Two Sum

Problem: [2, 7, 11, 15], target=9 → [0, 1] (2+7=9).

Brute Force: O(n²)

def two_sum(nums, target):
    for i in range(len(nums)):
        for j in range(i + 1, len(nums)):
            if nums[i] + nums[j] == target:
                return [i, j]

Hash Map: O(n)

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

Neden Hash Map O(n)?

Her eleman için:

n eleman için toplam O(n).

Varyantlar

Three Sum: [a, b, c] öyle ki a+b+c=0.

def three_sum(nums):
    nums.sort()
    result = []
    for i in range(len(nums) - 2):
        if i > 0 and nums[i] == nums[i-1]: continue
        left, right = i + 1, len(nums) - 1
        while left < right:
            s = nums[i] + nums[left] + nums[right]
            if s < 0: left += 1
            elif s > 0: right -= 1
            else:
                result.append([nums[i], nums[left], nums[right]])
                while left < right and nums[left] == nums[left+1]: left += 1
                while left < right and nums[right] == nums[right-1]: right -= 1
                left += 1; right -= 1
    return result