# Find all pairs with given a difference

0
78 ## Find all pair of elements in the given array with given difference. (array without duplicates).

### Example

Input array: [90, 70, 20, 80, 50, 25, 35, 15, 100, 150]
Pair of elements with difference 20 will print
90, 70 and 35, 15

## Algorithm 1

Time Complexity: O(N^2)

Run two loops such that select one element from the array. Starting from the index 0, and look forward in the array to get a pair with given difference.

a)    If we find the element then print them.

## C++ Program

``````#include <bits/stdc++.h>
using namespace std;

int main()
{

int arr[] = {90, 70, 20, 80, 50, 25, 35 , 15, 100, 150};//array
int N = sizeof(arr)/sizeof(arr);

int difference;
cin>>difference; // difference required between two elements

for(int i=0; i<N; i++)//select an element
{
for(int j=i+1; j<N; j++) // look forward in the array
if(abs(arr[i] - arr[j]) == difference)//if we get a pair with given difference then success
{
cout << "Pair exists with the given difference and the elements are : " << arr[i] <<"  "<<arr[j];
return 0;
}
}

cout <<"No such pair exists with given difference";

}``````

## Algorithm 2:

Time Complexity: O(NlogN)

Step 1 : First sort the given array. It takes O(NlogN).

Step 2 : In the same way as the first algorithm, for every element starting from first element, find the matching pair.

i)    Here we will use binary search to find the element.
ii)    for element a and given difference n we search for a + n.
Binary search takes O(NlogN), therefore time complexity is O(NlogN).

### Algorithm 2 working Example

Input array: Find pair of elements with difference = 20

Apply Algorithm on the input array,

Step 1 :

Sort the array,
Sorted array: Step 2 :

a = 15, difference (n) = 20
so, a + n = 35,
binary_search(35, input array) = True.
Print 15, 35. Move forward

Step 3 :
a = 20, difference (n) = 20
so, a + n = 40,
binary_search(40, input array) = False.
Move forward

Step 4 :
a = 35, difference (n) = 20
so, a + n = 55,
binary_search(55, input array) = False.
Move forward

Step 5 :
a = 70, difference (n) = 20
so, a + n = 90,
binary_search(90, input array) = True.
Print 70, 90. Move forward

Step 6 :
a = 90, difference (n) = 20
so, a + n = 110,
binary_search(110, input array) = False.
End Here.
Therefore, for the given input it prints:
70, 90 and 15, 35.

## C++ Program

``````#include <bits/stdc++.h>
using namespace std;

int main()
{

int arr[] = {90, 70, 20, 80, 50, 25, 35 , 15, 100, 150};//array
int N = sizeof(arr)/sizeof(arr);

int difference;
cin>>difference; // difference required between two elements

sort(arr,arr+N); //sort the array in ascending order

for(int i = 0; i < N; i++)
{
if(binary_search(arr,arr+N,abs(difference + arr[i]))) // if b - a = X and both b and a exists in array so simply we can search for X + a to obtain b.
{
cout<<"Pair exists\n";
return 0;
}
}

cout <<"No such pair exists with given difference";

}``````