Sum of subsets nearest to K possible from two given arrays
Last Updated :
13 May, 2021
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Given two arrays A[] and B[] consisting of N and M integers respectively, and an integer K, the task is to find the sum nearest to K possible by selecting exactly one element from the array A[] and an element from the array B[], at most twice.
Examples:
Input: A[] = {1, 7}, B[] = {3, 4}, K = 10
Output: 10
Explanation:
Sum obtained by selecting A[0] and A[1] = 3 + 7 = 10, which is closest to the value K(= 10).Input: A[] = {2, 3}, B[] = {4, 5, 30}, K = 18
Output: 17
Approach: The given problem can be solved by using recursion, by finding the sum of elements of the subsets of the array B[] having sum closest to (K - A[i]) for each array element A[i]. Follow the steps below to solve the problem:
- Initialize two variables, say mini as INT_MAX and ans as INT_MAX to store the minimum absolute difference and the value closest to K.
- Define a recursive function, say findClosest(arr, i, currSum) to find the subset-sum of the array closest to K, where i is the index in the array B[] and currSum stores the sum of the subset.
- If the value of i is at least M, then return from the function.
- If the absolute value of (currSum - K) is less than mini, then update the value of mini as abs(currSum - K) and update the value of ans as currSum.
- If the absolute value of (currSum - K) is equal to mini then, update the value of ans as the minimum of ans and currSum.
- Call the recursive function excluding the element B[i] as findClosest(i + 1, currSum).
- Call the recursive function including the element B[i] once as findClosest(i + 1, currSum + B[i]).
- Call the recursive function including the element B[i] twice as findClosest(i + 1, currSum + 2*B[i]).
- Traverse the given array A[] and for every element call the function findClosest(0, A[i]).
- After completing the above steps, print the value of ans as the resultant sum.
Below is the implementation of the above approach:
// C++ program of the above approach
#include <bits/stdc++.h>
using namespace std;
// Stores the sum closest to K
int ans = INT_MAX;
// Stores the minimum absolute difference
int mini = INT_MAX;
// Function to choose the elements
// from the array B[]
void findClosestTarget(int i, int curr,
int B[], int M,
int K)
{
// If absolute difference is less
// then minimum value
if (abs(curr - K) < mini) {
// Update the minimum value
mini = abs(curr - K);
// Update the value of ans
ans = curr;
}
// If absolute difference between
// curr and K is equal to minimum
if (abs(curr - K) == mini) {
// Update the value of ans
ans = min(ans, curr);
}
// If i is greater than M - 1
if (i >= M)
return;
// Includes the element B[i] once
findClosestTarget(i + 1, curr + B[i],
B, M, K);
// Includes the element B[i] twice
findClosestTarget(i + 1, curr + 2 * B[i],
B, M, K);
// Excludes the element B[i]
findClosestTarget(i + 1, curr, B, M, K);
}
// Function to find a subset sum
// whose sum is closest to K
int findClosest(int A[], int B[],
int N, int M, int K)
{
// Traverse the array A[]
for (int i = 0; i < N; i++) {
// Function Call
findClosestTarget(0, A[i], B,
M, K);
}
// Return the ans
return ans;
}
// Driver Code
int main()
{
// Input
int A[] = { 2, 3 };
int B[] = { 4, 5, 30 };
int N = sizeof(A) / sizeof(A[0]);
int M = sizeof(B) / sizeof(B[0]);
int K = 18;
// Function Call
cout << findClosest(A, B, N, M, K);
return 0;
}
// java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
public class GFG {
// Stores the sum closest to K
static int ans = Integer.MAX_VALUE;
// Stores the minimum absolute difference
static int mini = Integer.MAX_VALUE;
// Function to choose the elements
// from the array B[]
static void findClosestTarget(int i, int curr, int B[],
int M, int K)
{
// If absolute difference is less
// then minimum value
if (Math.abs(curr - K) < mini) {
// Update the minimum value
mini = Math.abs(curr - K);
// Update the value of ans
ans = curr;
}
// If absolute difference between
// curr and K is equal to minimum
if (Math.abs(curr - K) == mini) {
// Update the value of ans
ans = Math.min(ans, curr);
}
// If i is greater than M - 1
if (i >= M)
return;
// Includes the element B[i] once
findClosestTarget(i + 1, curr + B[i], B, M, K);
// Includes the element B[i] twice
findClosestTarget(i + 1, curr + 2 * B[i], B, M, K);
// Excludes the element B[i]
findClosestTarget(i + 1, curr, B, M, K);
}
// Function to find a subset sum
// whose sum is closest to K
static int findClosest(int A[], int B[], int N, int M,
int K)
{
// Traverse the array A[]
for (int i = 0; i < N; i++) {
// Function Call
findClosestTarget(0, A[i], B, M, K);
}
// Return the ans
return ans;
}
// Driver Code
public static void main(String[] args)
{
// Input
int A[] = { 2, 3 };
int B[] = { 4, 5, 30 };
int N = A.length;
int M = B.length;
int K = 18;
// Function Call
System.out.print(findClosest(A, B, N, M, K));
}
}
// This code is contributed by Kingash.
# Python3 program of the above approach
# Stores the sum closest to K
ans = 10**8
# Stores the minimum absolute difference
mini = 10**8
# Function to choose the elements
# from the array B[]
def findClosestTarget(i, curr, B, M, K):
global ans, mini
# If absolute difference is less
# then minimum value
if (abs(curr - K) < mini):
# Update the minimum value
mini = abs(curr - K)
# Update the value of ans
ans = curr
# If absolute difference between
# curr and K is equal to minimum
if (abs(curr - K) == mini):
# Update the value of ans
ans = min(ans, curr)
# If i is greater than M - 1
if (i >= M):
return
# Includes the element B[i] once
findClosestTarget(i + 1, curr + B[i], B, M, K)
# Includes the element B[i] twice
findClosestTarget(i + 1, curr + 2 * B[i], B, M, K)
# Excludes the element B[i]
findClosestTarget(i + 1, curr, B, M, K)
# Function to find a subset sum
# whose sum is closest to K
def findClosest(A, B, N, M, K):
# Traverse the array A[]
for i in range(N):
# Function Call
findClosestTarget(0, A[i], B, M, K)
# Return the ans
return ans
# Driver Code
if __name__ == '__main__':
# Input
A = [2, 3]
B = [4, 5, 30]
N = len(A)
M = len(B)
K = 18
# Function Call
print (findClosest(A, B, N, M, K))
# This code is contributed by mohit kumar 29.
// C# program of the above approach
using System;
class GFG
{
// Stores the sum closest to K
static int ans = Int32.MaxValue;
// Stores the minimum absolute difference
static int mini = Int32.MaxValue;
// Function to choose the elements
// from the array B[]
static void findClosestTarget(int i, int curr, int[] B,
int M, int K)
{
// If absolute difference is less
// then minimum value
if (Math.Abs(curr - K) < mini) {
// Update the minimum value
mini = Math.Abs(curr - K);
// Update the value of ans
ans = curr;
}
// If absolute difference between
// curr and K is equal to minimum
if (Math.Abs(curr - K) == mini) {
// Update the value of ans
ans = Math.Min(ans, curr);
}
// If i is greater than M - 1
if (i >= M)
return;
// Includes the element B[i] once
findClosestTarget(i + 1, curr + B[i], B, M, K);
// Includes the element B[i] twice
findClosestTarget(i + 1, curr + 2 * B[i], B, M, K);
// Excludes the element B[i]
findClosestTarget(i + 1, curr, B, M, K);
}
// Function to find a subset sum
// whose sum is closest to K
static int findClosest(int[] A, int[] B, int N, int M,
int K)
{
// Traverse the array A[]
for (int i = 0; i < N; i++) {
// Function Call
findClosestTarget(0, A[i], B, M, K);
}
// Return the ans
return ans;
}
// Driver Code
public static void Main()
{
// Input
int[] A = { 2, 3 };
int[] B = { 4, 5, 30 };
int N = A.Length;
int M = B.Length;
int K = 18;
// Function Call
Console.WriteLine(findClosest(A, B, N, M, K));
}
}
// This code is contributed by ukasp.
<script>
// Javascript program for the above approach
// Stores the sum closest to K
let ans = Number.MAX_SAFE_INTEGER
// Stores the minimum absolute difference
let mini = Number.MAX_SAFE_INTEGER
// Function to choose the elements
// from the array B[]
function findClosestTarget(i, curr, B,
M, K) {
// If absolute difference is less
// then minimum value
if (Math.abs(curr - K) < mini) {
// Update the minimum value
mini = Math.abs(curr - K);
// Update the value of ans
ans = curr;
}
// If absolute difference between
// curr and K is equal to minimum
if (Math.abs(curr - K) == mini) {
// Update the value of ans
ans = Math.min(ans, curr);
}
// If i is greater than M - 1
if (i >= M)
return;
// Includes the element B[i] once
findClosestTarget(i + 1, curr + B[i], B, M, K);
// Includes the element B[i] twice
findClosestTarget(i + 1, curr + 2 * B[i], B, M, K);
// Excludes the element B[i]
findClosestTarget(i + 1, curr, B, M, K);
}
// Function to find a subset sum
// whose sum is closest to K
function findClosest(A, B, N, M, K) {
// Traverse the array A[]
for (let i = 0; i < N; i++) {
// Function Call
findClosestTarget(0, A[i], B, M, K);
}
// Return the ans
return ans;
}
// Driver Code
// Input
let A = [2, 3];
let B = [4, 5, 30];
let N = A.length;
let M = B.length;
let K = 18;
// Function Call
document.write(findClosest(A, B, N, M, K));
//This code is contributed by Hritik.
</script>
Output:
17
Time Complexity: O(N * 3M)
Auxiliary Space: O(1)
