Background
This problem statement is a part of the Introduction to Data Structures Fun with arrays-101 on leetcode. Under the sub-heading Inserting the Items into an Array.
Problem Statement
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]
Solution approach 1
Case 1: nums1[i] < nums2[j]
- Compare one element each from both the arrays. If element in 1st array is less then the element in 2nd array. Insert the element to array 3. Increment the counter i of array 1 by 1.
Case 2: nums2[j] < nums1[i]
- If element in array 2 is less than element in array 1. Then insert the element in array 3. Increment the counter j of array 2 to the next element.
Case 3: nums1[i] == nums2[j]
- Insert nums2[j] in array 3. Increment both counter i and j by 1.
As a result, array 3 would be the final merge sorted array.
Things to observe
- We are using an extra array. Array 3, that means extra space. As asked in the question. The solution needs to be an in-place solution.
- Notice that the skeleton code below mentions no return type. Hence, even though nums3 is the merged sorted array. It cannot be returned.
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.
"""
i = 0
j = 0
nums3 = []
while i < m and j < n:
if nums1[i] < nums2[j]:
nums3.append(nums1[i])
i += 1
elif nums1[i] > nums2[j]:
nums3.append(nums2[j])
j += 1
elif nums1[i] == nums2[j]:
nums3.append(nums1[i])
i += 1
nums3.append(nums2[j])
j += 1
while i < m:
nums3.append(nums1[i])
i += 1
while j < n:
nums3.append(nums2[j])
j += 1
Solution approach 2
- Keep the merging logic same.
- Instead of returning nums3. We can do item assignment. And, copy all the elements of nums3 to nums1.
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.
"""
i = 0
j = 0
nums3 = []
while i < m and j < n:
if nums1[i] < nums2[j]:
nums3.append(nums1[i])
i += 1
elif nums1[i] > nums2[j]:
nums3.append(nums2[j])
j += 1
elif nums1[i] == nums2[j]:
nums3.append(nums1[i])
i += 1
nums3.append(nums2[j])
j += 1
while i < m:
nums3.append(nums1[i])
i += 1
while j < n:
nums3.append(nums2[j])
j += 1
for i in range(0, len(nums1)):
nums1[i] = nums3[i]
Learnings
- We have been able to merge the two sorted array.
- Time complexity - O(n)
- Time complexity of append - O(1)
- As all values of nums1 itself are changed. So, there is no need to return anything. Hence, the solution is accepted.
- List.insert might not be the way out. The reason being it is a built-in function. Secondly, insertion in at defined index adds on to the time complexity.
Things to figure out
- The solution needs to be an in-place solution. That is no extra memory needs to be used. How to solve it without using any extra space?
- Will optimizing on space complexity lead to increased time complexity?
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