Can generating pairs be done faster than O(N^2)?
Yes. For problems like Two Sum, utilizing a Hash Map allows you to check for complementary pairs in O(N) linear time.
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More FAQs in Using Nested Loops for Generating Pairs
Use a nested loop where the outer loop tracks index 'i', and the inner loop starts at index 'j = i + 1'.
It ensures you do not pair an element with itself, and it prevents generating redundant, mirrored pairs (e.g., checking [0,1] and then [1,0]).
Although the inner loop shrinks, the overall time complexity is still bounded by O(N^2) Quadratic Time.
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