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

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}

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 :

abs( ai + aj – K) > smallest : do nothing as this pair will not count in minimal possible value.
abs(ai + aj – K) = smallest : increment the count of pair resulting minimal possible value.
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. 

Python3

# Python3 program to find number of pairs  
# and minimal possible value  
  
# function for finding pairs and min value  
def pairs(arr, n, k): 
      
    # initialize smallest and count  
    smallest = 999999999999
    count = 0
  
    # iterate over all pairs  
    for i in range(n): 
        for j in range(i + 1, n): 
              
            # 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  
            elif abs(arr[i] + arr[j] - k) == smallest:  
                count += 1
  
    # print result  
    print("Minimal Value = ", smallest)  
    print("Total Pairs = ", count) 
  
# Driver Code 
if __name__ == '__main__': 
    arr = [3, 5, 7, 5, 1, 9, 9]  
    k = 12
    n = len(arr)  
    pairs(arr, n, k) 
      
# This code is contributed by PranchalK 

C#

// C# program to find number  
// of pairs and minimal  
// possible value 
using System; 
  
class GFG  
{ 
  
// function for finding 
// pairs and min value 
static void pairs(int []arr,  
                  int n, int k) 
{ 
    // initialize  
    // smallest and count 
    int smallest = 0; 
    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 
    Console.WriteLine("Minimal Value = " +  
                                smallest); 
    Console.WriteLine("Total Pairs = " + 
                                 count);  
} 
  
// Driver Code 
public static void Main()  
{ 
    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 anuj_67. 

PHP

<?php 
// PHP program to find number of 
// pairs and minimal possible value 
  
// function for finding pairs 
// and min value 
function pairs($arr, $n, $k) 
{ 
      
    // initialize smallest and count 
    $smallest = PHP_INT_MAX; 
    $count = 0; 
  
    // iterate over all pairs 
    for ($i = 0; $i < $n; $i++) 
        for($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 
        echo "Minimal Value = " , $smallest , "\n"; 
        echo "Total Pairs = ", $count , "\n";  
}  
  
    // Driver Code 
    $arr = array (3, 5, 7, 5, 1, 9, 9); 
    $k = 12; 
    $n = sizeof($arr); 
    pairs($arr, $n, $k); 
  
// This code is contributed by aj_36  
?> 

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.

C++

// C++ 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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My Personal Notes

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

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