"""Leetcode 88. Merge Sorted Array
Easy

URL: https://leetcode.com/problems/merge-sorted-array/

Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as 
one sorted array.

Note:
The number of elements initialized in nums1 and nums2 are m and n respectively.
You may assume that nums1 has enough space (size that is greater or equal to m + n) 
to hold additional elements from nums2.

Example:
Input:
nums1 = [1,2,3,0,0,0], m = 3
nums2 = [2,5,6],       n = 3
Output: [1,2,2,3,5,6]
"""

class Solution(object):
    def merge(self, nums1, m, nums2, n):
        """
        :type nums1: List[int]
        :type m: int
        :type nums2: List[int]
        :type n: int
        :rtype: None Do not return anything, modify nums1 in-place instead.

        Time complexity: O(m+n).
        Space complexity: O(1).
        """
        i = m - 1
        j = n - 1
        for k in reversed(range(len(nums1))):
            if i >= 0 and j >= 0:
                if nums1[i] > nums2[j]:
                    nums1[k] = nums1[i]
                    i -= 1
                else:
                    nums1[k] = nums2[j]
                    j -= 1
            elif i >= 0 and j < 0:
                nums1[k] = nums1[i]
                i -= 1
            elif i < 0 and j >= 0:
                nums1[k] = nums2[j]
                j -= 1


def main():
    # Ans: [1,2,2,3,5,6]
    nums1 = [1,2,3,0,0,0]
    m = 3
    nums2 = [2,5,6]
    n = 3
    Solution().merge(nums1, m, nums2, n)
    print nums1


if __name__ == '__main__':
    main()

"""Leetcode 88. Merge Sorted Array
Easy

URL: https://leetcode.com/problems/merge-sorted-array/

Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as 
one sorted array.

Note:
The number of elements initialized in nums1 and nums2 are m and n respectively.
You may assume that nums1 has enough space (size that is greater or equal to m + n) 
to hold additional elements from nums2.

Example:
Input:
nums1 = [1,2,3,0,0,0], m = 3
nums2 = [2,5,6],       n = 3
Output: [1,2,2,3,5,6]
"""

class Solution(object):
    def merge(self, nums1, m, nums2, n):
        """
        :type nums1: List[int]
        :type m: int
        :type nums2: List[int]
        :type n: int
        :rtype: None Do not return anything, modify nums1 in-place instead.

        Time complexity: O(m+n).
        Space complexity: O(1).
        """
        i = m - 1
        j = n - 1

        # Start from the last element of nums1 to replace.
        for k in reversed(range(len(nums1))):
            # Compare the last elements of nums1 and nums2.
            if i >= 0 and j >= 0:
                if nums1[i] > nums2[j]:
                    nums1[k] = nums1[i]
                    i -= 1
                else:
                    nums1[k] = nums2[j]
                    j -= 1
            elif i >= 0 and j < 0:
                nums1[k] = nums1[i]
                i -= 1
            elif i < 0 and j >= 0:
                nums1[k] = nums2[j]
                j -= 1


def main():
    # Ans: [1,2,2,3,5,6]
    nums1 = [1,2,3,0,0,0]
    m = 3
    nums2 = [2,5,6]
    n = 3
    Solution().merge(nums1, m, nums2, n)
    print nums1


if __name__ == '__main__':
    main()

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