sum is less // than K int sumlessthanK = 0; for (int i = 0; i <= lessThanK; ++i) { sumlessthanK += dp[i]; } // A set less than k will be part of // either of the two groups, hence we // multiply by 2 for the same and // remove it from the power set ans -= 2 * sumlessthanK; return ans; } // Driver code static public void Main (){ int[] arr = {1, 2, 3, 4}; int K = 4; // Function call Console.Write(cntWays(arr, K)); }} |
Output
6
Time Complexity: O(n*K)
Auxiliary Space: O(sum)
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