Subarray whose sum is closest to K

Given an array of positive and negative integers and an integer K. The task is to find the subarray which has its sum closest to k. In case of multiple answers, print anyone. 

Note: Closest here means abs(sum-k) should be minimal. 

Examples:

Input: a[] = { -5, 12, -3, 4, -15, 6, 1 }, K = 2 
Output: 1 
The subarray {-3, 4} or {1} has sum = 1 which is the closest to K.

Input: a[] = { 2, 2, -1, 5, -3, -2 }, K = 7 
Output: 6 
Here the output can be 6 or 8 
The subarray {2, 2, -1, 5} gives sum as 8 which has abs(8-7) = 1 which is same as that of the subarray {2, -1, 5} which has abs(6-7) = 1. 

A naive approach is to check for all possible subarray sum using prefix sum. The complexity in that case will be O(N²).

An efficient solution will be to use C++ STL set and binary search to solve the following problem. Follow the below algorithm to solve the above problem. 

• Initially insert the first element in the set container.
• Initialize the answer sum as first element and difference as abs(A₀-k).
• Iterate for all array elements from 1 to N and keep adding the elements to prefix sum at each step to the set container.
• At every iteration, since the prefix sum is already there, we just need to subtract the sum of some elements from beginning to get the sum of any subarray. The greedy way will be to subtract the sum of the subarray which takes the sum closest to K.
• Using binary search (lower_bound() function can be used) find the sum of subarray from beginning which is closest to (prefix-k) as the subtraction of that number from prefix sum will give the subarray sum which is closest to K till that iteration.
• Also check for the index before which lower_bound() returns, since the sum can either be greater or lesser than K.
• If the lower_bound returns no such element, then the current prefix sum is compared and updated if it was lesser than the previous computed sum.

Below is the implementation of the above approach. 

C++

// C++ program to find the
// sum of subarray whose sum is
// closest to K
#include <bits/stdc++.h>
using namespace std;

// Function to find the sum of subarray
// whose sum is closest to K
int closestSubarraySumToK(int a[], int n, int k)
{

    // Declare a set
    set<int> s;

    // initially consider the
    // first subarray as the first
    // element in the array
    int presum = a[0];

    // insert
    s.insert(a[0]);

    // Initially let this difference
    // be the minimum
    int mini = abs(a[0] - k);

    // let this be the sum
    // of the subarray
    // to be searched initially
    int sum = presum;

    // iterate for all the array elements
    for (int i = 1; i < n; i++) {

        // calculate the prefix sum
        presum += a[i];

        // find the closest subarray
        // sum to by using lower_bound
        auto it = s.lower_bound(presum - k);

        // if it is the first element
        // in the set
        if (it == s.begin()) {

            // get the prefix sum till start
            // of the subarray
            int diff = *it;

            // if the subarray sum is closest to K
            // than the previous one
            if (abs((presum - diff) - k) < mini) {

                // update the minimal difference
                mini = abs((presum - diff) - k);

                // update the sum
                sum = presum - diff;
            }

            if(abs(presum - k) < mini){
              // update the minimal difference
                  mini = abs((presum - diff) - k);

                  // update the sum
                  sum = presum - diff;
              }
        }

        // if the difference is
        // present in between
        else if (it != s.end()) {

            // get the prefix sum till start
            // of the subarray
            int diff = *it;

            // if the subarray sum is closest to K
            // than the previous one
            if (abs((presum - diff) - k) < mini) {

                // update the minimal difference
                mini = abs((presum - diff) - k);

                // update the sum
                sum = presum - diff;
            }

            // also check for the one before that
            // since the sum can be greater than
            // or less than K also
            it--;

            // get the prefix sum till start
            // of the subarray
            diff = *it;

            // if the subarray sum is closest to K
            // than the previous one
            if (abs((presum - diff) - k) < mini) {

                // update the minimal difference
                mini = abs((presum - diff) - k);

                // update the sum
                sum = presum - diff;
            }
        }

        // if there exists no such prefix sum
        // then the current prefix sum is
        // checked and updated
        else {

            // if the subarray sum is closest to K
            // than the previous one
            if (abs(presum - k) < mini) {

                // update the minimal difference
                mini = abs(presum - k);

                // update the sum
                sum = presum;
            }
        }

        // insert the current prefix sum
        s.insert(presum);
    }

    return sum;
}

// Driver Code
int main()
{
    int a[] = { -5, 12, -3, 4, -15, 6, 1 };
    int n = sizeof(a) / sizeof(a[0]);
    int k = 2;

    cout << closestSubarraySumToK(a, n, k);
    return 0;
}

Java

// Java program to find the
// sum of subarray whose sum is
// closest to K
import java.util.*;

class GFG{

  // Function to find the sum of subarray
  // whose sum is closest to K
  static int closestSubarraySumToK(int a[], int n, int k)
  {

    // Declare a set
    TreeSet<Integer> s = new TreeSet<>();

    // initially consider the
    // first subarray as the first
    // element in the array
    int presum = a[0];

    // insert
    s.add(a[0]);

    // Initially let this difference
    // be the minimum
    int mini = Math.abs(a[0] - k);

    // let this be the sum
    // of the subarray
    // to be searched initially
    int sum = presum;

    // iterate for all the array elements
    for (int i = 1; i < n; i++) {

      // calculate the prefix sum
      presum += a[i];

      // find the closest subarray
      // sum to by using lower_bound
      Integer it = s.lower(presum - k);

      // if it is the first element
      // in the set
      if (it == s.first()) {

        // get the prefix sum till start
        // of the subarray
        int diff = it;

        // if the subarray sum is closest to K
        // than the previous one
        if (Math.abs((presum - diff) - k) < mini) {

          // update the minimal difference
          mini = Math.abs((presum - diff) - k);

          // update the sum
          sum = presum - diff;
        }
      }

      // if the difference is
      // present in between
      else if (it == s.last()) {

        // get the prefix sum till start
        // of the subarray
        int diff = it;

        // if the subarray sum is closest to K
        // than the previous one
        if (Math.abs((presum - diff) - k) < mini) {

          // update the minimal difference
          mini = Math.abs((presum - diff) - k);

          // update the sum
          sum = presum - diff;
        }

        // also check for the one before that
        // since the sum can be greater than
        // or less than K also
        it--;

        // get the prefix sum till start
        // of the subarray
        diff = it;

        // if the subarray sum is closest to K
        // than the previous one
        if (Math.abs((presum - diff) - k) < mini) {

          // update the minimal difference
          mini = Math.abs((presum - diff) - k);

          // update the sum
          sum = presum - diff;
        }
      }

      // if there exists no such prefix sum
      // then the current prefix sum is
      // checked and updated
      else {

        // if the subarray sum is closest to K
        // than the previous one
        if (Math.abs(presum - k) < mini) {

          // update the minimal difference
          mini = Math.abs(presum - k);

          // update the sum
          sum = presum+1;
        }
      }

      // insert the current prefix sum
      s.add(presum);
    }

    return sum;
  }

  // Driver Code
  public static void main(String[] args)
  {
    int a[] = { -5, 12, -3, 4, -15, 6, 1 };
    int n = a.length;
    int k = 2;

    System.out.print(closestSubarraySumToK(a, n, k));
  }
}

// This code contributed by Rajput-Ji

Python3

# Python3 program to find the
# sum of subarray whose sum is
# closest to K
import bisect

# Function to find the sum of subarray
# whose sum is closest to K
def closestSubarraySumToK(a , n , k):

    # Declare a set
    s = []

    # initially consider the
    # first subarray as the first
    # element in the array
    presum = a[0]

    # insert
    s.append(a[0])

    # Initially let this difference
    # be the minimum
    mini = abs(a[0] - k)

    # let this be the sum
    # of the subarray
    # to be searched initially
    sum = presum

    # iterate for all the array elements
    for i in range(1, n):

        # calculate the prefix sum
        presum += a[i]

        # find the closest subarray
        # sum to by using lower_bound
        it = bisect.bisect_left(s,presum - k)

        if(it == -1):
            continue

        #if it is the first element
        # in the set
        if (it == 0):

            #get the prefix sum till start
            #of the subarray
            diff = s[it]

            # if the subarray sum is closest to K
            # than the previous one
            if (abs((presum - diff) - k) < mini):

                # update the minimal difference
                mini = abs((presum - diff) - k)

                # update the sum
                sum = presum - diff

            if (abs(presum - k) < mini):
                #update the minimal difference
                mini = abs((presum - diff) - k)

                #update the sum
                sum = presum - diff



        # if the difference is
        # present in between
        elif (it != len(s)):

            # get the prefix sum till start
            # of the subarray
            diff = s[it]

            # if the subarray sum is closest to K
            # than the previous one
            if (abs((presum - diff) - k) < mini):

                # update the minimal difference
                mini = abs((presum - diff) - k)

                # update the sum
                sum = presum - diff


            # also check for the one before that
            # since the sum can be greater than
            # or less than K also
            it -= 1

            # get the prefix sum till start
            # of the subarray
            diff = s[it]

            # if the subarray sum is closest to K
            # than the previous one
            if (abs((presum - diff) - k) < mini):

                # update the minimal difference
                mini = abs((presum - diff) - k)

                # update the sum
                sum = presum - diff;


        # if there exists no such prefix sum
        # then the current prefix sum is
        # checked and updated
        else :

            # if the subarray sum is closest to K
            # than the previous one
            if (abs(presum - k) < mini):

                # update the minimal difference
                mini = abs(presum - k)

                # update the sum
                sum = presum + 1;

        # insert the current prefix sum
        bisect.insort(s, presum)

    return sum



# Driver Code
a = [ -5, 12, -3, 4, -15, 6, 1 ]
n = len(a)
k = 2

print(closestSubarraySumToK(a, n, k))

#This code is contributed by phasing17

C#

// C# program to find the
// sum of subarray whose sum is
// closest to K
using System;
using System.Linq;
using System.Collections.Generic;

public class GFG {

  // Function to find the sum of subarray
  // whose sum is closest to K
  static int closestSubarraySumToK(int []a, int n, int k) {

    // Declare a set
    SortedSet<int> s = new SortedSet<int>();

    // initially consider the
    // first subarray as the first
    // element in the array
    int presum = a[0];

    // insert
    s.Add(a[0]);

    // Initially let this difference
    // be the minimum
    int mini = Math.Abs(a[0] - k);

    // let this be the sum
    // of the subarray
    // to be searched initially
    int sum = presum;

    // iterate for all the array elements
    for (int i = 1; i < n; i++) {

      // calculate the prefix sum
      presum += a[i];

      // find the closest subarray
      // sum to by using lower_bound
      int it = lower_bound(s,presum - k);

      // if it is the first element
      // in the set
      if (it == s.First()) {

        // get the prefix sum till start
        // of the subarray
        int diff = it;

        // if the subarray sum is closest to K
        // than the previous one
        if (Math.Abs((presum - diff) - k) < mini) {

          // update the minimal difference
          mini = Math.Abs((presum - diff) - k);

          // update the sum
          sum = presum - diff;
        }
      }

      // if the difference is
      // present in between
      else if (it == s.Last()) {

        // get the prefix sum till start
        // of the subarray
        int diff = it;

        // if the subarray sum is closest to K
        // than the previous one
        if (Math.Abs((presum - diff) - k) < mini) {

          // update the minimal difference
          mini = Math.Abs((presum - diff) - k);

          // update the sum
          sum = presum - diff;
        }

        // also check for the one before that
        // since the sum can be greater than
        // or less than K also
        it--;

        // get the prefix sum till start
        // of the subarray
        diff = it;

        // if the subarray sum is closest to K
        // than the previous one
        if (Math.Abs((presum - diff) - k) < mini) {

          // update the minimal difference
          mini = Math.Abs((presum - diff) - k);

          // update the sum
          sum = presum - diff;
        }
      }

      // if there exists no such prefix sum
      // then the current prefix sum is
      // checked and updated
      else {

        // if the subarray sum is closest to K
        // than the previous one
        if (Math.Abs(presum - k) < mini) {

          // update the minimal difference
          mini = Math.Abs(presum - k);

          // update the sum
          sum = presum + 1;
        }
      }

      // insert the current prefix sum
      s.Add(presum);
    }

    return sum-1;
  }
  public static int lower_bound(SortedSet<int> s, int val)
  {
    List<int> temp = new List<int>();
    temp.AddRange(s);
    temp.Sort();
    temp.Reverse();

    if (temp.IndexOf(val) + 1 == temp.Count)
      return -1;
    return temp[temp.IndexOf(val) + 1];
  }

  // Driver Code
  public static void Main(String[] args) {
    int []a = { -5, 12, -3, 4, -15, 6, 1 };
    int n = a.Length;
    int k = 2;

    Console.Write(closestSubarraySumToK(a, n, k));
  }
}

// This code is contributed by Rajput-Ji

Javascript

<script>
// javascript program to find the
// sum of subarray whose sum is
// closest to K

    // Function to find the sum of subarray
    // whose sum is closest to K
    function closestSubarraySumToK(a , n , k) {

        // Declare a set
        var s = [];

        // initially consider the
        // first subarray as the first
        // element in the array
        var presum = a[0];

        // insert
        s.push(a[0]);

        // Initially let this difference
        // be the minimum
        var mini = Math.abs(a[0] - k);

        // let this be the sum
        // of the subarray
        // to be searched initially
        var sum = presum;

        // iterate for all the array elements
        for (var i = 1; i < n; i++) {
            s.sort();

            // calculate the prefix sum
            presum += a[i];

            // find the closest subarray
            // sum to by using lower_bound
            var it = lower_bound(s,presum - k);
            if(it == -1)
                continue;

            // if it is the first element
            // in the set
            if (it == s[0]) {

                // get the prefix sum till start
                // of the subarray
                var diff = it;

                // if the subarray sum is closest to K
                // than the previous one
                if (Math.abs((presum - diff) - k) < mini) {

                    // update the minimal difference
                    mini = Math.abs((presum - diff) - k);

                    // update the sum
                    sum = presum - diff;
                }
            }

            // if the difference is
            // present in between
            else if (it == s[s.length-1]) {

                // get the prefix sum till start
                // of the subarray
                var diff = it;

                // if the subarray sum is closest to K
                // than the previous one
                if (Math.abs((presum - diff) - k) < mini) {

                    // update the minimal difference
                    mini = Math.abs((presum - diff) - k);

                    // update the sum
                    sum = presum - diff;
                }

                // also check for the one before that
                // since the sum can be greater than
                // or less than K also
                it--;

                // get the prefix sum till start
                // of the subarray
                diff = it;

                // if the subarray sum is closest to K
                // than the previous one
                if (Math.abs((presum - diff) - k) < mini) {

                    // update the minimal difference
                    mini = Math.abs((presum - diff) - k);

                    // update the sum
                    sum = presum - diff;
                }
            }

            // if there exists no such prefix sum
            // then the current prefix sum is
            // checked and updated
            else {

                // if the subarray sum is closest to K
                // than the previous one
                if (Math.abs(presum - k) < mini) {

                    // update the minimal difference
                    mini = Math.abs(presum - k);

                    // update the sum
                    sum = presum + 1;
                }
            }

            // insert the current prefix sum
            s.push(presum);
        }

        return sum;
    }
    function lower_bound(s, val) {
        let temp = [...s];
        temp.sort((a, b) => a - b);
        return temp[temp.indexOf(val) + 1];
    }

    // Driver Code
        var a = [ -5, 12, -3, 4, -15, 6, 1 ];
        var n = a.length;
        var k = 2;

        document.write(closestSubarraySumToK(a, n, k));

// This code is contributed by Rajput-Ji
</script>

Output

1

Complexity Analysis:

• Time Complexity: O(N log N), where N represents the size of the given array.
• Auxiliary Space: O(N), where N represents the size of the given array.