Two Sum
You are given an array of integers 'ARR' of length 'N' and an integer Target. Your task is to return all pairs of elements such that they add up to Target.

Note:

We cannot use the element at a given index twice.

Follow Up:

Try to do this problem in O(N) time complexity.

Input Format:

The first line of input contains an integer ‘T’ denoting the number of test cases to run. Then the test case follows. The first line of each test case contains two single space-separated integers ‘N’ and ‘Target’ denoting the number of elements in an array and the Target, respectively. The second line of each test case contains ‘N’ single space-separated integers, denoting the elements of the array.

Output Format :

For each test case, print a single line containing space-separated integers denoting all pairs of elements such that they add up to the target. A pair (a, b) and (b, a) is the same, so you can print it in any order. Each pair must be printed in a new line. If no valid pair exists, print a pair of (-1, -1). Refer to sample input/output for more clarity.

Note:

You do not need to print anything; it has already been taken care of. Just implement the given function.

Constraints:

1

Question

Two Sum

You are given an array of integers 'ARR' of length 'N' and an integer Target. Your task is to return all pairs of elements such that they add up to Target.

Note:

We cannot use the element at a given index twice.

Follow Up:

Try to do this problem in O(N) time complexity.

Input Format:

The first line of input contains an integer ‘T’ denoting the number of test cases to run. Then the test case follows.

The first line of each test case contains two single space-separated integers ‘N’ and ‘Target’ denoting the number of elements in an array and the Target, respectively.

The second line of each test case contains ‘N’ single space-separated integers, denoting the elements of the array.

Output Format :

For each test case, print a single line containing space-separated integers denoting all pairs of elements such that they add up to the target. A pair (a, b) and (b, a) is the same, so you can print it in any order.

Each pair must be printed in a new line. If no valid pair exists, print a pair of (-1, -1). Refer to sample input/output for more clarity.

Note:

You do not need to print anything; it has already been taken care of. Just implement the given function.

Constraints:

1 <= T <= 100
1 <= N <= 5000
-10 ^ 9 <= TARGET <=10 ^ 9
-10 ^ 9 <= ARR[i] <=10 ^ 9

Where 'T' denotes the number of test cases, 'N' represents the size of the array, 'TARGET' represents the sum required, and 'ARR[i]' represents array elements.

Time Limit: 1 sec.

Accepted Answer

The brute force approach is simple. Loop through each element x and find if there is another value that equals to target - x.

To improve our run time complexity, we need a more efficient way to check if the complement exists in the array. If the complement exists, we need to look up its index. What is the best way to maintain a mapping of each element in the array to its index? A hash table.

We reduce the look-up time from O(n) to O(1) by trading space for speed. A hash table is built exactly for this purpose, it supports fast lookup in near-constant time. I say "near" because if a collision occurred, a look-up could degenerate to O(n) time. But look up in hash table should be amortized O(1) time as long as the hash function was chosen carefully.

A simple implementation uses two iterations. In the first iteration, we add each element's value and its index to the table. Then, in the second iteration, we check if each element's complement (target - nums[i]) exists in the table. Beware that the complement must not be nums[i] itself.

Accepted Answer

Hashing Solution

We can store the frequency of every element in the array in a hashmap.
We will loop over every index i, and check the frequency of (Target - ARR[i]) is the hashmap:
- If (Target - ARR[i]) is equal to ARR[i], we will check if frequency of ARR[i] . If it is greater than 1 then we will decrease the frequency of ARR[i] by 2 and add a pair (ARR[i] , ARR[i]) to our answer.
- Else, if the frequency of ARR[i] and Target - ARR[i] is greater than equal to 1 then we add pair (ARR[i], Target - ARR[i]) to our answer and decrease the frequency of both by 1.
If no valid pairs exist, we will return [[-1,-1]].

Space Complexity: O(n)Explanation:

O(N), where ‘N’ is the total number of elements in an array.

As every element is stored in a hashmap, so it will take O(N) space.

Time Complexity: O(n)Explanation:

O(N), where ‘N’ is the total number of elements in an array.

As every element is visited at most two times and hashmap operations take constant time, the time complexity will be O(N).

You are given an array of integers 'ARR' of length 'N' and an integer Target. Your task is to return all pairs of elements such that they add up to Target.

Note:

Follow Up:

Input Format:

Output Format :

Note:

Constraints:

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