LeetCode #2460 — EASY

Apply Operations to an Array

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given a 0-indexed array nums of size n consisting of non-negative integers.

You need to apply n - 1 operations to this array where, in the i^th operation (0-indexed), you will apply the following on the i^th element of nums:

If nums[i] == nums[i + 1], then multiply nums[i] by 2 and set nums[i + 1] to 0. Otherwise, you skip this operation.

After performing all the operations, shift all the 0's to the end of the array.

For example, the array [1,0,2,0,0,1] after shifting all its 0's to the end, is [1,2,1,0,0,0].

Return the resulting array.

Note that the operations are applied sequentially, not all at once.

Example 1:

Input: nums = [1,2,2,1,1,0]
Output: [1,4,2,0,0,0]
Explanation: We do the following operations:
- i = 0: nums[0] and nums[1] are not equal, so we skip this operation.
- i = 1: nums[1] and nums[2] are equal, we multiply nums[1] by 2 and change nums[2] to 0. The array becomes [1,4,0,1,1,0].
- i = 2: nums[2] and nums[3] are not equal, so we skip this operation.
- i = 3: nums[3] and nums[4] are equal, we multiply nums[3] by 2 and change nums[4] to 0. The array becomes [1,4,0,2,0,0].
- i = 4: nums[4] and nums[5] are equal, we multiply nums[4] by 2 and change nums[5] to 0. The array becomes [1,4,0,2,0,0].
After that, we shift the 0's to the end, which gives the array [1,4,2,0,0,0].

Example 2:

Input: nums = [0,1]
Output: [1,0]
Explanation: No operation can be applied, we just shift the 0 to the end.

Constraints:

2 <= nums.length <= 2000
0 <= nums[i] <= 1000

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed array nums of size n consisting of non-negative integers. You need to apply n - 1 operations to this array where, in the ith operation (0-indexed), you will apply the following on the ith element of nums: If nums[i] == nums[i + 1], then multiply nums[i] by 2 and set nums[i + 1] to 0. Otherwise, you skip this operation. After performing all the operations, shift all the 0's to the end of the array. For example, the array [1,0,2,0,0,1] after shifting all its 0's to the end, is [1,2,1,0,0,0]. Return the resulting array. Note that the operations are applied sequentially, not all at once.

Baseline thinking

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

Pattern signal: Array · Two Pointers

Example 1

[1,2,2,1,1,0]

Example 2

[0,1]

Core Insight

What unlocks the optimal approach

Iterate over the array and simulate the described process.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
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 #2460: Apply Operations to an Array
class Solution {
    public int[] applyOperations(int[] nums) {
        int n = nums.length;
        for (int i = 0; i < n - 1; ++i) {
            if (nums[i] == nums[i + 1]) {
                nums[i] <<= 1;
                nums[i + 1] = 0;
            }
        }
        int[] ans = new int[n];
        int i = 0;
        for (int x : nums) {
            if (x > 0) {
                ans[i++] = x;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2460: Apply Operations to an Array
func applyOperations(nums []int) []int {
	n := len(nums)
	for i := 0; i < n-1; i++ {
		if nums[i] == nums[i+1] {
			nums[i] <<= 1
			nums[i+1] = 0
		}
	}
	ans := make([]int, n)
	i := 0
	for _, x := range nums {
		if x > 0 {
			ans[i] = x
			i++
		}
	}
	return ans
}

# Accepted solution for LeetCode #2460: Apply Operations to an Array
class Solution:
    def applyOperations(self, nums: List[int]) -> List[int]:
        n = len(nums)
        for i in range(n - 1):
            if nums[i] == nums[i + 1]:
                nums[i] <<= 1
                nums[i + 1] = 0
        ans = [0] * n
        i = 0
        for x in nums:
            if x:
                ans[i] = x
                i += 1
        return ans

// Accepted solution for LeetCode #2460: Apply Operations to an Array
impl Solution {
    pub fn apply_operations(nums: Vec<i32>) -> Vec<i32> {
        let mut nums = nums;

        for i in 0..nums.len() - 1 {
            if nums[i] == nums[i + 1] {
                nums[i] <<= 1;
                nums[i + 1] = 0;
            }
        }

        let mut cur = 0;
        for i in 0..nums.len() {
            if nums[i] != 0 {
                nums.swap(i, cur);
                cur += 1;
            }
        }

        nums
    }
}

// Accepted solution for LeetCode #2460: Apply Operations to an Array
function applyOperations(nums: number[]): number[] {
    const n = nums.length;
    for (let i = 0; i < n - 1; ++i) {
        if (nums[i] === nums[i + 1]) {
            nums[i] <<= 1;
            nums[i + 1] = 0;
        }
    }
    const ans: number[] = Array(n).fill(0);
    let i = 0;
    for (const x of nums) {
        if (x !== 0) {
            ans[i++] = x;
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

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

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

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.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.