<reponame>bhimeshchauhan/competitive_programming
"""

Merge Sorted Array

You are given two integer arrays nums1 and nums2, sorted in 
non-decreasing order, and two integers m and n, representing 
the number of elements in nums1 and nums2 respectively.

Merge nums1 and nums2 into a single array sorted in non-decreasing order.

The final sorted array should not be returned by the function, but 
instead be stored inside the array nums1. To accommodate this, 
nums1 has a length of m + n, where the first m elements denote the 
elements that should be merged, and the last n elements are set to 
0 and should be ignored. nums2 has a length of n.

Example 1:

Input: nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3
Output: [1,2,2,3,5,6]
Explanation: The arrays we are merging are [1,2,3] and [2,5,6].
The result of the merge is [1,2,2,3,5,6] with the underlined elements 
coming from nums1.

Example 2:

Input: nums1 = [1], m = 1, nums2 = [], n = 0
Output: [1]
Explanation: The arrays we are merging are [1] and [].
The result of the merge is [1].

Example 3:

Input: nums1 = [0], m = 0, nums2 = [1], n = 1
Output: [1]
Explanation: The arrays we are merging are [] and [1].
The result of the merge is [1].
Note that because m = 0, there are no elements in nums1. The 0 is only 
there to ensure the merge result can fit in nums1.
 

Constraints:

nums1.length == m + n
nums2.length == n
0 <= m, n <= 200
1 <= m + n <= 200
-109 <= nums1[i], nums2[j] <= 109
 

Follow up: Can you come up with an algorithm that runs in O(m + n) time?

"""

# Merge and sort
"""
Time complexity : O((n+m)log(n+m)).
Space complexity : O(n), but it can vary.
"""

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.
        """
        # Write the elements of num2 into the end of nums1.
        for i in range(n):
            nums1[i + m] = nums2[i]
        
        # Sort nums1 list in-place.
        nums1.sort()
        
        
# Three Pointers (Start From the Beginning)
"""
Time complexity : O(n+m).
Space complexity : O(m).
"""

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.
        """
        # Make a copy of the first m elements of nums1.
        nums1_copy = nums1[:m] 
        
        # Read pointers for nums1Copy and nums2 respectively.
        p1 = 0
        p2 = 0
        
        # Compare elements from nums1Copy and nums2 and write the smallest to nums1.
        for p in range(n + m):
            # We also need to ensure that p1 and p2 aren't over the boundaries
            # of their respective arrays.
            if p2 >= n or (p1 < m and nums1_copy[p1] <= nums2[p2]):
                nums1[p] = nums1_copy[p1] 
                p1 += 1
            else:
                nums1[p] = nums2[p2]
                p2 += 1


# Three Pointers (Start From the End)
"""
Time complexity : O(n+m).
Space complexity : O(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
                



"""

Merge Sorted Array

You are given two integer arrays nums1 and nums2, sorted in 
non-decreasing order, and two integers m and n, representing 
the number of elements in nums1 and nums2 respectively.

Merge nums1 and nums2 into a single array sorted in non-decreasing order.

The final sorted array should not be returned by the function, but 
instead be stored inside the array nums1. To accommodate this, 
nums1 has a length of m + n, where the first m elements denote the 
elements that should be merged, and the last n elements are set to 
0 and should be ignored. nums2 has a length of n.

Example 1:

Input: nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3
Output: [1,2,2,3,5,6]
Explanation: The arrays we are merging are [1,2,3] and [2,5,6].
The result of the merge is [1,2,2,3,5,6] with the underlined elements 
coming from nums1.

Example 2:

Input: nums1 = [1], m = 1, nums2 = [], n = 0
Output: [1]
Explanation: The arrays we are merging are [1] and [].
The result of the merge is [1].

Example 3:

Input: nums1 = [0], m = 0, nums2 = [1], n = 1
Output: [1]
Explanation: The arrays we are merging are [] and [1].
The result of the merge is [1].
Note that because m = 0, there are no elements in nums1. The 0 is only 
there to ensure the merge result can fit in nums1.
 

Constraints:

nums1.length == m + n
nums2.length == n
0 <= m, n <= 200
1 <= m + n <= 200
-109 <= nums1[i], nums2[j] <= 109
 

Follow up: Can you come up with an algorithm that runs in O(m + n) time?

"""

# Merge and sort
"""
Time complexity : O((n+m)log(n+m)).
Space complexity : O(n), but it can vary.
"""

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.
        """
        # Write the elements of num2 into the end of nums1.
        for i in range(n):
            nums1[i + m] = nums2[i]
        
        # Sort nums1 list in-place.
        nums1.sort()
        
        
# Three Pointers (Start From the Beginning)
"""
Time complexity : O(n+m).
Space complexity : O(m).
"""

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.
        """
        # Make a copy of the first m elements of nums1.
        nums1_copy = nums1[:m] 
        
        # Read pointers for nums1Copy and nums2 respectively.
        p1 = 0
        p2 = 0
        
        # Compare elements from nums1Copy and nums2 and write the smallest to nums1.
        for p in range(n + m):
            # We also need to ensure that p1 and p2 aren't over the boundaries
            # of their respective arrays.
            if p2 >= n or (p1 < m and nums1_copy[p1] <= nums2[p2]):
                nums1[p] = nums1_copy[p1] 
                p1 += 1
            else:
                nums1[p] = nums2[p2]
                p2 += 1


# Three Pointers (Start From the End)
"""
Time complexity : O(n+m).
Space complexity : O(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
Merge Sorted Array  You are given two integer arrays nums1 and nums2, sorted in  non-decreasing order, and two integers m and n, representing  the number of elements in nums1 and nums2 respectively.  Merge nums1 and nums2 into a single array sorted in non-decreasing order.  The final sorted array should not be returned by the function, but  instead be stored inside the array nums1. To accommodate this,  nums1 has a length of m + n, where the first m elements denote the  elements that should be merged, and the last n elements are set to  0 and should be ignored. nums2 has a length of n.  Example 1:  Input: nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3 Output: [1,2,2,3,5,6] Explanation: The arrays we are merging are [1,2,3] and [2,5,6]. The result of the merge is [1,2,2,3,5,6] with the underlined elements  coming from nums1.  Example 2:  Input: nums1 = [1], m = 1, nums2 = [], n = 0 Output: [1] Explanation: The arrays we are merging are [1] and []. The result of the merge is [1].  Example 3:  Input: nums1 = [0], m = 0, nums2 = [1], n = 1 Output: [1] Explanation: The arrays we are merging are [] and [1]. The result of the merge is [1]. Note that because m = 0, there are no elements in nums1. The 0 is only  there to ensure the merge result can fit in nums1.    Constraints:  nums1.length == m + n nums2.length == n 0 <= m, n <= 200 1 <= m + n <= 200 -109 <= nums1[i], nums2[j] <= 109    Follow up: Can you come up with an algorithm that runs in O(m + n) time? # Merge and sort Time complexity : O((n+m)log(n+m)). Space complexity : O(n), but it can vary. Do not return anything, modify nums1 in-place instead. # Write the elements of num2 into the end of nums1. # Sort nums1 list in-place. # Three Pointers (Start From the Beginning) Time complexity : O(n+m). Space complexity : O(m). Do not return anything, modify nums1 in-place instead. # Make a copy of the first m elements of nums1. # Read pointers for nums1Copy and nums2 respectively. # Compare elements from nums1Copy and nums2 and write the smallest to nums1. # We also need to ensure that p1 and p2 aren't over the boundaries # of their respective arrays. # Three Pointers (Start From the End) Time complexity : O(n+m). Space complexity : O(1). Do not return anything, modify nums1 in-place instead. # Set p1 and p2 to point to the end of their respective arrays. # And move p backwards through the array, each time writing # the smallest value pointed at by p1 or p2.