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

Last Updated : 19 Sep, 2023

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 than 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.

Below is the implementation of the above approach:

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# 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  ?>

JavaScript

<script>  // Javascript 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     var smallest = 1000000000;     var count=0;      // iterate over all pairs     for (var i=0; i<n; i++)         for(var j=i+1; j<n; j++)         {             // is Math.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 Math.abs value is equal to smallest             // increment count value             else if (Math.abs(arr[i] + arr[j] - k) == smallest)                 count++;         }          // print result         document.write( "Minimal Value = " + smallest + "<br>");         document.write( "Total Pairs = " + count + "<br>");     }   // driver program var arr = [3, 5, 7, 5, 1, 9, 9]; var k = 12; var n = arr.length; pairs(arr, n, k);  </script>

Output

Minimal Value = 0  Total Pairs = 4

Time Complexity: O(n²) where n is the number of elements in the array.
Auxiliary Space : O(1)

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; }

Python3

# Python program to find number of pairs # and minimal possible value  from sys import maxsize from bisect import bisect_left, bisect_right  # function for finding pairs and min value def pairs(arr, n, k):     # initialize smallest and count     smallest = maxsize     count = 0     s = set()      # iterate over all pairs     s.add(arr[0])     for i in range(1, n):         # Find the closest elements to k - arr[i]         sorted_s = sorted(s)         index = bisect_left(sorted_s, k - arr[i])         if index == len(sorted_s):             lower = sorted_s[index - 1]         else:             lower = sorted_s[index]         index = bisect_right(sorted_s, k - arr[i])         if index == len(sorted_s):             upper = sorted_s[index - 1]         else:             upper = sorted_s[index]          # Find absolute value of the pairs formed         # with closest greater and smaller elements.         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         elif curr_min == smallest:             count += 1         s.add(arr[i])      # print result     print("Minimal Value = ", smallest)     print("Total Pairs = ", count)  # driver program arr = [3, 5, 7, 5, 1, 9, 9] k = 12 n = len(arr) pairs(arr, n, k)  # This code is contributed by vikramshirsath177.

Java

import java.util.*;  class Main {     // function for finding pairs and min value     static void pairs(final int[] arr,final int n,final int k) {         // initialize smallest and count         int smallest = Integer.MAX_VALUE, count = 0;         Set<Integer> s = new TreeSet<>();          // iterate over all pairs         s.add(arr[0]);         for (int i = 1; i < n; i++) {             // Find the closest elements to  k - arr[i]             int lower = Integer.MIN_VALUE;             int upper = Integer.MAX_VALUE;             for (Integer x : s) {                 if (x <= (k - arr[i]) && x >= lower) {                    lower = x;                  }                 if (x >= (k - arr[i]) && x <= upper) {                     upper = x;                  }              }              // Find absolute value of the pairs formed             // with closest greater and smaller elements.             int curr_min = Math.min(Math.abs(lower + arr[i] - k), Math.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.add(arr[i]);          } // print result          System.out.println("Minimal Value = " + smallest);         System.out.println("Total Pairs = " + count);     }      // driver program     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);     } }

JavaScript

function pairs(arr, n, k) {     // initialize smallest and count     let smallest = Number.MAX_SAFE_INTEGER;     let count = 0;     let s = new Set();       // iterate over all pairs     s.add(arr[0]);     for (let i = 1; i < n; i++) {         // Find the closest elements to  k - arr[i]         let lower = [...s].find((element) => element >= k - arr[i]);         let upper = [...s].find((element) => element >= k - arr[i]);           // Find absolute value of the pairs formed         // with closest greater and smaller elements.         let curr_min = Math.min(Math.abs(lower + arr[i] - k), Math.abs(upper + arr[i] - k));           // if 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.add(arr[i]);     }       // print result     console.log(`Minimal Value = ${smallest}`);     console.log(`Total Pairs = ${count}`); }   // driver program let arr = [3, 5, 7, 5, 1, 9, 9]; let k = 12; let n = arr.length; pairs(arr, n, k);

using System; using System.Collections.Generic; using System.Linq;  class Program {     // function for finding pairs and min value     static void pairs(int[] arr, int n, int k)     {         // initialize smallest and count         int smallest = int.MaxValue, count = 0;         SortedSet<int> s = new SortedSet<int>();          // iterate over all pairs         s.Add(arr[0]);         for (int i = 1; i < n; i++)         {             // Find the closest elements to k - arr[i]             int lower = s.Where(e => e >= k - arr[i]).DefaultIfEmpty(int.MinValue).First();             int upper = s.Where(e => e > k - arr[i]).DefaultIfEmpty(int.MaxValue).First();              // Find absolute value of the pairs formed             // with closest greater and smaller elements.             int curr_min = Math.Min(Math.Abs(lower + arr[i] - k), Math.Abs(upper + arr[i] - k));              // if 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.Add(arr[i]);         }          // print result         Console.WriteLine("Minimal Value = " + smallest);         Console.WriteLine("Total Pairs = " + count);     }      // driver program     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);     } }

Output

Minimal Value = 0  Total Pairs = 4

Time Complexity : O(n Log n)
Auxiliary Space: O(n)

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

Shivam.Pradhan

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

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