LEETCODE 88:FILL FROM RIGHT PATTERN:Efficient Merge of Two Sorted Arrays | Detailed Solution

Welcome to our comprehensive solution for LeetCode Problem 88: "Merge Sorted Array." In this video, we will walk you through the problem statement, provide a step-by-step explanation of the approach, and discuss various strategies to solve the problem efficiently.
Problem Statement:
Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array. The number of elements initialized in nums1 and nums2 are m and n respectively. You may assume that nums1 has enough space to hold additional elements from nums2.
Example:
Let's consider an example to understand the problem better:
Input:
nums1 = [1,2,3,0,0,0], m = 3
nums2 = [2,5,6], n = 3
Output:
nums1 = [1,2,2,3,5,6]
Explanation:
In the example, nums1 has three initialized elements and three extra spaces for elements from nums2. The merged array should be sorted, resulting in nums1 being [1,2,2,3,5,6].
Approach:
To solve this problem, we need to merge the two arrays in a way that maintains their sorted order. Here is a step-by-step approach:
Initialize Pointers:
We will use three pointers: p1, p2, and p. Pointer p1 will track the current element in nums1 (initialized elements), p2 will track the current element in nums2, and p will track the position to insert the next element in nums1.
Comparison and Insertion:
Start comparing elements from the end of nums1 and nums2. Insert the larger element at the end of nums1. Move the pointers accordingly.
Fill Remaining Elements:
If there are any remaining elements in nums2, they will be placed in the beginning of nums1.
Detailed Steps:
Set p1 to m - 1 (last initialized element in nums1), p2 to n - 1 (last element in nums2), and p to m + n - 1 (last position in nums1).
While p1 gt -1 and p2 gt -1, compare nums1[p1] and nums2[p2]. If nums1[p1] is greater, place nums1[p1] at position p and decrement p1. Otherwise, place nums2[p2] at position p and decrement p2. Decrement p after each operation.
If there are remaining elements in nums2, copy them to the beginning of nums1. This is because nums1 might already be in place.
Edge Cases:
If nums2 is empty, nums1 is already sorted.
If nums1 is empty or has only zeros, simply copy nums2 into nums1.
Complexity Analysis:
Time Complexity: O(m + n), where m is the number of initialized elements in nums1 and n is the number of elements in nums2. We traverse both arrays once.
Space Complexity: O(1), as we are not using any extra space.
Conclusion:
This approach efficiently merges two sorted arrays by starting from the end and moving backwards. It ensures that we don't need extra space, making the solution optimal in terms of both time and space complexity.
Why This Approach Works:
By starting from the end, we avoid the need to shift elements in nums1 multiple times. This significantly reduces the time complexity and makes the algorithm more efficient. The comparison of elements from the end ensures that we place the largest element in the correct position first, thereby maintaining the sorted order without additional operations.
Practical Applications:
Merging data from two sorted lists or databases.
Combining results from two different sorted queries in a database.
Merging sorted logs from different servers in a chronological order.
Tips for Solving Similar Problems:
Always consider edge cases such as empty arrays or arrays with zeros.
Understand the constraints and optimize your solution accordingly.
Think about the most efficient way to merge without using extra space.
In the next section of the video, we will go through the implementation of this approach in a programming language, step-by-step. Make sure to follow along and try coding it yourself to better understand the logic.
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