# 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 + arr + arr = 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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Improved By : jit_t, vt_m, AnkitRai01