Find the Sub-array with sum closest to 0

Given an array of both positive and negative numbers, the task is to find out the subarray whose sum is closest to 0.
There can be multiple such subarrays, we need to output just 1 of them.
Examples:

Input : arr[] = {-1, 3, 2, -5, 4}
Output : 1, 3
Subarray from index 1 to 3 has sum closest to 0 i.e.
3 + 2 + -5 = 0

Input : {2, -5, 4, -6, 3} 
Output : 0, 2
2 + -5 + 4 = 1 closest to 0

Asked in : Microsoft

A Naive approach is to consider all subarrays one by one and update indexes of subarray with sum closest to 0.

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// C++ program to find subarray with
// sum closest to 0
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the subarray
pair<int, int> findSubArray(int arr[], int n)
{
  
    int start, end, min_sum = INT_MAX;
  
    // Pick a starting point
    for (int i = 0; i < n; i++) {
  
        // Consider current starting point
        // as a subarray and update minimum
        // sum and subarray indexes
        int curr_sum = arr[i];
        if (min_sum > abs(curr_sum)) {
            min_sum = abs(curr_sum);
            start = i;
            end = i;
        }
  
        // Try all subarrays starting with i
        for (int j = i + 1; j < n; j++) {
            curr_sum = curr_sum + arr[j];
  
            // update minimum sum
            // and subarray indexes
            if (min_sum > abs(curr_sum)) {
                min_sum = abs(curr_sum);
                start = i;
                end = j;
            }
        }
    }
  
    // Return starting and ending indexes
    pair<int, int> p = make_pair(start, end);
    return p;
}
  
// Drivers code
int main()
{
    int arr[] = { 2, -5, 4, -6, -3 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    pair<int, int> point = findSubArray(arr, n);
    cout << "Subarray starting from ";
    cout << point.first << " to " << point.second;
    return 0;
}
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// Java program to find subarray with
// sum closest to 0
  
class GFG
{
  
    static class Pair 
    {
  
        int first, second;
        public Pair(int first, int second)
        {
            this.first = first;
            this.second = second;
        }
  
    }
      
    // Function to find the subarray
    static Pair findSubArray(int arr[], int n)
    {
  
        int start = 0, end = 0, min_sum = Integer.MAX_VALUE;
  
        // Pick a starting point
        for (int i = 0; i < n; i++)
        {
  
            // Consider current starting point
            // as a subarray and update minimum
            // sum and subarray indexes
            int curr_sum = arr[i];
            if (min_sum > Math.abs(curr_sum)) 
            {
                min_sum = Math.abs(curr_sum);
                start = i;
                end = i;
            }
  
            // Try all subarrays starting with i
            for (int j = i + 1; j < n; j++) 
            {
                curr_sum = curr_sum + arr[j];
  
                // update minimum sum
                // and subarray indexes
                if (min_sum > Math.abs(curr_sum)) 
                {
                    min_sum = Math.abs(curr_sum);
                    start = i;
                    end = j;
                }
            }
        }
  
        // Return starting and ending indexes
        Pair p = new Pair(start, end);
        return p;
    }
  
    // Drivers code
    public static void main(String[] args) 
    {
        int arr[] = {2, -5, 4, -6, -3};
        int n = arr.length;
  
        Pair point = findSubArray(arr, n);
        System.out.println("Subarray starting from "
                + point.first + " to " + point.second);
    }
}
  
// This code has been contributed by 29AjayKumar
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# Python 3 program to find subarray with
# sum closest to 0
import sys
  
# Function to find the subarray
def findSubArray(arr, n):
    min_sum = sys.maxsize
  
    # Pick a starting point
    for i in range(n):
          
        # Consider current starting point
        # as a subarray and update minimum
        # sum and subarray indexes
        curr_sum = arr[i]
        if (min_sum > abs(curr_sum)):
            min_sum = abs(curr_sum)
            start = i
            end = i
  
        # Try all subarrays starting with i
        for j in range(i + 1, n, 1):
            curr_sum = curr_sum + arr[j]
  
            # update minimum sum
            # and subarray indexes
            if (min_sum > abs(curr_sum)):
                min_sum = abs(curr_sum)
                start = i
                end = j
  
    # Return starting and ending indexes
    p = [start, end]
    return p
  
# Driver Code
if __name__ == '__main__':
    arr = [2, -5, 4, -6, -3]
    n = len(arr)
  
    point = findSubArray(arr, n)
    print("Subarray starting from ", end = "")
    print(point[0], "to", point[1])
  
# This code is contributed by
# Surendra_Gangwar
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// C# program to find subarray with
// sum closest to 0
using System;
      
class GFG
{
  
    public class Pair 
    {
  
        public int first, second;
        public Pair(int first, int second)
        {
            this.first = first;
            this.second = second;
        }
  
    }
      
    // Function to find the subarray
    static Pair findSubArray(int []arr, int n)
    {
  
        int start = 0, end = 0, min_sum = int.MaxValue;
  
        // Pick a starting point
        for (int i = 0; i < n; i++)
        {
  
            // Consider current starting point
            // as a subarray and update minimum
            // sum and subarray indexes
            int curr_sum = arr[i];
            if (min_sum > Math.Abs(curr_sum)) 
            {
                min_sum = Math.Abs(curr_sum);
                start = i;
                end = i;
            }
  
            // Try all subarrays starting with i
            for (int j = i + 1; j < n; j++) 
            {
                curr_sum = curr_sum + arr[j];
  
                // update minimum sum
                // and subarray indexes
                if (min_sum > Math.Abs(curr_sum)) 
                {
                    min_sum = Math.Abs(curr_sum);
                    start = i;
                    end = j;
                }
            }
        }
  
        // Return starting and ending indexes
        Pair p = new Pair(start, end);
        return p;
    }
  
    // Drivers code
    public static void Main(String[] args) 
    {
        int []arr = {2, -5, 4, -6, -3};
        int n = arr.Length;
  
        Pair point = findSubArray(arr, n);
        Console.WriteLine("Subarray starting from "
                + point.first + " to " + point.second);
    }
}
  
// This code is contributed by Princi Singh
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Output:
Subarray starting from 0 to 2

Time Complexity: O(n2)

An Efficient method is to perform following steps:-

  1. Maintain a Prefix sum array . Also maintain indexes in the prefix sum array.
  2. Sort the prefix sum array on the basis of sum.
  3. Find the two elements in a prefix sum array with minimum difference.
    i.e.  Find min(pre_sum[i] - pre_sum[i-1]) 
  4. Return indexes of pre_sum with minimum difference.
  5. Subarray with (lower_index+1, upper_index) will have the sum closest to 0.
  6. Taking lower_index+1 because on subtracting value at lower_index we get the sum closest to 0. That’s why lower_index need not to be included.
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// C++ program to find subarray with sum 
// closest to 0
#include <bits/stdc++.h>
using namespace std;
  
struct prefix {
    int sum;
    int index;
};
  
// Sort on the basis of sum
bool comparison(prefix a, prefix b)
{
    return a.sum < b.sum;
}
  
// Returns subarray with sum closest to 0. 
pair<int, int> findSubArray(int arr[], int n)
{
    int start, end, min_diff = INT_MAX;
  
    prefix pre_sum[n + 1];
  
    // To consider the case of subarray starting
    // from beginning of the array
    pre_sum[0].sum = 0;
    pre_sum[0].index = -1;
  
    // Store prefix sum with index
    for (int i = 1; i <= n; i++) {
        pre_sum[i].sum = pre_sum[i-1].sum + arr[i-1];
        pre_sum[i].index = i - 1;
    }
  
    // Sort on the basis of sum
    sort(pre_sum, pre_sum + (n + 1), comparison);
  
    // Find two consecutive elements with minimum difference
    for (int i = 1; i <= n; i++) {
        int diff = pre_sum[i].sum - pre_sum[i-1].sum;
  
        // Update minimum difference
        // and starting and ending indexes
        if (min_diff > diff) {
            min_diff = diff;
            start = pre_sum[i-1].index;
            end = pre_sum[i].index;
        }
    }
  
    // Return starting and ending indexes
    pair<int, int> p = make_pair(start + 1, end);
    return p;
}
  
// Drivers code
int main()
{
    int arr[] = { 2, 3, -4, -1, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    pair<int, int> point = findSubArray(arr, n);
    cout << "Subarray starting from ";
    cout << point.first << " to " << point.second;
  
    return 0;
}
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# Python3 program to find subarray 
# with sum closest to 0 
class prefix:
      
    def __init__(self, sum, index):
        self.sum = sum
        self.index = index
  
# Returns subarray with sum closest to 0. 
def findSubArray(arr, n): 
  
    start, end, min_diff = None, None, float('inf'
  
    pre_sum = [None] * (n + 1)
  
    # To consider the case of subarray 
    # starting from beginning of the array 
    pre_sum[0] = prefix(0, -1)
  
    # Store prefix sum with index 
    for i in range(1, n + 1): 
        pre_sum[i] = prefix(pre_sum[i - 1].sum + 
                                arr[i - 1], i - 1
  
    # Sort on the basis of sum 
    pre_sum.sort(key = lambda x: x.sum)
  
    # Find two consecutive elements 
    # with minimum difference 
    for i in range(1, n + 1): 
        diff = pre_sum[i].sum - pre_sum[i - 1].sum
  
        # Update minimum difference 
        # and starting and ending indexes 
        if min_diff > diff: 
            min_diff = diff 
            start = pre_sum[i - 1].index 
            end = pre_sum[i].index 
          
    # Return starting and ending indexes 
    return (start + 1, end) 
  
# Driver code 
if __name__ == "__main__"
  
    arr = [2, 3, -4, -1, 6
    n = len(arr) 
  
    point = findSubArray(arr, n) 
    print("Subarray starting from"
           point[0], "to", point[1]) 
  
# This code is contributed by Rituraj Jain
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Output:

Subarray starting from 0 to 3

Time Complexity: O(n log n)

Reference:
https://www.careercup.com/question?id=14583859

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