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two_sum.py

# Two Sum
# Find two numbers that add to target
# return their indices
# hash map for linear time
# Arrays & Hashing

PythonRun

Coco

Coach

Sample

This solution passes. Which line keeps one array element from being used twice?

Check for the complement before storing the current index.

Try a code-reading rep

The important line is rarely the cleverest one.

Read a working solution, identify the decision that keeps it correct, then test the plausible inverse.

Why is this check first?

This solution passes. Find the decision that keeps one array element from being used twice, then predict what its inverse breaks.

def two_sum(nums, target):
seen = {}
for i, num in enumerate(nums):
complement = target - num
if complement in seen:
return [seen[complement], i]
seen[num] = i
return []

1. Which decision protects the invariant?

Check for the complement before storing the current index.Compute the complement once before the condition.Return the earlier index before the current index.

2. Suppose line 7 moves above line 5. What happens?

It can use the current element twice and return [0, 0].It stays correct, but the lookup becomes slower.It can no longer find pairs made from duplicate values.

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Do the first rep before you decide. Hear one thing worth tightening.

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