LeetCode link: 88. Merge Sorted Array, difficulty: Easy.
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 + nnums2.length == n0 <= m, n <= 2001 <= m + n <= 200-10^9 <= nums1[i], nums2[j] <= 10^9
Follow up: Can you come up with an algorithm that runs in O(m + n) time?
Intuition
- This problem can be solved by using the built-in
sort()method of a programming language to sort the array, which is very simple to implement. However, the time complexity of this approach isO(n log n). - That’s why this problem usually restricts the time complexity to
O(n). How can we achieveO(n)? - We need to consider the given conditions. Both arrays are already sorted, and this condition must be leveraged.
- When reorganizing the arrays, we can either place the numbers in order from left to right, or from right to left. Which direction should we choose?
- If you don’t ask yourself this question, you’d normally just go left to right. But in that case, a problem arises: if the current number in
nums1is not the smallest, it can’t be directly replaced and needs to be stored somewhere temporarily. A simple workaround is to create a new empty array to hold the final result, and then copy that array back intonums1. But this increases the amount of code you need to write. - Therefore, it’s better to place the numbers in order from right to left.
Complexity
Time complexity
m + n
Space complexity
m + n
Python #
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 good example first.
# nums1: [1, 5, 7, 0, 0, 0, 0]
# nums2: [2, 3, 4, 8]
# result:[1, 2, 3, 4, 5, 7, 8]
i = m - 1
j = n - 1
for k in range(len(nums1) - 1, -1, -1):
if i < 0 or (j >= 0 and nums2[j] > nums1[i]):
nums1[k] = nums2[j]
j -= 1
else:
nums1[k] = nums1[i]
i -= 1
Ruby #
# @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)
# Make a good example first.
# nums1: [1, 5, 7, 0, 0, 0, 0]
# nums2: [2, 3, 4, 8]
# result:[1, 2, 3, 4, 5, 7, 8]
i = m - 1
j = n - 1
(nums1.size - 1).downto(0) do |k|
if i < 0 || (j >= 0 and nums2[j] > nums1[i])
nums1[k] = nums2[j]
j -= 1
else
nums1[k] = nums1[i]
i -= 1
end
end
end
Java #
class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
// Make a good example first.
// nums1: [1, 5, 7, 0, 0, 0, 0]
// nums2: [2, 3, 4, 8]
// result:[1, 2, 3, 4, 5, 7, 8]
var i = m - 1;
var j = n - 1;
for (var k = nums1.length - 1; k >= 0; k--) {
if (i < 0 || (j >= 0 && nums2[j] > nums1[i])) {
nums1[k] = nums2[j];
j--;
} else {
nums1[k] = nums1[i];
i--;
}
}
}
}