###### tags: `Week_1`, `LeetCode 75 Plan`
# Recommended Topics ..

## [1. Two Sum](https://leetcode.com/problems/two-sum/)
harry:

function twoSum(nums: number[], target: number): number[] {
    let seenIndex = {};
    for(let i = 0; i < nums.length; i++) {
        let remain = target - nums[i];
        let foundIndex = seenIndex[remain]; 
        if(foundIndex!==undefined) {
            return [foundIndex, i];
        } else {                             
            seenIndex[nums[i]] = i;
        }
    }
};

sam:

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        
        diff_map = dict()
        
        for i, num in enumerate(nums):
            if num not in diff_map:
                diff_map[target-num] = i
            else:
                return [i, diff_map[num]]

## [88. Merge Sorted Array](https://leetcode.com/problems/merge-sorted-array/)
harry:

# @param {Integer[]} nums1
# @param {Integer} m
# @param {Integer[]} nums2
# @param {Integer} n
# @return {Void} Do not return anything, modify nums1 in-place instead.
def merge(nums1, m, nums2, n)
    return nums1 if n.zero?
    
    if m.zero?
        n.times.each {|i| nums1[i] = nums2[i] }
        return nums1
    end
    
    n.times do |i|
        nums1[i+m] = nums2[i]
    end
    
    nums1.sort!
end

# M+N
def merge(nums1, m, nums2, n)
  while n > 0
    nums1[m + n - 1], m, n = if m > 0 && nums1[m - 1] > nums2[n - 1]
                               [nums1[m - 1], m - 1, n]
                             else 
                               [nums2[n - 1], m, n - 1]
                             end
  end
end

# M+N
# @param {Integer[]} nums1
# @param {Integer} m
# @param {Integer[]} nums2
# @param {Integer} n
# @return {Void} Do not return anything, modify nums1 in-place instead.
def merge(nums1, m, nums2, n)
    while n.positive?
        if m.positive? && nums1[m-1] > nums2[n-1]
            nums1[m+n-1] = nums1[m-1]
            m-=1
        else
            nums1[m+n-1] = nums2[n-1]
            n-=1
        end
    end
end

Sam:

1. Time = O(N+Nlog(N))
class Solution:
    def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
        """
        Do not return anything, modify nums1 in-place instead.
        """

        start_replace_index = 0
        for i in range(len(nums1)-1, -1, -1):
            # print(i, nums1[i])
            if nums1[i] != 0:
                start_replace_index = i + 1
                break
        
        # print(start_replace_index)
        for num in nums2:
            nums1[start_replace_index] = num
            start_replace_index += 1

        nums1.sort()

2. Time = O(M+N), Space (1)
class Solution:
    def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
        """
        Do not return anything, modify nums1 in-place instead.
        """
        
        # Set p1 and p2 to point to the end of their respective arrays.
        p1 = m - 1
        p2 = n - 1
    
        # And move p backwards through the array, each time writing
        # the smallest value pointed at by p1 or p2.
        for p in range(n + m - 1, -1, -1):
            if p2 < 0:
                break
            if p1 >= 0 and nums1[p1] > nums2[p2]:
                nums1[p] = nums1[p1]
                p1 -= 1
            else:
                nums1[p] = nums2[p2]
                p2 -= 1