What is the brute force approach and complexity for Partition Array According to Given Pivot?

Give each number a group (0/1/2) and stable-sort by it with time complexity O(N log N) and space complexity O(N).

What is the optimal approach and complexity for Partition Array According to Given Pivot?

Front reader writes smaller numbers at lo++, back reader writes bigger numbers at hi-- with optimal time complexity O(N) and space complexity O(1).

How do you identify the pattern for Partition Array According to Given Pivot?

Look for key signals in the prompt: "smaller, then equal, then bigger than pivot" indicates Three fixed groups in a fixed order → partition (like Sort Colors).; "relative order must be kept" indicates The partition must be **stable**. Swapping across the array is not allowed → copy in reading order instead.; "return the rearranged array" indicates We may build a new array, so we can write each group straight into its final place from either end..

Partition Array According to Given Pivot

Medium·LeetCode #2161 Pro OnlyTricky / Edge Case

LeetCode

Problem Statement

You are given an integer array nums and a number pivot (which appears in nums at least once).

Rearrange the numbers so that:

every number smaller than pivot comes first,
then every number equal to pivot,
then every number bigger than pivot.

The catch: inside the smaller group, the numbers must stay in the same order as in nums. The same goes for the bigger group. Return the rearranged array.

Input

An integer array nums and an integer pivot that is one of the values in nums.

Output

A new array: the smaller numbers in their original order, then all the pivots, then the bigger numbers in their original order.

Example 1

Input: nums = [9, 12, 5, 10, 14, 3, 10], pivot = 10

Output: [9, 5, 3, 10, 10, 12, 14]

Smaller than 10: 9, 5, 3 (in that order in nums). Equal: 10, 10. Bigger: 12, 14 (12 comes before 14 in nums). [9, 5, 3, 10, 10, 14, 12] would be wrong, because 14 jumped ahead of 12.

Example 2

Input: nums = [-3, 4, 3, 2], pivot = 2

Output: [-3, 2, 4, 3]

Smaller: -3. Equal: 2. Bigger: 4, 3 — kept in their original order, even though 4 > 3.

Constraints

• 1 ≤ nums.length ≤ 10^5
• −10^6 ≤ nums[i] ≤ 10^6
• pivot equals at least one element of nums

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“smaller, then equal, then bigger than pivot”

→ Three fixed groups in a fixed order → partition (like Sort Colors).

“relative order must be kept”

→ The partition must be **stable**. Swapping across the array is not allowed → copy in reading order instead.

“return the rearranged array”

→ We may build a new array, so we can write each group straight into its final place from either end.

Approach Complexity Ladder

Brute Force — stable sort by groupO(N log N)O(N)

Give each number a group (0/1/2) and stable-sort by it

Better — three lists, then joinO(N)O(N)

One pass drops each number into its group's list; glue the lists

Optimal — fill the answer from both endsO(N)O(1)

Front reader writes smaller numbers at lo++, back reader writes bigger numbers at hi--

Brute Force Intuition

Give each number a group: 0 if it is smaller than pivot, 1 if equal, 2 if bigger. Sort by the group with stable_sort, which never changes the order of numbers that compare equal. So each group keeps its original order.

Recognition rule for this sub-pattern:

A few fixed categories (0/1/2, even/odd, <pivot/=pivot/>pivot). Keep each finished region behind its own pointer.

Remember this · Two Pointers

Trigger

Sorted array, pairs/triplets with a target, comparing from both ends, compacting an array in place.

Think

If I compare the elements at two positions, can I safely discard one side?

Key question

Which pointer moves, and why can the skipped elements never be the answer?

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