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‎day-26/README.md‎

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# Day 26
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## ⭐️ Move Zeroes – 26.1
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### 🔗 Problem
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[LeetCode #283 – Move Zeroes](https://leetcode.com/problems/move-zeroes/)
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### 🧠 Core Idea
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Use a two-pointer approach to shift all non-zero elements forward while maintaining their relative order.
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One pointer (last_non_zero) marks the next position to place a non-zero, while the other scans through the array.
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After placing all non-zero values, fill the remaining positions with zeros.
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This ensures the operation is in-place and preserves order.
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### 📊 Example
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Input: `nums = [0,1,0,3,12]`
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Output: [1,3,12,0,0]
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### ⏱️ Complexity
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- Time: O(n) – Single pass through the array plus one pass to fill zeros
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- Space: O(1) – In-place, no extra structures
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👉 See full code in [move_zeroes.py](https://github.com/lyushher/LeetCode-Python-Easy-DSA/blob/main/day-26/move_zeroes.py)
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---
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## ⭐️ Find All Numbers Disappeared in an Array – 26.2
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### 🔗 Problem
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[LeetCode #448 – Find All Numbers Disappeared in an Array](https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/)
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### 🧠 Core Idea
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Since numbers are within the range [1, n], we can mark visited indices by flipping the sign of the element at abs(x) - 1.
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After one pass, the indices that remain positive indicate the missing numbers.
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Collect those indices + 1 as the result.
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This avoids extra memory usage and leverages in-place marking.
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### 📊 Example
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Input: `nums = [4,3,2,7,8,2,3,1]`
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Output: [5,6] -> Because 5 and 6 are not present in the array
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### ⏱️ Complexity
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- Time: O(n) – Single pass to mark and one pass to collect
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-
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- Space: O(1) – In-place marking (excluding output list)
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👉 See full code in [find_all_numbers_disappeared_in_an_array.py](https://github.com/lyushher/LeetCode-Python-Easy-DSA/blob/main/day-26/find_all_numbers_disappeared_in_an_array.py)

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