## Pair Sum with Python O(n)

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Pair Sum with Python O(n)

Given a list of n integers arr[0..(n-1)], determine the number of different pairs of elements within it which sum to k. If an integer appears in the list multiple times, each copy is considered to be different; that is, two pairs are considered different if one pair includes at least one array index which the other doesn’t, even if they include the same values.

# Signature

int numberOfWays(int[] arr, int k)

# Input

n is in the range [1, 100,000]. Each value arr[i] is in the range [1, 1,000,000,000]. k is in the range [1, 1,000,000,000].

# Output

Return the number of different pairs of elements which sum to k.

# Example 1

n = 5 k = 6 arr = [1, 2, 3, 4, 3] output = 2The valid pairs are 2+4 and 3+3.

# Example 2

n = 5 k = 6 arr = [1, 5, 3, 3, 3] output = 4There’s one valid pair 1+5, and three different valid pairs 3+3 (the 3rd and 4th elements, 3rd and 5th elements, and 4th and 5th elements).

Solution using Python:

Complexity:

• Time: O(n)
• Space: Array of n and Hash of n (Who cares about space?? really ??)

Execution:

Case 1:

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[1, 2, 3, 4, 3]
{1: 1, 2: 1, 3: 2, 4: 1}
Total Pairs is: 2.0

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Case 2:

[1, 5, 3, 3, 3]
{1: 1, 5: 1, 3: 3}
Total Pairs is: 4.0

Complexity:

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• Time: O(n)
• Space: Array of n and Hash of n (Who cares about space?? really ??)

Execution:

Case 1:

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[1, 2, 3, 4, 3]
{1: 1, 2: 1, 3: 2, 4: 1}
Total Pairs is: 2.0

Case 2:

[1, 5, 3, 3, 3]
{1: 1, 5: 1, 3: 3}
Total Pairs is: 4.0

Complexity:

• Time: O(n)
• Space: Array of n and Hash of n (Who cares about space?? really ??)

Execution:

Case 1:

[1, 2, 3, 4, 3]
{1: 1, 2: 1, 3: 2, 4: 1}
Total Pairs is: 2.0

Case 2:

[1, 5, 3, 3, 3]
{1: 1, 5: 1, 3: 3}
Total Pairs is: 4.0

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