Building software that solves real problems. Software Engineer | Frontend | Backend Shipping ideas to the internet.
Joined March 2026
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Yesterday I wore an article on how my last week went
Its is posted on LinkedIn
linkedin.com/posts/ogochukwu…
Real inventory systems maintain item → available quantity and decrement as demand arrives. That's exactly what this hash-table solution is doing. #softwareengineering Watch: youtube.com/watch?v=bLHsVKCY…
Required character missing or count already zero? Return False. No full magazine scan needed. #hashtables #optimization Watch: youtube.com/watch?v=bLHsVKCY…
Magazine = inventory. Letter = demand. Every required character simply decrements its available count. #algorithms #hashtables Watch: youtube.com/watch?v=bLHsVKCY…
M='aabc' becomes inventory: a→2, b→1, c→1. We no longer care where the characters are — only how many remain. #hashtables #python Watch: youtube.com/watch?v=bLHsVKCY…
Brute force asks: where is another unused character? Optimized asks: how many do I have left? That shift changes the whole solution. #hashtables #algorithms Watch: youtube.com/watch?v=bLHsVKCY…
Bad inventory flow: search stock records again for every unit. Better flow: item → available count → decrement. Same optimization idea. #softwareengineering Watch: youtube.com/watch?v=nK0kLzU4…
We track used positions, but what we really need is available quantity by character. That is the clue toward hashing. #hashtables #optimization Watch: youtube.com/watch?v=nK0kLzU4…
Worst case, used stores every magazine index. That gives O(M) auxiliary space. #python #bigO Watch: youtube.com/watch?v=nK0kLzU4…
If L ≈ M ≈ n, then O(L × M) becomes O(n²). Repeated scans create the quadratic growth. #algorithms Watch: youtube.com/watch?v=nK0kLzU4…
Why O(L × M)? Every required character in L may trigger a full scan through M. #algorithms #bigO Watch: youtube.com/watch?v=nK0kLzU4…
The code is correct, but every requirement can trigger another scan. Correctness and efficiency are two different questions. #algorithms Watch: youtube.com/watch?v=NC2-2qC-…
Second 'a': index 0 matches but is already used, so the search continues to index 1. #python #algorithms Watch: youtube.com/watch?v=NC2-2qC-…
Next requirement, inner loop starts again. The debugger makes the repeated magazine scan obvious. #algorithms #optimization Watch: youtube.com/watch?v=NC2-2qC-…
First 'a': match M[0] → add 0 to used → found=True → break. One requirement consumes one copy. #python #debugging Watch: youtube.com/watch?v=NC2-2qC-…
Debugger focus: char_l, j, used, found. Those 4 values show the entire brute-force search at runtime. #python #debugging Watch: youtube.com/watch?v=NC2-2qC-…
The code is correct, but every requirement can trigger another scan. Correctness and efficiency are two different questions. #algorithms Watch: youtube.com/watch?v=ZHrTOxom…
Second 'a': index 0 matches but is already used, so the search continues to index 1. #python #algorithms Watch: youtube.com/watch?v=ZHrTOxom…
Next requirement, inner loop starts again. The debugger makes the repeated magazine scan obvious. #algorithms #optimization Watch: youtube.com/watch?v=ZHrTOxom…
Yes a frequency map will help if we are only concerned about count
But that is where the hash comes in for the brute force iteration is the key to showing waste
Debugger focus: char_l, j, used, found. Those 4 values show the entire brute-force search at runtime. #python #debugging Watch: youtube.com/watch?v=ZHrTOxom…