The simplest method is to run two loops, the outer loop picks the first element (smaller element) and the inner loop looks for the element picked by outer loop plus n. Time complexity of this method is O(n^2). We can use sorting and Binary Search to improve time complexity to O(nLogn). The first step is to sort the array in ascending order. Once the array is sorted, traverse the array from left to right, and for each element arr[i], binary search for arr[i] + n in arr[i+1..n-1]. If the element is found, return the pair. Both first and second steps take O(nLogn). So overall complexity is O(nLogn). The second step of the above algorithm can be improved to O(n). The first step remain same. The idea for second step is take two index variables i and j, initialize them as 0 and 1 respectively. Now run a linear loop. If arr[j] – arr[i] is smaller than n, we need to look for greater arr[j], so increment j. If arr[j] – arr[i] is greater than n, we need to look for greater arr[i], so increment i. Thanks to Aashish Barnwal for suggesting this approach. The following code is only for the second step of the algorithm, it assumes that the array is already sorted.
C++
// C++ program to find a pair with the given difference
# Python program to find a pair with the given difference
# The function assumes that the array is sorted
deffindPair(arr,n):
size =len(arr)
# Initialize positions of two elements
i,j =0,1
# Search for a pair
whilei < size andj < size:
ifi !=j andarr[j]-arr[i] ==n:
print"Pair found (",arr[i],",",arr[j],")"
returnTrue
elifarr[j] -arr[i] < n:
j+=1
else:
i+=1
print"No pair found"
returnFalse
# Driver function to test above function
arr =[1, 8, 30, 40, 100]
n =60
findPair(arr, n)
# This code is contributed by Devesh Agrawal
C#
// C# program to find a pair with the given difference
usingSystem;
classGFG {
// The function assumes that the array is sorted
staticboolfindPair(int[]arr, intn)
{
intsize = arr.Length;
// Initialize positions of two elements
inti = 0, j = 1;
// Search for a pair
while(i < size && j < size)
{
if(i != j && arr[j] - arr[i] == n)
{
Console.Write("Pair Found: "
+ "( "+ arr[i] + ", "+ arr[j] +" )");
returntrue;
}
elseif(arr[j] - arr[i] < n)
j++;
else
i++;
}
Console.Write("No such pair");
returnfalse;
}
// Driver program to test above function
publicstaticvoidMain ()
{
int[]arr = {1, 8, 30, 40, 100};
intn = 60;
findPair(arr, n);
}
}
// This code is contributed by Sam007.
PHP
<?php
// PHP program to find a pair with
// the given difference
// The function assumes that the
// array is sorted
functionfindPair(&$arr, $size, $n)
{
// Initialize positions of
// two elements
$i= 0;
$j= 1;
// Search for a pair
while($i< $size&& $j< $size)
{
if($i!= $j&& $arr[$j] -
$arr[$i] == $n)
{
echo"Pair Found: ". "(".
$arr[$i] . ", ". $arr[$j] . ")";
returntrue;
}
elseif($arr[$j] - $arr[$i] < $n)
$j++;
else
$i++;
}
echo"No such pair";
returnfalse;
}
// Driver Code
$arr= array(1, 8, 30, 40, 100);
$size= sizeof($arr);
$n= 60;
findPair($arr, $size, $n);
// This code is contributed
// by ChitraNayal
?>
Javascript
<script>
// JavaScript program for the above approach
// The function assumes that the array is sorted
functionfindPair(arr, size, n) {
// Initialize positions of two elements
let i = 0;
let j = 1;
// Search for a pair
while(i < size && j < size) {
if(i != j && arr[j] - arr[i] == n) {
document.write("Pair Found: ("+ arr[i] + ", "+
arr[j] + ")");
returntrue;
}
elseif(arr[j] - arr[i] < n)
j++;
else
i++;
}
document.write("No such pair");
returnfalse;
}
// Driver program to test above function
let arr = [1, 8, 30, 40, 100];
let size = arr.length;
let n = 60;
findPair(arr, size, n);
// This code is contributed by Potta Lokesh
</script>
Output
Pair Found: (100, 40)
Time Complexity: O(n*log(n)), Where n is number of element in given array Auxiliary Space: O(1)
Hashing can also be used to solve this problem. Create an empty hash table HT. Traverse the array, use array elements as hash keys and enter them in HT. Traverse the array again look for value n + arr[i] in HT.
C++
// C++ program to find a pair with the given difference
#include <bits/stdc++.h>
usingnamespacestd;
// The function assumes that the array is sorted
boolfindPair(intarr[], intsize, intn)
{
unordered_map<int, int> mpp;
for(inti = 0; i < size; i++) {
mpp[arr[i]]++;
// Check if any element whose frequency
// is greater than 1 exist or not for n == 0
if(n == 0 && mpp[arr[i]] > 1)
returntrue;
}
// Check if difference is zero and
// we are unable to find any duplicate or
// element whose frequency is greater than 1
// then no such pair found.
if(n == 0)
returnfalse;
for(inti = 0; i < size; i++) {
if(mpp.find(n + arr[i]) != mpp.end()) {
cout << "Pair Found: ("<< arr[i] << ", "
<< n + arr[i] << ")";
returntrue;
}
}
cout << "No Pair found";
returnfalse;
}
// Driver program to test above function
intmain()
{
intarr[] = { 1, 8, 30, 40, 100 };
intsize = sizeof(arr) / sizeof(arr[0]);
intn = -60;
findPair(arr, size, n);
return0;
}
Java
// Java program for the above approach
importjava.io.*;
importjava.util.*;
classGFG
{
// The function assumes that the array is sorted
staticbooleanfindPair(int[] arr, intsize, intn)
{
HashMap<Integer,
Integer> mpp = newHashMap<Integer,
Integer>();
// Traverse the array
for(inti = 0; i < size; i++)
{
// Update frequency
// of arr[i]
mpp.put(arr[i],
mpp.getOrDefault(arr[i], 0) + 1);
// Check if any element whose frequency
// is greater than 1 exist or not for n == 0
if(n == 0&& mpp.get(arr[i]) > 1)
returntrue;
}
// Check if difference is zero and
// we are unable to find any duplicate or
// element whose frequency is greater than 1
// then no such pair found.
if(n == 0)
returnfalse;
for(inti = 0; i < size; i++) {
if(mpp.containsKey(n + arr[i])) {
System.out.print("Pair Found: ("+ arr[i] + ", "+
+ (n + arr[i]) + ")");
returntrue;
}
}
System.out.print("No Pair found");
returnfalse;
}
// Driver Code
publicstaticvoidmain(String[] args)
{
int[] arr = { 1, 8, 30, 40, 100};
intsize = arr.length;
intn = -60;
findPair(arr, size, n);
}
}
// This code is contributed by code_hunt.
Python3
# Python program to find a pair with the given difference
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