Samri-A

Day 26 : Monday Lately, I was struggling with backtracking problems so I went back to practice recursion 1, Find 3 even digits

Approach: applied permutation backtracking method with conditions that check the first number if it is zero and the third digit if it’s even Time : O(n^3) excluding the output sorting requirement Space : O(n)

2, swap nodes in pair

Approach: recursively move 2 steps at a time and swap the value of the next and the current node value Time : O(n) Space: O(1)

3, Find Kth char in string game

Approach: the hint says simulate the game , so i did pure simulation with recursion recursively build a string by iterating on pervious string with the base case length equals or greater than k Time : O( log k) Space : O(k)

4. Unique 3 digit even : same as find 3 even , the return type is the number of those digits I also joined today’s div2 contest and as usual , I only solved the easy greedy problem

How ppl perceive me as so young is insane …. Every year, freshman students think I am freshman too … Today, I was መለከፍ by highschool student 🤦‍♀️ Literally I am experiencing backpain as old person eko 😭 endee 😁

እንኳን አደረሳችሁ🤗

ወምድርኒ ትገብር ፋሲካ ተሐጺባ በደመ ክርስቶስ ❤️ መልካም በዓል 🤍 @byte_philosopher

Join the A2SV Remote Education G7 Program! The A2SV Remote Education G7 community conversation has officially started! We are excited to invite brilliant members of the A2SV community to join our remote program and take part in this unique learning experience. 🎒Who Can Apply: Students from all universities are welcome, except the following A2SV partner universities who are not eligible for the remote program (since A2SV provides in-person education at these institutions) : AAiT, AASTU, ASTU, AUCA and University of Rwanda Requirements: Solved ≥60 problems across LeetCode & Codeforces (combined) ≥30 active days across LeetCode and Codeforces combined. Application Deadline: April 13, 2026 👉Apply here: Application Form

This week didn’t go planned 🤦‍♀️ Little excuse 😁 I was sick 🤷‍♀️

I started on Monday but failed to post each day 😕 I became mother of unaccountability 😁

Day 21: Monday 1, Heaters

approach : Sort both heater and houses list , loop over the houses and find the nearest heater and calculate it's distance and store it if it is greater than the previous calculated radius time : O(nlogn) space : O(1)

2, Capacity to ship in D days

approach : wrote a function that check if for a certain capacity we can finish shipping the weight below or on that day then binary search between maximum weight and the total weight to get the minimum capacity time : O(nlog(totalweight - maximumweight) space : O(1)

3, Binary subarray with sum

approch : initate hasmap with zero key with one value. loop over the list of nums and compute prefix sum and if the sum minus the goal is in the hashmap add the value to count and then add the prefix sum to hashmap time : O(n) space : O(n)

4, maximum sum obtained from any permutation

approch : create a lengnth of nums plus one list and mark requests range then compute prefix sum of the range and store it in the freq list then sort both freq and nums and sum the multiplity of the same index freq by nums. finnaly return sum time : O(nlogn) space: O(n)

Day 22: Tuesday 1, Design circular queue

approach : Used doubly linked list and keep track of empty the head and the tail time : O(1) for all methods space : O(1)

2, Min operation

Approach: Iterate and Check if the first number in window is zero if it’s we flip all numbers in the window and count it. finally if there any zero in the modified list return -1 else return the count Time : O(n) Space : O(1)

Day 23: Wednesday 1, Magnetic forces between two

Approach: wrote a function that take a gap and place the balls and return true if we can place all balls in a given baskets. Then , sort and binary search between 1 and the valid gap ( range between the first basket and the last basket divided by the number of balls ) using that function get the maximum true gap Time : O(nlogn) Space : O(1)

2, Minimum in rotated sorted array

Approach: using binary search if the mid is greater than the last element the smallest must be to the right else it must be to the left Time : O(logn) Space : O(1)

Day 24 : Thursday Went to home and did some stuff and generally it was wasted day , solved 1 problem for shake of consistency but I don’t remember which one it’s Day 25: Friday 1, Maximum candies allocated

Approach: wrote a function that returns true if we can allocate for all with the given number of candies. Then binary search on left 1 to the right maximum of candies using the function , we will find the maximum true value Time : O(logn) Space :O(1)

2, Search 2D matrix

Approach: use binary search to find the row the target might be found and then binary search on that row to find the target if found return True Time : O(log(row * col) Space: O(1)

3, Codeforce : 1719

Approach : sort both arrays using bubble sort and at each switch save the swaps in the list and then compare both sorted arrays and if a is greater, save the swap Finally loop over saved swaps and print all time : O(n^2 ) space: O(1)

yesterday I joined codeforce div2 contest , I solved A and then ran 😂 very hot news 😂

Back with full energy, 0 stress and 100% focus 🤗

መልካም በዓል መልካም ሰንበት 💛

በእግዚአብሔር ስም የሚመጣ ቡሩክ ነው መዝሙረ ዳዊት 117:26

@byte_philosopher

The last minute

hackathon in alx

let that sink in!!

some smart friends make your life easier 👌

One thing I learned: if you feel where the boat will take you, where you don’t want to be, don’t hope it won’t , save yourself early. And I’m not talking about a boat. 🙂

For some personal reason , for 1 or 2 weeks I m taking a break from my 100 days (weekdays) of self development challenge ✌️

Day 20: Friday leetcode: 1, NQueen 2, Minimum Depth Codeforce : div3 Contest

Happy Monday @tech_world_o1