Smallest Difference pair of values between two unsorted Arrays
Given two arrays of integers, compute the pair of values (one value in each array) with the smallest (non-negative) difference. Return the difference.
Examples :
Input : A[] = {1, 3, 15, 11, 2} B[] = {23, 127, 235, 19, 8} Output : 3 That is, the pair (11, 8) Input : A[] = {10, 5, 40} B[] = {50, 90, 80} Output : 10 That is, the pair (40, 50)
A simple solution is to Brute Force using two loops with Time Complexity O(n2) and Space Complexity (1)
A better solution is to sort the arrays. Once the arrays are sorted, we can find the minimum difference by iterating through the arrays using the approach discussed in below post.
Find the closest pair from two sorted arrays
Consider the following two arrays:
- A: {1, 2, 11, 15}
- B: {4, 12, 19, 23, 127, 235}
- Suppose a pointer a points to the beginning of A and a pointer b points to the beginning of B. The current difference between a and b is 3. Store this as the min.
- How can we (potentially) make this difference smaller? Well, the value at b is bigger than the value at a, so moving b will only make the difference larger. Therefore, we want to move a.
- Now a points to 2 and b (still) points to 4. This difference is 2, so we should update min. Move a, since it is smaller.
- Now a points to 11 and b points to 4. Move b.
- Now a points to 11 and b points to 12. Update min to 1. Move b. And so on.
Below is the implementation of the idea.
C++
// C++ Code to find Smallest // Difference between two Arrays #include <bits/stdc++.h> using namespace std; // function to calculate Small // result between two arrays int findSmallestDifference( int A[], int B[], int m, int n) { // Sort both arrays using // sort function sort(A, A + m); sort(B, B + n); int a = 0, b = 0; // Initialize result as max value int result = INT_MAX; // Scan Both Arrays upto // sizeof of the Arrays while (a < m && b < n) { if ( abs (A[a] - B[b]) < result) result = abs (A[a] - B[b]); // Move Smaller Value if (A[a] < B[b]) a++; else b++; } // return final sma result return result; } // Driver Code int main() { // Input given array A int A[] = {1, 2, 11, 5}; // Input given array B int B[] = {4, 12, 19, 23, 127, 235}; // Calculate size of Both arrays int m = sizeof (A) / sizeof (A[0]); int n = sizeof (B) / sizeof (B[0]); // Call function to print // smallest result cout << findSmallestDifference(A, B, m, n); return 0; } |
Java
// Java Code to find Smallest // Difference between two Arrays import java.util.*; class GFG { // function to calculate Small // result between two arrays static int findSmallestDifference( int A[], int B[], int m, int n) { // Sort both arrays // using sort function Arrays.sort(A); Arrays.sort(B); int a = 0 , b = 0 ; // Initialize result as max value int result = Integer.MAX_VALUE; // Scan Both Arrays upto // sizeof of the Arrays while (a < m && b < n) { if (Math.abs(A[a] - B[b]) < result) result = Math.abs(A[a] - B[b]); // Move Smaller Value if (A[a] < B[b]) a++; else b++; } // return final sma result return result; } // Driver Code public static void main(String[] args) { // Input given array A int A[] = { 1 , 2 , 11 , 5 }; // Input given array B int B[] = { 4 , 12 , 19 , 23 , 127 , 235 }; // Calculate size of Both arrays int m = A.length; int n = B.length; // Call function to // print smallest result System.out.println(findSmallestDifference (A, B, m, n)); } } // This code is contributed // by Arnav Kr. Mandal. |
Python3
# Python 3 Code to find # Smallest Difference between # two Arrays import sys # function to calculate # Small result between # two arrays def findSmallestDifference(A, B, m, n): # Sort both arrays # using sort function A.sort() B.sort() a = 0 b = 0 # Initialize result as max value result = sys.maxsize # Scan Both Arrays upto # sizeof of the Arrays while (a < m and b < n): if ( abs (A[a] - B[b]) < result): result = abs (A[a] - B[b]) # Move Smaller Value if (A[a] < B[b]): a + = 1 else : b + = 1 # return final sma result return result # Driver Code # Input given array A A = [ 1 , 2 , 11 , 5 ] # Input given array B B = [ 4 , 12 , 19 , 23 , 127 , 235 ] # Calculate size of Both arrays m = len (A) n = len (B) # Call function to # print smallest result print (findSmallestDifference(A, B, m, n)) # This code is contributed by # Smitha Dinesh Semwal |
C#
// C# Code to find Smallest // Difference between two Arrays using System; class GFG { // function to calculate Small // result between two arrays static int findSmallestDifference( int []A, int []B, int m, int n) { // Sort both arrays using // sort function Array.Sort(A); Array.Sort(B); int a = 0, b = 0; // Initialize result as max value int result = int .MaxValue; // Scan Both Arrays upto // sizeof of the Arrays while (a < m && b < n) { if (Math.Abs(A[a] - B[b]) < result) result = Math.Abs(A[a] - B[b]); // Move Smaller Value if (A[a] < B[b]) a++; else b++; } // return final sma result return result; } // Driver Code public static void Main() { // Input given array A int []A = {1, 2, 11, 5}; // Input given array B int []B = {4, 12, 19, 23, 127, 235}; // Calculate size of Both arrays int m = A.Length; int n = B.Length; // Call function to // print smallest result Console.Write(findSmallestDifference (A, B, m, n)); } } // This code is contributed // by nitin mittal. |
PHP
<?php // PHP Code to find Smallest // Difference between two Arrays // function to calculate Small // result between two arrays function findSmallestDifference( $A , $B , $m , $n ) { // Sort both arrays // using sort function sort( $A ); sort( $A , $m ); sort( $B ); sort( $B , $n ); $a = 0; $b = 0; $INT_MAX = 1; // Initialize result // as max value $result = $INT_MAX ; // Scan Both Arrays upto // sizeof of the Arrays while ( $a < $m && $b < $n ) { if ( abs ( $A [ $a ] - $B [ $b ]) < $result ) $result = abs ( $A [ $a ] - $B [ $b ]); // Move Smaller Value if ( $A [ $a ] < $B [ $b ]) $a ++; else $b ++; } // return final sma result return $result ; } // Driver Code { // Input given array A $A = array (1, 2, 11, 5); // Input given array B $B = array (4, 12, 19, 23, 127, 235); // Calculate size of Both arrays $m = sizeof( $A ) / sizeof( $A [0]); $n = sizeof( $B ) / sizeof( $B [0]); // Call function to print // smallest result echo findSmallestDifference( $A , $B , $m , $n ); return 0; } // This code is contributed by nitin mittal. ?> |
Javascript
<script> // JavaScript Code to find Smallest // Difference between two Arrays // function to calculate Small // result between two arrays function findSmallestDifference(A, B, m, n) { // Sort both arrays using // sort function A.sort((a, b) => a - b); B.sort((a, b) => a - b); let a = 0, b = 0; // Initialize result as max value let result = Number.MAX_SAFE_INTEGER; // Scan Both Arrays upto // sizeof of the Arrays while (a < m && b < n) { if (Math.abs(A[a] - B[b]) < result) result = Math.abs(A[a] - B[b]); // Move Smaller Value if (A[a] < B[b]) a++; else b++; } // Return final sma result return result; } // Driver Code // Input given array A let A = [ 1, 2, 11, 5 ]; // Input given array B let B = [ 4, 12, 19, 23, 127, 235 ]; // Calculate size of Both arrays let m = A.length; let n = B.length; // Call function to print // smallest result document.write(findSmallestDifference( A, B, m, n)); // This code is contributed by Surbhi Tyagi. </script> |
1
Time Complexity: O(m log m + n log n)
This algorithm takes O(m log m + n log n) time to sort and O(m + n) time to find the minimum difference. Therefore, the overall runtime is O(m log m + n log n).
Auxiliary Space: O(1)
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