Abstract / Summary

Given a sorted list of numbers and a target value find where the target should be placed to keep the list sorted.

Note

The full problem description, including helpful examples, can be found on LeetCode.

This problem is ranked as

and has an acceptance rate of about 52% (9+ million submissions).

Typical patterns to solve this problem:

• Array
• Binary Search

Every single line of code was written by me, 100% from scratch, between August 2023 and August 2024.
However, two years later I set up this website and used various AI tools to generate most of the walkthrough / explanations below.

See this dedicated page for more details.

Overview

The provided implementation defines a single public method searchInsert inside a class named Solution. The method receives a reference to a vector of integers, nums, which is assumed to be sorted in non-decreasing order, and an integer target. Its purpose is to determine the index at which target should be inserted to maintain the sorted order. The method accomplishes this by delegating the core work to the standard library algorithm lower_bound and then converting the resulting iterator into a zero-based index.

Key Steps

Invocation of lower_bound

The first logical step is a call to lower_bound, passing the beginning and ending iterators of the vector along with the target value.

auto it = lower_bound(nums.begin(), nums.end(), target);

lower_bound performs a binary search over the range [nums.begin(), nums.end()) and returns an iterator pointing to the first element that is not less than target. If every element in the vector is smaller than target, the returned iterator equals nums.end().

Conversion of iterator to index

The second step translates the iterator returned by lower_bound into an integer index by subtracting the beginning iterator of the vector.

int index = it - nums.begin();

The subtraction yields the distance between the two iterators, which corresponds to the position where target either already exists or should be inserted.

Return of the computed index

Finally, the method returns the computed index directly to the caller.

return index;

Edge Cases Handled

When the vector is empty, nums.begin() and nums.end() are identical. lower_bound immediately returns nums.begin(), and the subtraction yields zero, correctly indicating that the insertion point for any target is at index 0.

If target is smaller than the smallest element in the vector, lower_bound returns an iterator to the first element. The subtraction yields zero, which is the correct insertion position at the front of the vector.

When target exceeds the largest element, lower_bound returns nums.end(). Subtracting nums.begin() from nums.end() yields the size of the vector, indicating that the insertion point is immediately after the last element.

If target matches one or more elements in the vector, lower_bound points to the first occurrence of that value. The subtraction therefore returns the index of the first matching element, satisfying the requirement to insert before any duplicates.

Performance Optimizations Observed

The implementation leverages the highly optimized lower_bound algorithm from the C++ Standard Library. This algorithm is implemented with careful attention to iterator categories and branch prediction, providing a fast binary search without manual loop constructs. By using the library function, the code benefits from potential inlining, cache-friendly access patterns, and compiler-level optimizations that are difficult to replicate manually.

The subtraction of iterators to obtain the index is a constant-time operation, avoiding any additional traversal of the container. The overall approach eliminates the need for explicit loops or conditional checks, resulting in a concise and efficient solution that operates directly on iterator arithmetic.

Imagine a long row of numbered toy blocks lined up from smallest to largest on the floor. There is a specific number written on a card that needs a spot in this line. If the number is already there, the task is to find exactly where it sits. If the number is missing, the task is to find the perfect place to put it so the line stays in the right order.

The code uses a special helper called lower bound that scans through the blocks very quickly. It looks at the row starting from the very first block and keeps going until it finds the first spot where a block is either the same as the target number or bigger than it. Once that spot is found, the computer calculates how far away that spot is from the very beginning of the line. This distance tells the computer exactly which position the number belongs in so that everything stays perfectly organized.

Several LLMs ranked my code quality on a scale from 1 (very poor) to 10 (outstanding):

Time: logarithmic
As a rule of thumb, the maximum input size for one second execution time is

• factorial: up to 10 elements
• exponential: up to 20 elements
• polynomial: a few thousand elements
• loglinear: up to 100,000 elements
• linear: a few million elements
• logarithmic: a billion elements
• constant: practically unlimited

I wrote a basic crawler to find other solutions to that problem.
Obviously, I did that after I solved the problem myself.

In general, AlgoMonster and WalkCCC solutions are high-quality and always worth a look.
I haven't checked most of the other solutions, but they might be helpful as well.

And of course, the LeetCode forum is always a good place (or even the best) to look for alternative approaches.

(click on a column header to sort the table)

	author
author	solutions
www	AlgoMonster	C++	Python	Java									TS
	WalkCCC	C++	Python	Java
	leetcode.ca	C++	Python	Java		Rust		JS	Go			PHP	TS
	LeetcodeHelp	C++	Python	Java
	MemoryLimitExceeded	C++	Python
	Progiez	C++	Python	Java
github	aylei					Rust
	begeekmyfriend						C
	BigEggStudy				C#
	black-shadows	C++
	bujosa					Rust
	chandansgowda		Python
	charles-wangkai			Java
	chenyanzhe	C++
	Dijkstra-Edu			Java
	DimitrisJim						C
	everthis							JS
	fartem									Ruby
	fengvyi	C++
	fpdjsns	C++
	gaosq0604		Python
	gouthampradhan			Java
	halfrost								Go
	haoel	C++
	JoshCrozier							JS
	keineahnung2345	C++
	Krazune				C#
	lennylxx						C
	linouk23	C++
	lkwq007		Python
	luangong	C++
	MacJim	C++
	muhammadfarhankt								Go
	pruthvikp		Python
	qiyuangong		Python
	ruoru					Rust
	Shamik07										Dart
	shreymehul				C#
	sonugiri1043	C++
	suxrobGM				C#
	varunu28			Java
	vthwang			Java
	XiaotaoGuo	C++
	YuriSpiridonov		Python
	zfang399	C++
AlgoMonster	C++, Python, Java, TypeScript
WalkCCC	C++, Python, Java
leetcode.ca	C++, Go, Java, JavaScript, PHP, Python, Rust, TypeScript
LeetcodeHelp	C++, Java, Python
MemoryLimitExceeded	Python, C++
Progiez	C++, Java, Python
aylei	Rust
begeekmyfriend	C
BigEggStudy	C#
black-shadows	C++, C++
bujosa	Rust
chandansgowda	Python
charles-wangkai	Java
chenyanzhe	C++
Dijkstra-Edu	Java
DimitrisJim	C
everthis	JavaScript
fartem	Ruby
fengvyi	C++
fpdjsns	C++
gaosq0604	Python
gouthampradhan	Java
halfrost	Go
haoel	C++
JoshCrozier	JavaScript
keineahnung2345	C++
Krazune	C#
lennylxx	C
linouk23	C++
lkwq007	Python
luangong	C++
MacJim	C++
muhammadfarhankt	Go
pruthvikp	Python
qiyuangong	Python
ruoru	Rust
Shamik07	Dart
shreymehul	C#
sonugiri1043	C++
suxrobGM	C#
varunu28	Java
vthwang	Java
XiaotaoGuo	C++
YuriSpiridonov	Python
zfang399	C++

There are a few extra pages giving you some insights into my setup:

• Tech Stack - my development environment and tools
• AI - how AI helped me create this website (NOT solving the problems)
• Progress - progression over time
• Quality - code quality