Day 7 – Squares of Sorted Array & Best Time to Buy/Sell Stock | Two Pointer + Greedy | LeetCode Challenge

🗓️ Date: August 1, 2025
📌 Challenge:
🧩 Problem 1: Squares of a Sorted Array – LeetCode #977 (Easy)
🧩 Problem 2: Best Time to Buy and Sell Stock – LeetCode #121 (Easy)
💻 Topic: Two Pointer, Greedy, Arrays

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✅ Problem Statement (Summary):

🔹 Problem 1: Squares of a Sorted Array
Goal: Given a sorted array (non-decreasing), return a new array of squares sorted in non-decreasing order.

Example:
Input: nums = [-4, -1, 0, 3, 10]
→ Output: [0, 1, 9, 16, 100]
Explanation: Squared → [16, 1, 0, 9, 100] → Sorted → [0, 1, 9, 16, 100]

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🧠 What I Learned (Problem 1):

• Use two pointers (start and end) to compare absolute values.

• Square the larger value, insert it at the end of result array (back to front).

• Helps handle negatives efficiently without sorting again.

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🧪 Code (C++):

class Solution {
public:
    vector<int> sortedSquares(vector<int>& nums) {
        int i = 0, j = nums.size() - 1, idx = nums.size() - 1;
        vector<int> result(nums.size());
        while (j >= i) {
            int i2 = nums[i] * nums[i];
            int j2 = nums[j] * nums[j];
            if (i2 < j2) {
                result[idx] = j2;
                j--;
            } else {
                result[idx] = i2;
                i++;
            }
            idx--;
        }
        return result;
    }
};

🕒 Time Complexity: O(n)
📦 Space Complexity: O(n)

📸 LeetCode Submission Screenshot:

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✅ Key Takeaways:

• Square and compare from both ends.

• Insert larger square from end to start.

• Avoid using extra sort for efficiency.

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🔹 Problem 2: Best Time to Buy and Sell Stock
Goal: Find max profit by choosing one day to buy and one future day to sell.

Example:
Input: prices = [7,1,5,3,6,4]
→ Output: 5
Explanation: Buy at 1, sell at 6 → profit = 6 - 1 = 5

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🧠 What I Learned (Problem 2):

• Use greedy approach with two pointers.

• Track min price so far using left pointer.

• At each step, compute potential profit and store max.

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🧪 Code (C++):

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int i = 0, j = 1, maxBuy = 0, profit, n = prices.size();
        while (j < n) {
            if (prices[i] <= prices[j]) {
                profit = prices[j] - prices[i];
                maxBuy = max(profit, maxBuy);
            } else {
                i = j;
            }
            j++;
        }
        return maxBuy;
    }
};

🕒 Time Complexity: O(n)
📦 Space Complexity: O(1)

📸 LeetCode Submission Screenshot:

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✅ Key Takeaways:

• Buy low (track min so far), sell high (future price).

• Shift buy pointer only when a lower price is found.

• One-pass greedy is optimal.

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📈 Progress Tracker:
✅ Day: 7 / 30
🧩 Total Problems Solved: 2
🧭 Focus Today: Two Pointer + Greedy

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🗂️ Tags:
#leetcode #30DaysChallenge #twopointers #greedy #cpp #dsa #arrays #squaresorted #buysellstock #codingjourney