Open In App

Subset with sum closest to zero

Improve
Improve
Like Article
Like
Save
Share
Report

Given an array ‘arr’ consisting of integers, the task is to find the non-empty subset such that its sum is closest to zero i.e. absolute difference between zero and the sum is minimum.
Examples: 
 

Input : arr[] = {2, 2, 2, -4} 
Output :
arr[0] + arr[1] + arr[3] = 0 
That’s why answer is zero.
Input : arr[] = {1, 1, 1, 1} 
Output :
 

 

One simple approach is to generate all possible subsets recursively and find the one with the sum closest to zero. Time complexity of this approach will be O(2^n).
A better approach will be using Dynamic programming in Pseudo Polynomial Time Complexity
Let’s suppose sum of all the elements we have selected upto index ‘i-1’ is ‘S’. So, starting from index ‘i’, we have to find a subset with sum closest to -S. 
Let’s define dp[i][S] first. It means sum of the subset of the subarray{i, N-1} of array ‘arr’ with sum closest to ‘-S’. 
Now, we can include ‘i’ in the current sum or leave it. Thus we, have two possible paths to take. If we include ‘i’, current sum will be updated as S+arr[i] and we will solve for index ‘i+1’ i.e. dp[i+1][S+arr[i]] else we will solve for index ‘i+1’ directly. Thus, the required recurrence relation will be.
 

dp[i][s] = RetClose(arr[i]+dp[i][s+arr[i]], dp[i+1][s], -s); 
where RetClose(a, b, c) returns a if |a-c|<|b-c| else it returns b 
 

Below is the implementation of the above approach:
 

C++




#include <bits/stdc++.h>
using namespace std;
 
#define arrSize 51
#define maxSum 201
#define MAX 100
#define inf 999999
  
// Variable to store states of dp
int dp[arrSize][maxSum];
bool visit[arrSize][maxSum];
  
// Function to return the number closer to integer s
int RetClose(int a, int b, int s)
{
    if (abs(a - s) < abs(b - s))
        return a;
    else
        return b;
}
  
// To find the sum closest to zero
// Since sum can be negative, we will add MAX
// to it to make it positive
int MinDiff(int i, int sum, int arr[], int n)
{
  
    // Base cases
    if (i == n)
        return 0;
    // Checks if a state is already solved
    if (visit[i][sum + MAX])
        return dp[i][sum + MAX];
    visit[i][sum + MAX] = 1;
  
    // Recurrence relation
    dp[i][sum + MAX] =  RetClose(arr[i] +
                        MinDiff(i + 1, sum + arr[i], arr, n),
                        MinDiff(i + 1, sum, arr, n), -1 * sum);
  
    // Returning the value
    return dp[i][sum + MAX];
}
 
// Function to calculate the closest sum value
void FindClose(int arr[],int n)
{
    int ans=inf;
 
    // Calculate the Closest value for every
    // subarray arr[i-1:n]
    for (int i = 1; i <= n; i++)
        ans = RetClose(arr[i - 1] +
                MinDiff(i, arr[i - 1], arr, n), ans, 0);
 
    cout<<ans<<endl;
}
  
// Driver function
int main()
{
    // Input array
    int arr[] = { 25, -9, -10, -4, -7, -33 };
    int n = sizeof(arr) / sizeof(int);
    
    FindClose(arr,n);
    return 0;
}


Java




// Java Program for above approach
import java.io.*;
 
class GFG
{
     
static int arrSize = 51;
static int maxSum = 201;
static int MAX = 100;
static int inf = 999999;
 
// Variable to store states of dp
static int dp[][] = new int [arrSize][maxSum];
static int visit[][] = new int [arrSize][maxSum];
 
// Function to return the number
// closer to integer s
static int RetClose(int a, int b, int s)
{
    if (Math.abs(a - s) < Math.abs(b - s))
        return a;
    else
        return b;
}
 
// To find the sum closest to zero
// Since sum can be negative, we will add MAX
// to it to make it positive
static int MinDiff(int i, int sum,
                   int arr[], int n)
{
 
    // Base cases
    if (i == n)
        return 0;
         
    // Checks if a state is already solved
    if (visit[i][sum + MAX] > 0 )
        return dp[i][sum + MAX];
    visit[i][sum + MAX] = 1;
 
    // Recurrence relation
    dp[i][sum + MAX] = RetClose(arr[i] +
                        MinDiff(i + 1, sum + arr[i], arr, n),
                        MinDiff(i + 1, sum, arr, n), -1 * sum);
 
    // Returning the value
    return dp[i][sum + MAX];
}
 
// Function to calculate the closest sum value
static void FindClose(int arr[], int n)
{
    int ans = inf;
 
    // Calculate the Closest value for every
    // subarray arr[i-1:n]
    for (int i = 1; i <= n; i++)
        ans = RetClose(arr[i - 1] +
            MinDiff(i, arr[i - 1],
                       arr, n), ans, 0);
 
        System.out.println(ans);
}
 
// Driver Code
public static void main (String[] args)
{
 
    // Input array
    int arr[] = { 25, -9, -10, -4, -7, -33 };
    int n = arr.length;
     
    FindClose(arr,n);
}
}
 
// This code is contributed by ajit_00023@


Python3




# Python3 Code for above implementation
import numpy as np
 
arrSize = 51
maxSum = 201
MAX = 100
inf = 999999
 
# Variable to store states of dp
dp = np.zeros((arrSize,maxSum));
visit = np.zeros((arrSize,maxSum));
 
# Function to return the number closer to integer s
def RetClose(a, b, s) :
 
    if (abs(a - s) < abs(b - s)) :
        return a;
    else :
        return b;
 
 
# To find the sum closest to zero
# Since sum can be negative, we will add MAX
# to it to make it positive
def MinDiff(i, sum, arr, n) :
 
    # Base cases
    if (i == n) :
        return 0;
         
    # Checks if a state is already solved
    if (visit[i][sum + MAX]) :
        return dp[i][sum + MAX];
         
    visit[i][sum + MAX] = 1;
 
    # Recurrence relation
    dp[i][sum + MAX] = RetClose(arr[i] +
                        MinDiff(i + 1, sum + arr[i], arr, n),
                        MinDiff(i + 1, sum, arr, n), -1 * sum);
 
    # Returning the value
    return dp[i][sum + MAX];
 
 
# Function to calculate the closest sum value
def FindClose(arr,n) :
 
    ans=inf;
 
    # Calculate the Closest value for every
    # subarray arr[i-1:n]
    for i in range(1, n + 1) :
        ans = RetClose(arr[i - 1] +
                MinDiff(i, arr[i - 1], arr, n), ans, 0);
 
    print(ans);
 
 
# Driver function
if __name__ == "__main__" :
 
    # Input array
    arr = [ 25, -9, -10, -4, -7, -33 ];
    n = len(arr);
     
    FindClose(arr,n);
     
    # This code is contributed by AnkitRai01


C#




// C# Program for above approach
using System;
 
class GFG
{
     
static int arrSize = 51;
static int maxSum = 201;
static int MAX = 100;
static int inf = 999999;
 
// Variable to store states of dp
static int [,]dp = new int [arrSize,maxSum];
static int [,]visit = new int [arrSize,maxSum];
 
// Function to return the number
// closer to integer s
static int RetClose(int a, int b, int s)
{
    if (Math.Abs(a - s) < Math.Abs(b - s))
        return a;
    else
        return b;
}
 
// To find the sum closest to zero
// Since sum can be negative, we will add MAX
// to it to make it positive
static int MinDiff(int i, int sum,
                int []arr, int n)
{
 
    // Base cases
    if (i == n)
        return 0;
         
    // Checks if a state is already solved
    if (visit[i,sum + MAX] > 0 )
        return dp[i,sum + MAX];
    visit[i,sum + MAX] = 1;
 
    // Recurrence relation
    dp[i,sum + MAX] = RetClose(arr[i] +
                        MinDiff(i + 1, sum + arr[i], arr, n),
                        MinDiff(i + 1, sum, arr, n), -1 * sum);
 
    // Returning the value
    return dp[i,sum + MAX];
}
 
// Function to calculate the closest sum value
static void FindClose(int []arr, int n)
{
    int ans = inf;
 
    // Calculate the Closest value for every
    // subarray arr[i-1:n]
    for (int i = 1; i <= n; i++)
        ans = RetClose(arr[i - 1] +
            MinDiff(i, arr[i - 1],
                    arr, n), ans, 0);
 
        Console.WriteLine(ans);
}
 
// Driver Code
public static void Main ()
{
 
    // Input array
    int []arr = { 25, -9, -10, -4, -7, -33 };
    int n = arr.Length;
     
    FindClose(arr,n);
}
}
 
// This code is contributed by  anuj_67..


Javascript




<script>
    // Javascript Program for above approach
     
    let arrSize = 51;
    let maxSum = 201;
    let MAX = 100;
    let inf = 999999;
 
    // Variable to store states of dp
    let dp = new Array(arrSize);
    let visit = new Array(arrSize);
     
    for(let i = 0; i < arrSize; i++)
    {
        dp[i] = new Array(maxSum);
        visit[i] = new Array(maxSum);
        for(let j = 0; j < maxSum; j++)
        {
            dp[i][j] = 0;
            visit[i][j] = 0;
        }
    }
     
 
    // Function to return the number
    // closer to integer s
    function RetClose(a, b, s)
    {
        if (Math.abs(a - s) < Math.abs(b - s))
            return a;
        else
            return b;
    }
 
    // To find the sum closest to zero
    // Since sum can be negative, we will add MAX
    // to it to make it positive
    function MinDiff(i, sum, arr, n)
    {
 
        // Base cases
        if (i == n)
            return 0;
 
        // Checks if a state is already solved
        if (visit[i][sum + MAX] > 0 )
            return dp[i][sum + MAX];
        visit[i][sum + MAX] = 1;
 
        // Recurrence relation
        dp[i][sum + MAX] = RetClose(arr[i] +
                            MinDiff(i + 1, sum + arr[i], arr, n),
                            MinDiff(i + 1, sum, arr, n), -1 * sum);
 
        // Returning the value
        return dp[i][sum + MAX];
    }
 
    // Function to calculate the closest sum value
    function FindClose(arr, n)
    {
        let ans = inf;
 
        // Calculate the Closest value for every
        // subarray arr[i-1:n]
        for (let i = 1; i <= n; i++)
            ans = RetClose(arr[i - 1] +
                MinDiff(i, arr[i - 1],
                           arr, n), ans, 0);
 
            document.write(ans);
    }
     
    // Input array
    let arr = [ 25, -9, -10, -4, -7, -33 ];
    let n = arr.length;
       
    FindClose(arr,n);
 
</script>


Output: 

-1

 

Time complexity: O(N*S), where N is the number of elements in the array and S is the sum of all the numbers in the array.
 



Last Updated : 11 May, 2021
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads