DEV Community: Oybek Rejametov

💡 [88] - Merge Sorted Arrays In-Place

Oybek Rejametov — Thu, 17 Jul 2025 12:10:34 +0000

💡 Merge Sorted Arrays In-Place

Merging two sorted arrays is a classic problem that helps you practice array manipulation, pointers, and in-place updates.

Normally, you might create a new array to store the result. But what if you must merge in-place, without using extra space?

This is exactly what this problem asks.

You're given:

nums1, a sorted array of size m + n, where the first m elements are valid and the rest are 0 placeholders.
nums2, a sorted array of size n.

Your task: merge nums2 into nums1, in-place, and keep the final result sorted.

Initial Arrays

nums1 = [1, 2, 3, 0, 0, 0]
nums2 = [2, 5, 6]

Pointers Initialization

i = 2   # Index of last non-zero element in nums1 (value: 3)
j = 2   # Index of last element in nums2 (value: 6)
k = 5   # Index of last position in nums1

We start comparing from the end and place the larger value at the end of nums1.

🔄 Step-by-Step Merge

Step 1

       i        k            j
[1, 2, 3, 0, 0, 0]    [2, 5, 6]

Compare nums1[i] = 3 and nums2[j] = 6
Since 6 > 3, place 6 at nums1[k]

✅ Result:

nums1 = [1, 2, 3, 0, 0, 6]

🔁 Move pointers:

j-- → 1 (move to next element in nums2)
k-- → 4 (move to next empty spot in nums1)

Step 2

       i     k            j 
[1, 2, 3, 0, 0, 6]    [2, 5, 6]

Compare nums1[i] = 3 and nums2[j] = 5
Since 5 > 3, place 5 at nums1[k]

✅ Result:

nums1 = [1, 2, 3, 0, 5, 6]

🔁 Move pointers:

j-- → 0
k-- → 3

Step 3

       i  k            j
[1, 2, 3, 0, 5, 6]    [2, 5, 6]

Compare nums1[i] = 3 and nums2[j] = 2
Since 3 > 2, place 3 at nums1[k]

✅ Result:

nums1 = [1, 2, 3, 3, 5, 6]

🔁 Move pointers:

i-- → 1
k-- → 2

Step 4

    i  k               j
[1, 2, 3, 3, 5, 6]    [2, 5, 6]

Compare nums1[i] = 2 and nums2[j] = 2
Since 2 <= 2, place 2 from nums2 at nums1[k]

✅ Result:

nums1 = [1, 2, 2, 3, 5, 6]

🔁 Move pointers:

j-- → -1 (finished nums2)
k-- → 1

✅ Done!

nums2 is fully merged into nums1.

Final result:

nums1 = [1, 2, 2, 3, 5, 6]

✅ Python Solution

def merge(nums1, m, nums2, n):
    i = m - 1         # last index of actual nums1 values
    j = n - 1         # last index of nums2
    k = m + n - 1     # last index of nums1 total space

    while j >= 0:
        if i >= 0 and nums1[i] > nums2[j]:
            nums1[k] = nums1[i]
            i -= 1
        else:
            nums1[k] = nums2[j]
            j -= 1
        k -= 1

🧠 "Two Pointers" ni tushunish: oddiy, ammo kuchli pattern - o'zbek tilida

Oybek Rejametov — Mon, 14 Jul 2025 13:02:49 +0000

Agar siz algoritmlarni o‘rganayotgan yoki interviewlarga tayyorlanayotgan bo‘lsangiz, albatta two pointers (ikki ko‘rsatkich) degan atamaga duch kelgansiz/kelasiz. Bu qiyin ko‘ringaniga qaramay, aslida eng oddiy va foydali patternlardan biri bo‘lib, array va string masalalarini samarali yechishda juda ko‘p ishlatiladi.

Ushbu postda sizga two pointers nima ekanini, qanday ishlashini bir nechta oddiy misollar orqali tushuntiraman.

📌 Two Pointers nima?

G‘oya juda oddiy: ikki o‘zgaruvchi (ko‘rsatkich) yordamida array yoki string bo‘ylab yuramiz.

Asosan 2 xil usul bor:

Biri boshidan, ikkinchisi oxiridan boshlanadi — va ular bir-biriga qarab harakat qiladi.
Ikkalasi ham bir yo‘nalishda harakat qiladi (masalan, sliding window yoki fast & slow pointerlar kabi).

🤓 Misol 1: Palindromni tekshirish (Valid Palindrome)

Kiritish1: "racecar"

Tushuntirish:
Chapdan o‘ngga: "racecar"
O‘ngdan chapga: "racecar"

✅ Ular bir xil → palindrom

Natija: True
Kiritish2: "hello"

Tushuntirish:
Chapdan o‘ngga: "hello"
O‘ngdan chapga: "olleh"

❌ Ular bir xil emas

Natija: False

❌ Oddiy (lekin sekin) yondashuv:

for i from 0 to length(s) - 1:
    reversed = ""

    for j from length(s) - 1 down to 0:
            reversed = reversed + s[j]

    if s != reversed:
            return false        

    return true

Vaqt murakkabligi: O(n^2), xotira: O(n)

✅ Two Pointers (tejamli)

left = 0
right = length(s) - 1

while left < right:
    if s[left] != s[right]:
        return false
    left = left + 1
    right = right - 1

return true

Qo‘shimcha xotira ishlatilmaydi
Vaqt: O(n), xotira: O(1)

Bu yerda ikki pointer stringning qarama-qarshi uchlaridan harakat qilib, belgilarni solishtiradi.

🤓 Misol 2: Takroriy elementlarni olib tashlash (Remove Duplicates from Sorted Array)

Vazifa:

Berilgan saralangan arraydan takroriy elementlarni in-place(berilgan arrayni o'zi bilan ishlash) olib tashlash va faqat noyob(unikal) elementlar bilan qolgan uzunlikni qaytarish.

Kiritish: [0, 0, 1, 1, 1, 2, 2, 3, 3, 4]
Natija: [0, 1, 2, 3, 4]

❌ Oddiy (yomon) yechim:

new_array = []

for i from 0 to length(arr) - 1:
   if arr[i] not in new_array:
   append arr[i] to new_array

return new_array

Yangi array yaratadi
In-place ishlamaydi
Qo‘shimcha O(n) xotira talab qiladi
Vaqt: O(n^2) (sababi: har safar in ishlatiladi)

✅ Two Pointers (samarali)

left = 0
right = 1

while right < len(nums):
      if nums[right] != nums[left]:
          left += 1
          nums[left] = nums[right]
      right += 1    

return left + 1

right arrayni skan qiladi
left noyob qiymatlar joylashadigan pozitsiyani ko‘rsatadi
Array o‘zida yangilanadi
Vaqt: O(n), xotira: O(1)

✅ Qachon ishlatish kerak?

Two pointers quyidagi holatlarda ayniqsa foydali:

Array saralangan bo‘lsa
Sizga juftliklar, subarraylar yoki oraliqlar kerak bo‘lsa
Sizga O(n) samaradorlikdagi algoritm kerak bo‘lsa (oddiy O(n²) o‘rniga)

🎯 Mashq uchun masalalar

Quyidagilarni yechib ko‘ring:

✍️ Yakuniy so‘z

Two pointers — intervyularda juda ko‘p uchraydigan, boshlovchilar uchun mos va samarali yondashuv.

U sizga soddaroq va tezroq kod yozishda yordam beradi.

Agar post foydali bo‘lgan bo‘lsa — albatta reaksiya qoldiring!

🧠 Understanding "Two Pointers": A Simple but Powerful Technique

Oybek Rejametov — Mon, 14 Jul 2025 12:49:02 +0000

If you're learning algorithms or preparing for coding interviews, you've probably come across the term two pointers. It sounds fancy, but it's actually one of the easiest and most useful tricks for solving array and string problems efficiently.

In this post, I'll explain what two pointers are, how they work, and walk through a few simple examples.

📌 What Are Two Pointers?

The idea is simple: use two variables (pointers) to iterate through an array or string.

There are two main styles:

Start one pointer at the beginning, the other at the end — and move them toward each other.
Move both pointers in the same direction, often at different speeds (like chasing/sliding windows).

🤓 Example 1: Valid Palindrome:

Input1: "racecar"

Explanation:
Left to right: "racecar"
Right to left: "racecar"
✅ They match → it's a palindrome

Output: True

Input2:“hello”

Explanation:
Left to right: "hello"
Right to left: "olleh"
❌ Not a match

Output: False

❌ Brute Force (not effective)

for i from 0 to length(s) - 1:
    reversed = ""

    for j from length(s) - 1 down to 0:
            reversed = reversed + s[j]

    if s != reversed:
            return false        

    return true

Time Complexity: O(n^2), Space: O(n)

✅ Two Pointers (effective)

left = 0
right = length(s) - 1

while left < right:
    if s[left] != s[right]:
        return false
    left = left + 1
    right = right - 1

return true

Without extra space
Time complexity: O(n), space: O(1)

Here, two pointers move from each end toward the middle and compare characters.

⏱ Time complexity: O(n)

🤓 Example 2: Remove Duplicates from Sorted Array

We want to remove duplicates in-place and return the new length of the array with only unique elements at the start.

Input1: [0, 0, 1, 1, 1, 2, 2, 3, 3, 4]
Output: [0, 1, 2, 3, 4]

❌ Brute Force (not effective)

new_array = []

for i from 0 to length(arr) - 1:
   if arr[i] not in new_array:
   append arr[i] to new_array

return new_array

Creates a new array
Does not modify in-place
Uses O(n) extra space
Time Complexity: O(n^2) (due to in check in array each time)

✅ Two Pointers (effective)

left = 0
right = 1

while right < len(nums):
      if nums[right] != nums[left]:
          left += 1
          nums[left] = nums[right]
      right += 1    

return left + 1

read scans through the array
write marks the next position for a unique value
Overwrites the input array in-place
Time Complexity: O(n), Space: O(1)

✅ When Should You Use It?

Use two pointers when:

The array is sorted
You need to find pairs, subarrays, or ranges
You want an efficient O(n) solution instead of brute-force O(n²)

🎯 Practice Problems

Try these to get started:

✍️ Final Thoughts

Two pointers is a beginner-friendly technique that you'll see in many interviews.

It helps you write simple and fast code.

If this post was helpful — let me know!

100 LeetCode Problems to Prepare for Your Next Coding Interview

Oybek Rejametov — Sun, 13 Jul 2025 18:14:11 +0000

DEV Community: Oybek Rejametov

💡 [88] - Merge Sorted Arrays In-Place

💡 Merge Sorted Arrays In-Place

Initial Arrays

Pointers Initialization

🔄 Step-by-Step Merge

Step 1

Step 2

Step 3

Step 4

✅ Done!

✅ Python Solution

🧠 "Two Pointers" ni tushunish: oddiy, ammo kuchli pattern - o'zbek tilida

📌 Two Pointers nima?

🤓 Misol 1: Palindromni tekshirish (Valid Palindrome)

❌ Oddiy (lekin sekin) yondashuv:

✅ Two Pointers (tejamli)

🤓 Misol 2: Takroriy elementlarni olib tashlash (Remove Duplicates from Sorted Array)

❌ Oddiy (yomon) yechim:

✅ Two Pointers (samarali)

✅ Qachon ishlatish kerak?

🎯 Mashq uchun masalalar

✍️ Yakuniy so‘z

🧠 Understanding "Two Pointers": A Simple but Powerful Technique

📌 What Are Two Pointers?

🤓 Example 1: Valid Palindrome:

❌ Brute Force (not effective)

✅ Two Pointers (effective)

🤓 Example 2: Remove Duplicates from Sorted Array

❌ Brute Force (not effective)

✅ Two Pointers (effective)

✅ When Should You Use It?

🎯 Practice Problems

✍️ Final Thoughts

100 LeetCode Problems to Prepare for Your Next Coding Interview

Two Pointers (Arrays, Strings)

HashMap/HashSet

Strings (others)

LinkedLists

Binary Search

Stack

Queues

Heap/PriorityQueue

Binary Tree

Graphs (DFS/BFS)

Graphs (Topological Sort)

Dynamic Programming