How to Build Logic to Solve Problems?

Before writing code, it’s essential to develop a logical approach. Here’s a structured way to think and solve problems systematically:

📌 Step-by-Step Thought Process for Problem Solving

1️⃣ Understand the Problem

Read the problem statement carefully.
Identify input(s) and expected output(s).
Think of edge cases (e.g., empty array, single-element array).

2️⃣ Break the Problem into Small Steps

Instead of thinking about the whole solution at once, break it down.
Consider how you'd manually solve it step by step.

3️⃣ Identify the Pattern

Most problems fall into common categories:
Sorting, Searching, Hashing, Two-Pointer, Recursion, Graphs, Trees.
Think if a common data structure or algorithm fits (e.g., stacks for "last in, first out").

4️⃣ Think of an Approach

Brute Force Approach: Solve the problem in the simplest way possible.
Optimized Approach: Reduce unnecessary computations (use HashMap, Binary Search, etc.).
Consider time complexity (O(n), O(log n), O(n^2), etc.).

5️⃣ Implement the Solution

Translate the logical steps into code.

6️⃣ Test with Edge Cases

Check if your solution works for small inputs, empty cases, and large inputs.

💡 Applying This Thought Process to Problems

Now, let’s break down the logic behind each problem.

1️⃣ Reverse an Array

🧠 Thinking Process

You need to reverse the order of elements.
Use two-pointer approach:
- One pointer at the start, one at the end.
- Swap elements, move both pointers towards the center.
Alternative: Use reverse() if allowed.

✅ Logic in Steps

Set two pointers: left = 0, right = arr.length - 1.
Swap elements at left and right.
Move left++ and right-- until they meet.

2️⃣ Find Maximum and Minimum in an Array

🧠 Thinking Process

Scan through the array, keeping track of min and max.
Brute force: Sort and pick first & last (not efficient).
Optimized: Use a single pass.

✅ Logic in Steps

Initialize min = arr[0], max = arr[0].
Loop through the array:
- If arr[i] > max, update max.
- If arr[i] < min, update min.
Return {min, max}.

3️⃣ Check if a String is a Palindrome

🧠 Thinking Process

A palindrome reads the same forward and backward.
Two-pointer method: Compare characters from the start and end.
Alternative: Reverse string and compare with the original.

✅ Logic in Steps

Set left = 0, right = str.length - 1.
Compare str[left] and str[right].
If they don’t match, return false.
Move left++ and right-- until they meet.

4️⃣ Implement a Stack Using an Object

🧠 Thinking Process

A stack follows Last In, First Out (LIFO).
Operations needed:
- Push: Add element at the end.
- Pop: Remove the last added element.
- Peek: Return the last element without removing.
Two approaches:
- Use array methods (push, pop).
- Use an object with a counter.

✅ Logic in Steps

Maintain a top index.
For push, increase top and store value.
For pop, return value at top, then decrease top.
For peek, return value at top without removing.

5️⃣ Implement a Queue Using an Object

🧠 Thinking Process

A queue follows First In, First Out (FIFO).
Operations needed:
- Enqueue: Add element at the end.
- Dequeue: Remove the first added element.
Two approaches:
- Use array methods (push, shift).
- Use an object with front and rear pointers.

✅ Logic in Steps

Maintain front and rear indices.
For enqueue, store value at rear, then increase rear.
For dequeue, return value at front, then increase front.

6️⃣ Find the First Non-Repeating Character

🧠 Thinking Process

Traverse the string and count occurrences.
Find the first character with count = 1.
Use HashMap (object) for quick lookup.

✅ Logic in Steps

Create a charCount object to count occurrences.
Loop through str and update charCount.
Loop again and return the first character with charCount[c] === 1.

7️⃣ Merge Two Sorted Arrays

🧠 Thinking Process

Since arrays are sorted, use a two-pointer approach:
- Compare elements at both pointers.
- Add the smaller element to the result array.
- Move the pointer forward.
Alternative: Merge both and sort (not optimal).

✅ Logic in Steps

Create two pointers i = 0, j = 0.
Compare arr1[i] and arr2[j].
Add the smaller one to the result and move the pointer.
Append remaining elements.

🔹 Final Tips for Building Logic

✅ Use Examples: Always test with small inputs ([1,2,3]).
✅ Draw It Out: Visualizing with diagrams helps.
✅ Break It Down: Think in small steps, solve manually first.
✅ Recognize Patterns: Many problems repeat structures (two-pointers, hash maps, recursion).
✅ Think About Edge Cases: Empty input, single element, very large input.

Would you like me to provide practice questions or go deeper into specific problem types? 🚀