Pair Sum Problem Statement

You are provided with an array ARR consisting of N distinct integers in ascending order and an integer TARGET. Your objective is to count all the distinct pairs in ARR whose sum equals TARGET.

Note:

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

If no such pair exists with a sum equal to 'TARGET', then return -1.

Example:

Input:

ARR = [1, 2, 3], TARGET = 4

Output:

1

Explanation:

There exists only one pair in ARR with a sum of 4, which is (1, 3). (1, 3) and (3, 1) are counted as a single pair.

Constraints:

1 ≤ T ≤ 100

2 ≤ N ≤ 5000

1 ≤ ARR[i], TARGET ≤ 10⁵

The array ARR contains distinct elements.

Question

Pair Sum Problem Statement

You are provided with an array ARR consisting of N distinct integers in ascending order and an integer TARGET. Your objective is to count all the distinct pairs in ARR whose sum equals TARGET.

Note:

Pair (x, y) and Pair (y, x) are considered as the same pair.
If no such pair exists with a sum equal to 'TARGET', then return -1.

Example:

Input:

ARR = [1, 2, 3], TARGET = 4

Output:

Explanation:

There exists only one pair in ARR with a sum of 4, which is (1, 3). (1, 3) and (3, 1) are counted as a single pair.

Constraints:

1 ≤ T ≤ 100
2 ≤ N ≤ 5000
1 ≤ ARR[i], TARGET ≤ 10⁵
The array ARR contains distinct elements.

Accepted Answer

Count distinct pairs in an array whose sum equals a given target.

Use two pointers approach to iterate through the array and find pairs with sum equal to target.
Keep track of visited pairs to avoid counting duplicates.
Return -1 if no such pair exists with the given target.

Accepted Answer

function pairSum(sortedArray , target){

    let count = 0
    let leftPointer = 0
    let rightPointer = sortedArray.length - 1

    while(leftPointer < rightPointer){
        let sum = sortedArray[leftPointer] + sortedArray[rightPointer]

        if(sum === target){
            count++;
            leftPointer ++
            rightPointer --
            

            //Remove duplicates from left
            while(leftPointer < rightPointer && sortedArray[leftPointer] === sortedArray[leftPointer -1]){
                leftPointer++
            }

            //Remove duplicates from right
            while(leftPointer < rightPointer && sortedArray[rightPointer] === sortedArray[leftPointer +1]){
                rightPointer--
            }
        }
        else if(sum < target){
            leftPointer ++;
        }else if(sum > target){
            rightPointer --;
        }
    }

    return count > 0 ? count : -1
}