Check if all K-length subset sums of first array greater than that of the second array

Given two arrays A[] and B[] of size N and an integer K, the task is to check if all possible subset-sums of subsets of size K of the array A[] are greater than that of the array B[] or not. If found to be true, then print “YES”. Otherwise, print “NO”.

Examples:

Input: A[] = {12, 11, 10, 13}, B[] = {7, 10, 6, 2}, K = 3
Output: YES
Explanation: All possible subset sum of size K(= 3) in A[] are {33, 36, 35, 34}.
All possible subset sum of size K(= 3) in B[] are {23, 19, 15, 18}.
Since all subset-sums of size K in the array A[] is greater than all possible subset-sums of size K in the array B[], the required output is “YES”.

Input : A[] = {5, 3, 3, 4, 4, 6, 1}, B[] = {9, 10, 9, 8, 4, 6, 2}, K = 6
Output : NO

Naive Approach: The simplest approach to solve this problem is to generate all possible subsets of size K from the arrays A[] and B[] and calculate their respective sums. Check if all the sums obtained in array A[] exceeds that of array B[] or not. If found to be true, then print “YES”. Otherwise, print “NO”.

Time Complexity: O(K × N^2K)
Auxiliary Space: O(N^K)

Efficient Approach: To optimize the above approach, the idea is based on the fact that if the smallest subset-sum of size K of the array A[] is greater than the largest subset-sum of size K of the array B[], then print “YES”. Otherwise, print “NO”. Follow the steps below to solve the problem:

• Sort the array A[] in ascending order.
• Sort the array B[] in descending order.
• Traverse the array and check if the sum of the first K elements of the array A[] is greater than the sum of the first K elements of the array B[] or not. If found to be true, then print “YES”. Otherwise, print “NO”.

Below is the implementation of the above approach;

C++

// C++ program to implement
// the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to check all subset-sums
// of K-length subsets in A[] is greater
// that in the array B[] or not
bool checkSubsetSum(int A[], int B[], int N,
                    int K)
{
    // Sort the array in
    // ascending order
    sort(A, A + N);

    // Sort the array in
    // descending order
    sort(B, B + N,
         greater<int>());

    // Stores sum of first K
    // elements of A[]
    int sum1 = 0;

    // Stores sum of first K
    // elements of B[]
    int sum2 = 0;

    // Traverse both the arrays
    for (int i = 0; i < K; i++) {

        // Update sum1
        sum1 += A[i];

        // Update sum2
        sum2 += B[i];
    }

    // If sum1 exceeds sum2
    if (sum1 > sum2) {
        return true;
    }

    return false;
}

// Driver Code
int main()
{
    int A[] = { 12, 11, 10, 13 };
    int B[] = { 7, 10, 6, 2 };

    int N = sizeof(A) / sizeof(A[0]);

    int K = 3;
    if (checkSubsetSum(A, B, N, K)) {
        cout << "YES";
    }
    else {
        cout << "NO";
    }

    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;

class GFG{

// Function reverses the elements of the array
static void reverse(int myArray[])
{
    Collections.reverse(Arrays.asList(myArray));
}

// Function to check all subset-sums
// of K-length subsets in A[] is greater
// that in the array B[] or not
static boolean checkSubsetSum(int A[], int B[],
                              int N, int K)
{

    // Sort the array in
    // ascending order
    Arrays.sort(A);

    // Sort the array in
    // descending order
    Arrays.sort(B);
    reverse(B);

    // Stores sum of first K
    // elements of A[]
    int sum1 = 0;

    // Stores sum of first K
    // elements of B[]
    int sum2 = 0;

    // Traverse both the arrays
    for(int i = 0; i < K; i++)
    {

        // Update sum1
        sum1 += A[i];

        // Update sum2
        sum2 += B[i];
    }

    // If sum1 exceeds sum2
    if (sum1 > sum2)
    {
        return true;
    }
    return false;
}

// Driver Code
public static void main(String[] args)
{
    int A[] = { 12, 11, 10, 13 };
    int B[] = { 7, 10, 6, 2 };

    int N = A.length;
    int K = 3;

    if (checkSubsetSum(A, B, N, K))
    {
        System.out.print("YES");
    }
    else
    {
        System.out.print("NO");
    }
}
}

// This code is contributed by susmitakundugoaldanga

Python3

# Python3 program to implement
# the above approach

# Function to check all subset-sums
# of K-length subsets in A[] is greater
# that in the array B[] or not
def checkSubsetSum(A, B, N, K):

    # Sort the array in
    # ascending order
    A.sort()

    # Sort the array in
    # descending order
    B.sort(reverse = True)

    # Stores sum of first K
    # elements of A[]
    sum1 = 0

    # Stores sum of first K
    # elements of B[]
    sum2 = 0

    # Traverse both the arrays
    for i in range(K):

        # Update sum1
        sum1 += A[i]

        # Update sum2
        sum2 += B[i]

    # If sum1 exceeds sum2
    if (sum1 > sum2):
        return True

    return False

# Driver Code
A = [ 12, 11, 10, 13 ]
B = [ 7, 10, 6, 2]

N = len(A)
K = 3

if (checkSubsetSum(A, B, N, K)):
    print("Yes")
else:
    print("No")

# This code is contributed by avanitrachhadiya2155

C#

// C# program to implement
// the above approach
using System;

public class GFG{

// Function reverses the elements of the array
static void reverse(int []myArray)
{
    Array.Sort(myArray);
    Array.Reverse(myArray);
}

// Function to check all subset-sums
// of K-length subsets in []A is greater
// that  in the array []B or not
static bool checkSubsetSum(int []A, int []B,
                              int N, int K)
{

    // Sort the array in
    // ascending order
    Array.Sort(A);

    // Sort the array in
    // descending order
    Array.Sort(B);
    reverse(B);

    // Stores sum of first K
    // elements of []A
    int sum1 = 0;

    // Stores sum of first K
    // elements of []B
    int sum2 = 0;

    // Traverse both the arrays
    for(int i = 0; i < K; i++)
    {

        // Update sum1
        sum1 += A[i];

        // Update sum2
        sum2 += B[i];
    }

    // If sum1 exceeds sum2
    if (sum1 > sum2)
    {
        return true;
    }
    return false;
}

// Driver Code
public static void Main(String[] args)
{
    int []A = { 12, 11, 10, 13 };
    int []B = { 7, 10, 6, 2 };

    int N = A.Length;
    int K = 3;

    if (checkSubsetSum(A, B, N, K))
    {
        Console.Write("YES");
    }
    else
    {
        Console.Write("NO");
    }
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// JavaScript program to implement
// the above approach

// Function to check all subset-sums
// of K-length subsets in A[] is greater
// that in the array B[] or not
function checkSubsetSum(A, B, N, K)
{
    // Sort the array in
    // ascending order
    A.sort((a,b)=> a-b);

    // Sort the array in
    // descending order
    B.sort((a,b)=>b-a);

    // Stores sum of first K
    // elements of A[]
    var sum1 = 0;

    // Stores sum of first K
    // elements of B[]
    var sum2 = 0;

    // Traverse both the arrays
    for (var i = 0; i < K; i++) {

        // Update sum1
        sum1 += A[i];

        // Update sum2
        sum2 += B[i];
    }

    // If sum1 exceeds sum2
    if (sum1 > sum2) {
        return true;
    }

    return false;
}

// Driver Code
var A = [12, 11, 10, 13];
var B = [7, 10, 6, 2];
var N = A.length;
var K = 3;
if (checkSubsetSum(A, B, N, K)) {
    document.write( "YES");
}
else {
    document.write( "NO");
}

</script>

Output:

YES

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