LeetCode #2826 — MEDIUM

Sorting Three Groups

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode
The Problem

Problem Statement

You are given an integer array nums. Each element in nums is 1, 2 or 3. In each operation, you can remove an element from nums. Return the minimum number of operations to make nums non-decreasing.

Example 1:

Input: nums = [2,1,3,2,1]

Output: 3

Explanation:

One of the optimal solutions is to remove nums[0], nums[2] and nums[3].

Example 2:

Input: nums = [1,3,2,1,3,3]

Output: 2

Explanation:

One of the optimal solutions is to remove nums[1] and nums[2].

Example 3:

Input: nums = [2,2,2,2,3,3]

Output: 0

Explanation:

nums is already non-decreasing.

Constraints:

  • 1 <= nums.length <= 100
  • 1 <= nums[i] <= 3
Follow-up: Can you come up with an algorithm that runs in O(n) time complexity?
Patterns Used
🔍Modified Binary Search→📊Dynamic Programming→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums. Each element in nums is 1, 2 or 3. In each operation, you can remove an element from nums. Return the minimum number of operations to make nums non-decreasing.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Dynamic Programming

Example 1

[2,1,3,2,1]

Example 2

[1,3,2,1,3,3]

Example 3

[2,2,2,2,3,3]
Step 02

Core Insight

What unlocks the optimal approach

  • The problem asks to change the array nums to make it sorted (i.e., all the 1s are on the left of 2s, and all the 2s are on the left of 3s.).
  • We can try all the possibilities to make nums indices range in [0, i) to 0 and [i, j) to 1 and [j, n) to 2. Note the ranges are left-close and right-open; each might be empty. Namely, 0 <= i <= j <= n.
  • Count the changes we need for each possibility by comparing the expected and original values at each index position.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Upper-end input sizes
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2826: Sorting Three Groups
class Solution {
    public int minimumOperations(List<Integer> nums) {
        int[] f = new int[3];
        for (int x : nums) {
            int[] g = new int[3];
            if (x == 1) {
                g[0] = f[0];
                g[1] = Math.min(f[0], f[1]) + 1;
                g[2] = Math.min(f[0], Math.min(f[1], f[2])) + 1;
            } else if (x == 2) {
                g[0] = f[0] + 1;
                g[1] = Math.min(f[0], f[1]);
                g[2] = Math.min(f[0], Math.min(f[1], f[2])) + 1;
            } else {
                g[0] = f[0] + 1;
                g[1] = Math.min(f[0], f[1]) + 1;
                g[2] = Math.min(f[0], Math.min(f[1], f[2]));
            }
            f = g;
        }
        return Math.min(f[0], Math.min(f[1], f[2]));
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(log n)
Space
O(1)

Approach Breakdown

LINEAR SCAN
O(n) time
O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH
O(log n) time
O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

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