easyDSA

How do you solve the Two Sum problem efficiently?

Q: How do you solve the Two Sum problem efficiently?

**Intuition.** For each number you need its *complement* ( target - num ). Instead of scanning the rest of the array for it (O(n²)), keep a **hash map** of values you've already seen, so the complement check is O(1). One pass, **O(n)**. target = 9; at 7, is (9-7)=2 in the map? yes → done 2 <rect x="80" y="40" width="56" height="40" rx="6" fill="currentColor" fill-opacity="0.05" stroke="currentColor" stroke-opacity="0.2

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Explain Two Sum.

arrayshashing

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One-pass hash map

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


# --- demo ---
print(two_sum([2, 7, 11, 15], 9))   # [0, 1]

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