Tutorialspoint.dev

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} 



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 << " ";
        cout << "Total Pairs = " << count << " ";    
  
// 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 [tabby title="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 , " ";
        echo "Total Pairs = ", $count , " "
  
    // 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 <<" ";
    cout << "Total Pairs = " << count <<" ";
}
  
// 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)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

You Might Also Like

leave a comment

code

0 Comments

load comments

Subscribe to Our Newsletter