Day 6 – Merge Sorted Array | Two Pointer Approach | LeetCode Challenge

🗓️ Date: July 30, 2025

📌 Challenge:

🧩 Problem: Merge Sorted Array – LeetCode #88 (Easy)

💻 Topic: Arrays, Two Pointers, In-place Merge

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✅ Problem Statement (Summary):

You are given two sorted arrays:

• nums1 of size m + n, where the last n elements are zeros for space.

• nums2 of size n.

Merge both arrays in-place into nums1, sorted in non-decreasing order.
Do not return a new array.

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

Constraints:

• Do it in-place using O(1) space

• Can you do it in O(m + n) time?

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🧠 What I Learned Today:

• Learned how to merge from the end of the arrays to avoid overwriting values in nums1.

• This is a good example of reverse two-pointer approach.

• Reduced the need for shifting elements by working backward.

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🧪 Code (C++):

class Solution {
public:
    void merge(vector<int>& a, int m, vector<int>& b, int n) {
        int l = m - 1, r = n - 1, c = m + n - 1;
        while (r >= 0 && l >= 0) {
            if (a[l] > b[r])
                a[c--] = a[l--];
            else
                a[c--] = b[r--];
        }
        while (r >= 0)
            a[c--] = b[r--];
    }
};

🕒 Time Complexity: O(m + n)

📦 Space Complexity: O(1)

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📸 LeetCode Submission Screenshot:

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📈 Progress Tracker:

✅ Day: 6 / 30

🧩 Total Problems Solved: 1

🧭 Focus Today: Two Pointers (Reverse Merge Strategy)

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🗂️ Tags:

#leetcode #30DaysChallenge #twopointers #inplace #dsa #c++ #arrays #merge