Minimum Possible value of |ai + aj – k| for given array and k.

3

You are given an array of n integer and an integer K. Find the number of total unordered pairs {i, j} such that absolute value of (ai + aj – K), i.e., |ai + aj – k| is minimal possible where i != j.

Examples:

Input : arr[] = {0, 4, 6, 2, 4}, 
            K = 7
Output : Minimal Value = 1
         Total  Pairs = 5 
Explanation : Pairs resulting minimal value are :
              {a1, a3}, {a2, a4}, {a2, a5}, {a3, a4}, {a4, a5} 

Input : arr[] = {4, 6, 2, 4}  , K = 9
Output : Minimal Value = 1
         Total Pairs = 4 
Explanation : Pairs resulting minimal value are :
              {a1, a2}, {a1, a4}, {a2, a3}, {a2, a4} 

A simple solution is iterate over all possible pairs and for each pair we will check whether the value of (ai + aj – K) is smaller then our current smallest value of not. So as per result of above condition we have total of three cases :

  1. abs( ai + aj – K) > smallest : do nothing as this pair will not count in minimal possible value.
  2. abs(ai + aj – K) = smallest : increment the count of pair resulting minimal possible value.
  3. abs( ai + aj – K) < smallest : update the smallest value and set count to 1.

C++

// CPP program to find number of pairs  and minimal 
// possible value
#include<bits/stdc++.h>
using namespace std;

// function for finding pairs and min value
void pairs(int arr[], int n, int k)
{
    // initialize smallest and count
    int smallest = INT_MAX;
    int count=0;

    // iterate over all pairs
    for (int i=0; i<n; i++)
        for(int j=i+1; j<n; j++)
        {
            // is abs value is smaller than smallest
            // update smallest and reset count to 1
            if ( abs(arr[i] + arr[j] - k) < smallest )
            { 
                smallest = abs(arr[i] + arr[j] - k);
                count = 1;
            }

            // if abs value is equal to smallest
            // increment count value
            else if (abs(arr[i] + arr[j] - k) == smallest)
                count++;
        }

        // print result
        cout << "Minimal Value = " << smallest << "\n";
        cout << "Total Pairs = " << count << "\n";    
} 

// driver program
int main()
{
    int arr[] = {3, 5, 7, 5, 1, 9, 9};
    int k = 12;
    int n = sizeof(arr) / sizeof(arr[0]);
    pairs(arr, n, k);
    return 0;
} 

Java

// Java program to find number of pairs 
// and minimal possible value
import java.util.*;

class GFG {
    
    // function for finding pairs and min value
    static void pairs(int arr[], int n, int k)
    {
        // initialize smallest and count
        int smallest = Integer.MAX_VALUE;
        int count=0;
     
        // iterate over all pairs
        for (int i=0; i<n; i++)
            for(int j=i+1; j<n; j++)
            {
                // is abs value is smaller than
                // smallest update smallest and
                // reset count to 1
                if ( Math.abs(arr[i] + arr[j] - k) <
                                        smallest )
                { 
                    smallest = Math.abs(arr[i] + arr[j]
                                             - k);
                    count = 1;
                }
     
                // if abs value is equal to smallest
                // increment count value
                else if (Math.abs(arr[i] + arr[j] - k)
                                    == smallest)
                    count++;
            }
     
            // print result
           System.out.println("Minimal Value = " + 
                                    smallest);
           System.out.println("Total Pairs = " +
                                       count);    
    }
    
    /* Driver program to test above function */
    public static void main(String[] args) 
    {
        int arr[] = {3, 5, 7, 5, 1, 9, 9};
        int k = 12;
        int n = arr.length;
        pairs(arr, n, k);
    }
}
// This code is contributed by Arnav Kr. Mandal.


Output:

Minimal Value = 0
Total Pairs = 4

An efficient solution is to use a self balancing binary search tree (which is implemented in set in C++ and TreeSet in Java). We can find closest element in O(log n) time in map.

// CPP program to find number of pairs
// and minimal possible value
#include<bits/stdc++.h>
using namespace std;

// function for finding pairs and min value
void pairs(int arr[], int n, int k)
{
    // initialize smallest and count
    int smallest = INT_MAX, count = 0;
    set<int> s;

    // iterate over all pairs
    s.insert(arr[0]);
    for (int i=1; i<n; i++)
    {
        // Find the closest elements to  k - arr[i]
        int lower = *lower_bound(s.begin(),
                                 s.end(),
                                 k - arr[i]);

        int upper = *upper_bound(s.begin(),
                                 s.end(),
                                 k - arr[i]);

        // Find absolute value of the pairs formed
        // with closest greater and smaller elements.
        int curr_min = min(abs(lower + arr[i] - k),
                           abs(upper + arr[i] - k));

        // is abs value is smaller than smallest
        // update smallest and reset count to 1
        if (curr_min < smallest)
        {
            smallest = curr_min;
            count = 1;
        }

        // if abs value is equal to smallest
        // increment count value
        else if (curr_min == smallest )
            count++;
        s.insert(arr[i]);

    }        // print result

    cout << "Minimal Value = " << smallest <<"\n";
    cout << "Total Pairs = " << count <<"\n";
}

// driver program
int main()
{
    int arr[] = {3, 5, 7, 5, 1, 9, 9};
    int k = 12;
    int n = sizeof(arr) / sizeof(arr[0]);
    pairs(arr, n, k);
    return 0;
}

Output:

Minimal Value = 0
Total Pairs = 4

Time Complexity : O(n Log n)

This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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