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easyBasicsSometimes asked

Left Rotate by One

Basics · 6 min

Given an integer array nums, rotate it left by one position: the first element moves to the end.

In plain English

Save the first element, shift everything one slot left, then drop the saved value at the end.

Examples

Example 1

Input: nums = [1, 2, 3, 4]
Output: [2, 3, 4, 1]
Note: First element wraps to the back.

Example 2

Input: nums = [5]
Output: [5]
Note: Single element unchanged.

Example 3

Input: nums = [7, 8]
Output: [8, 7]
Note: Two-element swap.

Constraints

1 ≤ nums.length ≤ 10^5

Clarifying questions2First question: Rotate by one, or a general k?Explore questions

Q1.Rotate by one, or a general k?
By one here; for k I'd use the three-reversal method.
Q2.Left or right rotation?
Left — first element moves to the end.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

Stash `nums[0]`, shift `nums[i] = nums[i+1]` for all i, then put the stashed value in the last slot. O(n) time, O(1) space.

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you preserve the clobbered value and avoid an out-of-bounds read.

Where candidates slip

Losing the head, or off-by-one at the upper bound.

How to stand out

Generalise to rotate-by-k via three reversals in O(1) space.

Choose

Compare the available approaches

Use constraints and trade-offs to select the route you can defend in an interview.

Learn & trace

Build each solution step by step

Move from the core idea to a visual trace, implementation, proof, and edge cases.

03.1

Approach 1 of 1 · Optimal

Shift with saved head

TimeO(n)SpaceO(1)

This is the standard, expected solution — lead with it.

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

Which element gets overwritten first, and how do you preserve it?

2Hint 2Explore hint

Save nums[0], shift everything left by one, then place it at the end.

3Hint 3Explore hint

Watch the upper bound so you don't read nums[n].

A · Core idea

Rotating left by one slides every element down one index and wraps the head to the tail. Save the first element, shift, then drop it at the end.

Why save the head first. The shift nums[i] = nums[i+1] overwrites nums[0] on the very first step, so you must capture the original head before the loop, then place it in the last slot afterward. Skipping the save loses it.

Why the loop stops at n−2. The last write is nums[n-2] = nums[n-1]; going to nums[n-1] = nums[n] reads off the end. After the loop, nums[n-1] is set to the saved head.

Rotating by k. A left rotation by k is best done with the three-reversal trick: reverse the first k, reverse the rest, reverse the whole — O(n) time, O(1) space, no big temp array. Mentioning that generalisation is the standout move.

B · Visual dry run

Interactive dry run

Select a frame to connect the decision with its visual state.

nums = [1, 2, 3, 4]

save

first = 1

Statefirst=1

Final result[2, 3, 4, 1]

C · Code with teaching comments

solution.javaJava

Shift with saved head — optimal

2 teaching notes

1
public void rotateLeft(int[] nums) {
2
    if (nums.length <= 1) return;
3
    // Save the head before it gets overwritten.
4
    int first = nums[0];
5
    for (int i = 0; i < nums.length - 1; i++) {
6
        // Slide each element one slot left.
7
        nums[i] = nums[i + 1];
8
    }
9
    nums[nums.length - 1] = first;
10
}

Yellow row = the algorithm's core logic. Amber `//` lines above it = what you'd say out loud.10 lines

D · Why it works and complexityO(n) timeO(1) spaceOpen the invariant and proofExplore the proof

ComplexityO(n) time, O(1) space — one shift per element with a single saved value — linear time, constant space.

E · Edge cases and pitfalls4nums = [5] — Single element — return unchanged.Explore edge cases

Edge cases handled

nums = [5]Single element — return unchanged.
nums = [7, 8]Becomes [8, 7].

Pitfalls

Overwriting nums[0] before saving it.
Looping to the last index and reading nums[n] out of bounds.

Practice

Try the solution yourself

Edit the optimal implementation, test another input, and observe the result.

Editable code previewpublic void rotateLeft(int[] nums) { if (nums.length <= 1) return; int first = nums[0]; for (int i = 0; i < nums.length - 1; i++) { nums[i] = nums[i + 1];Explore the playground

Try it — Java· run a complete program; method snippets need a driver

Avoid mistakes

Recognize the interview traps

Understand why common attempts fail before reviewing the corrected implementation.

Overwriting nums[0] before saving it.
Looping to the last index and reading out of bounds (nums[n]).

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Shift with saved head

Code pattern to memorize

t=a[0]; shift left; a[n-1]=t   (k: three reversals)

Rules to remember

1Save nums[0] first.
2Shift left in order.
3Drop saved value at the end.

Use this pattern when

Cyclic shifts
Rotations

Interview takeaway

Always snapshot the value an in-place shift will clobber.

Continue

Transfer the pattern to another problem

Use a direct variation or pattern sibling to prove the idea is reusable.

Transferable patternA core fresher pattern: Shift with saved head. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

Rotate Array by k
Coming soon
Three reversals: reverse[0..k), reverse[k..n), reverse all.
Reverse an Array
The primitive behind the rotation trick.

Same pattern

easyBasicsSometimes asked

Left Rotate by One

Basics · 6 min

Given an integer array nums, rotate it left by one position: the first element moves to the end.

In plain English

Save the first element, shift everything one slot left, then drop the saved value at the end.

Examples

Example 1

Input: nums = [1, 2, 3, 4]
Output: [2, 3, 4, 1]
Note: First element wraps to the back.

Example 2

Input: nums = [5]
Output: [5]
Note: Single element unchanged.

Example 3

Input: nums = [7, 8]
Output: [8, 7]
Note: Two-element swap.

Constraints

1 ≤ nums.length ≤ 10^5

Clarifying questions2First question: Rotate by one, or a general k?Explore questions

Q1.Rotate by one, or a general k?
By one here; for k I'd use the three-reversal method.
Q2.Left or right rotation?
Left — first element moves to the end.

Quick interview revision

Understand

30-second explanation

Build the interview-ready mental model before comparing implementations.

Stash `nums[0]`, shift `nums[i] = nums[i+1]` for all i, then put the stashed value in the last slot. O(n) time, O(1) space.

Interview contextWhat is testedWhere candidates slipHow to stand outExplore interview context

What's being tested

Whether you preserve the clobbered value and avoid an out-of-bounds read.

Where candidates slip

Losing the head, or off-by-one at the upper bound.

How to stand out

Generalise to rotate-by-k via three reversals in O(1) space.

Choose

Compare the available approaches

Use constraints and trade-offs to select the route you can defend in an interview.

Learn & trace

Build each solution step by step

Move from the core idea to a visual trace, implementation, proof, and edge cases.

03.1

Approach 1 of 1 · Optimal

Shift with saved head

TimeO(n)SpaceO(1)

This is the standard, expected solution — lead with it.

Stuck? Progressive hintsReveal one at a time

1Hint 1Explore hint

Which element gets overwritten first, and how do you preserve it?

2Hint 2Explore hint

Save nums[0], shift everything left by one, then place it at the end.

3Hint 3Explore hint

Watch the upper bound so you don't read nums[n].

A · Core idea

Rotating left by one slides every element down one index and wraps the head to the tail. Save the first element, shift, then drop it at the end.

Why the loop stops at n−2. The last write is nums[n-2] = nums[n-1]; going to nums[n-1] = nums[n] reads off the end. After the loop, nums[n-1] is set to the saved head.

B · Visual dry run

Interactive dry run

Select a frame to connect the decision with its visual state.

nums = [1, 2, 3, 4]

save

first = 1

Statefirst=1

Final result[2, 3, 4, 1]

C · Code with teaching comments

solution.javaJava

Shift with saved head — optimal

2 teaching notes

1
public void rotateLeft(int[] nums) {
2
    if (nums.length <= 1) return;
3
    // Save the head before it gets overwritten.
4
    int first = nums[0];
5
    for (int i = 0; i < nums.length - 1; i++) {
6
        // Slide each element one slot left.
7
        nums[i] = nums[i + 1];
8
    }
9
    nums[nums.length - 1] = first;
10
}

Yellow row = the algorithm's core logic. Amber `//` lines above it = what you'd say out loud.10 lines

D · Why it works and complexityO(n) timeO(1) spaceOpen the invariant and proofExplore the proof

ComplexityO(n) time, O(1) space — one shift per element with a single saved value — linear time, constant space.

E · Edge cases and pitfalls4nums = [5] — Single element — return unchanged.Explore edge cases

Edge cases handled

nums = [5]Single element — return unchanged.
nums = [7, 8]Becomes [8, 7].

Pitfalls

Overwriting nums[0] before saving it.
Looping to the last index and reading nums[n] out of bounds.

Practice

Try the solution yourself

Edit the optimal implementation, test another input, and observe the result.

Editable code previewpublic void rotateLeft(int[] nums) { if (nums.length <= 1) return; int first = nums[0]; for (int i = 0; i < nums.length - 1; i++) { nums[i] = nums[i + 1];Explore the playground

Try it — Java· run a complete program; method snippets need a driver

Avoid mistakes

Recognize the interview traps

Understand why common attempts fail before reviewing the corrected implementation.

Overwriting nums[0] before saving it.
Looping to the last index and reading out of bounds (nums[n]).

Revise

Carry the pattern into the interview

Quick interview revision

Pattern to remember

Shift with saved head

Code pattern to memorize

t=a[0]; shift left; a[n-1]=t   (k: three reversals)

Rules to remember

1Save nums[0] first.
2Shift left in order.
3Drop saved value at the end.

Use this pattern when

Cyclic shifts
Rotations

Interview takeaway

Always snapshot the value an in-place shift will clobber.

Continue

Transfer the pattern to another problem

Use a direct variation or pattern sibling to prove the idea is reusable.

Transferable patternA core fresher pattern: Shift with saved head. Master it on small inputs and it transfers to harder problems.

Practice queueDirect variations and pattern siblings

Direct follow-ups

Rotate Array by k
Coming soon
Three reversals: reverse[0..k), reverse[k..n), reverse all.
Reverse an Array
The primitive behind the rotation trick.

Same pattern