Subset with sum closest to zero

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 : 0
arr[0] + arr[1] + arr[3] = 0

Input : arr[] = {1, 1, 1, 1}
Output : 1

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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<

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..

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.

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