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 :
- 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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