🧠 Day 20: Search in a Sorted 2D Matrix
How I Grasped the Logic to Search Efficiently in a Sorted 2D Matrix
Hi, I’m Richa — a Senior Frontend Engineer with 5+ years of experience building scalable, production-grade web interfaces for enterprise and consumer applications. I work primarily with React, TypeScript, and modern frontend architectures, focusing on component systems, performance, and maintainability. Most of my experience comes from building real-world products in regulated domains like banking and insurance, where clarity, reliability, and long-term ownership matter more than quick demos. Through this blog, I write about frontend engineering fundamentals, scalable UI design, problem-solving, and the lessons I’ve learned working on large codebases. My goal is to share practical insights — not shortcuts — for developers who want to grow strong engineering foundations. I also mentor early-career developers and strongly believe that curiosity, asking the right questions, and understanding why something works are more important than memorizing tools. If you’re serious about improving as an engineer, you’re in the right place.
Yesterday, I solved this problem using brute force.
Today, I’m trying to understand the optimized solution—not memorize it.
And honestly?
At first glance, this code scared me.
So I decided to pause everything and trace it like a beginner.
🧩 The Matrix We’ll Use for Understanding
mat = [
[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12]
]
target = 8
Rows = 3
Cols = 4
🧠 Step 1: Pretend This Is a 1D Array
I did not flatten the matrix in code.
I only flattened it in my head.
Index: 0 1 2 3 4 5 6 7 8 9 10 11
Value: [1, 2, 3, 4 | 5, 6, 7, 8 | 9,10,11,12]
This is the key mental shift.
🧠 Step 2: Binary Search Setup
let left = 0;
let right = rows * cols - 1; // 11
So now:
left = 0
right = 11
🔁 WHILE LOOP — Iteration by Iteration
We now enter:
while (left <= right)
Let’s go iteration by iteration, like a beginner.
🔄 Iteration 1
Step A: Find mid
mid = (0 + 11) / 2 = 5
Step B: Convert mid → row & column
row = Math.floor(5 / 4) = 1
col = 5 % 4 = 1
ASCII view:
Matrix:
[
[1, 2, 3, 4],
[5, 6, 7, 8], ← row = 1
[9,10,11,12]
]
↑
col = 1
So:
midValue = mat[1][1] = 6
Step C: Compare
6 < 8
So target must be on the right side.
left = mid + 1 = 6
🔄 Iteration 2
Now:
left = 6
right = 11
Step A: Find mid
mid = (6 + 11) / 2 = 8
Step B: Convert mid → row & column
row = Math.floor(8 / 4) = 2
col = 8 % 4 = 0
ASCII:
[
[1, 2, 3, 4],
[5, 6, 7, 8],
[9,10,11,12] ← row = 2
↑
col = 0
So:
midValue = mat[2][0] = 9
Step C: Compare
9 > 8
So target must be on the left side.
right = mid - 1 = 7
🔄 Iteration 3
Now:
left = 6
right = 7
Step A: Find mid
mid = (6 + 7) / 2 = 6
Step B: Convert mid → row & column
row = Math.floor(6 / 4) = 1
col = 6 % 4 = 2
ASCII:
[
[1, 2, 3, 4],
[5, 6, 7, 8], ← row = 1
↑
col = 2
]
midValue = 7
Step C: Compare
7 < 8
So:
left = mid + 1 = 7
🔄 Iteration 4 (Final)
left = 7
right = 7
Step A: mid
mid = 7
Step B: Convert mid
row = Math.floor(7 / 4) = 1
col = 7 % 4 = 3
ASCII:
[
[1, 2, 3, 4],
[5, 6, 7, 8], ← row = 1
↑
col = 3
]
midValue = 8
🎯 FOUND IT
midValue === target
return true;
📊 Dry Run Table (Beginner Gold)
| Iter | Left | Right | Mid | Row | Col | Value | Action |
| 1 | 0 | 11 | 5 | 1 | 1 | 6 | Move Right |
| 2 | 6 | 11 | 8 | 2 | 0 | 9 | Move Left |
| 3 | 6 | 7 | 6 | 1 | 2 | 7 | Move Right |
| 4 | 7 | 7 | 7 | 1 | 3 | 8 | Found |
🧠 Final Beginner Realization
This problem is not about matrices.
It’s about:
Seeing sorted data
Applying binary search
Learning how index math maps back to 2D
Once you trace it like this once,
your brain stops panicking.
💬 One Honest Line Before Ending
If you couldn’t think of this solution on your own —
you’re normal.
Understanding > memorizing.
