LeetCode 35: Search Insert Position - Java Solution with Visualization

*Understanding the Binary Search Solution for Search Insert Position*

This Java method solves the "Search Insert Position" problem using the binary search algorithm.

Here's how it works:

1. Purpose:
The method determines the index where a target value should either be found or inserted in a sorted array called nums.

2. Initialization:
- `int s = 0` sets the starting index, s, to the beginning of the array.

3. Binary Search Loop:
The method uses a `while` loop, while s is less than or equal to e, to narrow the search space until the target is located or the correct insert position is found.

4. Middle Index Calculation:
`int mid = s + (e - s) / 2` calculates the middle index of the current search space.
This formula avoids potential integer overflow compared to (s + e) / 2.

5. Compare Mid-Value with Target:
- If nums at mid is equal to the target, the target is found at index mid, and the method returns mid.
- If nums at mid is greater than the target, the target lies in the left half of the array, so the ending index e is updated to mid - 1.
- If nums at mid is less than the target, the target lies in the right half of the array, so the starting index s is updated to mid + 1.

6. Return the Insert Position:
If the loop exits without finding the target, the variable s will point to the position where the target should be inserted. The method then returns s.

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### Key Features:
- Efficient: Binary search reduces the search space by half in each iteration, making it highly efficient with O(log n) time complexity.
- Flexible: Handles edge cases like inserting the target at the start or end of the array.
- Space Complexity: Uses O(1) additional space, requiring no extra memory.

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### Example Execution:
Input:
nums = 1, 3, 5, 6; target = 7

Execution Steps:
1. Starting index s = 0, ending index e = 3. Middle index mid = 1. nums at mid is 3, which is less than the target. Update s = 2.
2. Now, s = 2, e = 3. Middle index mid = 2. nums at mid is 5, which is less than the target. Update s = 3.
3. Next, s = 3, e = 3. Middle index mid = 3. nums at mid is 6, which is less than the target. Update s = 4.
4. Loop ends as s is greater than e. The method returns s = 4.

Output: The target value 7 should be inserted at index 4.

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