Pair Sum

Easy
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Average time to solve is 15m
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Problem statement

You are given an array/list ‘ARR’ consisting of ‘N’ distinct integers arranged in ascending order. You are also given an integer ‘TARGET’. Your task is to count all the distinct pairs in ‘ARR’ such that their sum is equal to ‘TARGET’.

Note:

1. Pair (x,y) and Pair(y,x) are considered as the same pair. 

2. If there exists no such pair with sum equals to 'TARGET', then return -1.

Example:

Let ‘ARR’ = [1 2 3] and ‘TARGET’ = 4. Then, there exists only one pair in ‘ARR’ with a sum of 4 which is (1, 3). (1, 3) and (3, 1) are counted as only one pair.
Detailed explanation ( Input/output format, Notes, Images )
Input Format: 
The first line of input contains an integer ‘T’ which denotes the number of test cases. 

The first line of each test case contains two single space-separated integers ‘N’ and ‘TARGET’ representing the number of elements in the array/list ‘ARR’ and the required pair-sum respectively.

The next line of each test case contains ‘N’ single space-separated integers denoting the elements of  ‘ARR’.
Output Format : 
For each test case, return the numbers of pairs in  ‘ARR’ whose sum is equal to ‘TARGET’.
Note: 
You don't need to print anything, it has already been taken care of. Just implement the given function.
Constraints: 
1 <= ‘T’ <= 100
2 <= ‘N’ <= 5000
1 <= ‘ARR[i]’, ‘TARGET’ <= 10^5

Where ARR[i]’ represents the elements of array/list ‘ARR’. 

Time Limit: 1 sec
Sample Input 1:
2
5 6
1 2 3 4 5
6 7
1 2 3 4 5 6
Sample Output 1:
2
3

Explanation for Sample Output 1:

In test case 1, there exist only 2 pairs whose sum is equal to ‘TARGET’ i.e (1, 5) and (2, 4).

In test case 2, there are 3 pairs whose sum is equal to ‘TARGET’ which are  (1, 6), (2, 5), and (3, 4).
Sample Input 2:
2
4 10
1 3 5 6
5 12
1 3 6 9 11
Sample Output 2:
-1
 2

Explanation for Sample Output 2:

In test case 1, there is not a pair whose sum is equal to ‘TARGET’. So we return -1.

In test case 2, there are 2 pairs whose sum is equal to ‘TARGET’, (1, 11) and (3, 9) respectively.
Hint

Try to traverse through every pair possible.

Approaches (2)
Brute Force

First, we declare a variable 'COUNTPAIR’ in which we store all pairs whose sum is equal to 'TARGET’. Then, we traverse the array ‘ARR’ and assume every element as the first element of the pair. Then we again traverse the remaining array and consider every element as a second element of the pair, and check whether the sum of the two elements is equal to 'TARGET' or not. If it is equal to 'TARGET',’ then we increase our ‘COUNTPAIR’ by 1.

 

The steps are as follows:

 

  1. We declare a variable ‘COUNTPAIR’ and initialize it with 0.
  2. We run a loop for ‘i’ = 0 to ‘N’ - 1:
    • We run a loop for ‘j’ = ‘i’ + 1 to ‘N’:
      • If ‘ARR[i]’ + ‘ARR[j]’ is equal to 'TARGET':
        • ‘COUNTPAIR’++
        • We break the loop as the elements are sorted and distinct. So, the pair sum will increase and can't be equal to ‘TARGET’.
  3. If ‘COUNTPAIR’ is equal to ‘0’:
    • Then, Return - 1.
  4. Else
    • Return ‘COUNTPAIR’.
Time Complexity

O(N ^ 2), Where ‘N’ represents the number of elements in the array/list ‘ARR’.

 

Since we are using two nested loops for finding a pair, thus the time complexity will be O(N ^ 2). 

Space Complexity

O(1).

 

Since we are not using any extra space for finding our resultant answer. Thus the space complexity will be O(1).

Code Solution
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Pair Sum
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