Many problems ask "is there an earlier item that goes with this one?" The obvious solution checks every earlier item for each item: two nested loops, O(n²). The pattern that fixes it: as you go, store what you've seen in a set or dictionary, so "have I seen the partner?" is an instant lookup.

The classic example is two sum: given a list and a target, find two positions whose values add up to the target.

def two_sum(nums, target):
    seen = {}
    for i, n in enumerate(nums):
        partner = target - n
        if partner in seen:
            return (seen[partner], i)
        seen[n] = i
    return None

print(two_sum([2, 7, 11, 15], 9))
print(two_sum([3, 2, 4], 6))
print(two_sum([1, 2], 10))

For each number we ask "have I already seen the number that would complete the pair?" The dictionary maps each value seen to its position. One pass: O(n).