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

Javascript

 ``

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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