Update pairs_with_given_sum.py

Dictionary Usage:

We use a dictionary seen to keep track of the number of times each number has appeared so far.
For each number num in the array, we calculate its complement (req_sum - num). If this complement has been seen before, then it means there are pairs that sum up to req_sum, and we add the count of such complements to our count.
Efficiency:

This approach processes each element of the array exactly once and performs operations in constant time for each element. Thus, the time complexity is 
𝑂
(
𝑛
)
O(n), where 
𝑛
n is the number of elements in the array.
Readability:

The code is more straightforward and easier to understand compared to generating combinations. It leverages hash maps for efficient lookups and counting.

Author: Tidimatso Anthony

data_structures/arrays/pairs_with_given_sum.py

@@ -7,23 +7,33 @@
 https://practice.geeksforgeeks.org/problems/count-pairs-with-given-sum5022/0
 """
 
-from itertools import combinations
-
-
-def pairs_with_sum(arr: list, req_sum: int) -> int:
+def pairs_with_sum(arr: list[int], req_sum: int) -> int:
     """
-    Return the no. of pairs with sum "sum"
+    Return the number of pairs with sum equal to req_sum.
+    
     >>> pairs_with_sum([1, 5, 7, 1], 6)
     2
     >>> pairs_with_sum([1, 1, 1, 1, 1, 1, 1, 1], 2)
     28
     >>> pairs_with_sum([1, 7, 6, 2, 5, 4, 3, 1, 9, 8], 7)
     4
     """
-    return len([1 for a, b in combinations(arr, 2) if a + b == req_sum])
-
+    count = 0
+    seen = {}
+    
+    for num in arr:
+        complement = req_sum - num
+        if complement in seen:
+            count += seen[complement]
+        
+        if num in seen:
+            seen[num] += 1
+        else:
+            seen[num] = 1
+    
+    return count
 
 if __name__ == "__main__":
     from doctest import testmod
-
     testmod()
+