LeetCode #1089 — EASY

Duplicate Zeros

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

The Problem

Problem Statement

Given a fixed-length integer array arr, duplicate each occurrence of zero, shifting the remaining elements to the right.

Note that elements beyond the length of the original array are not written. Do the above modifications to the input array in place and do not return anything.

Example 1:

Input: arr = [1,0,2,3,0,4,5,0]
Output: [1,0,0,2,3,0,0,4]
Explanation: After calling your function, the input array is modified to: [1,0,0,2,3,0,0,4]

Example 2:

Input: arr = [1,2,3]
Output: [1,2,3]
Explanation: After calling your function, the input array is modified to: [1,2,3]

Constraints:

1 <= arr.length <= 10⁴
0 <= arr[i] <= 9

Step 01

Brute Force Baseline

Problem summary: Given a fixed-length integer array arr, duplicate each occurrence of zero, shifting the remaining elements to the right. Note that elements beyond the length of the original array are not written. Do the above modifications to the input array in place and do not return anything.

Baseline thinking

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

Pattern signal: Array · Two Pointers

Example 1

[1,0,2,3,0,4,5,0]

Example 2

[1,2,3]

Step 02

Core Insight

What unlocks the optimal approach

This is a great introductory problem for understanding and working with the concept of in-place operations. The problem statement clearly states that we are to modify the array in-place. That does not mean we cannot use another array. We just don't have to return anything.
A better way to solve this would be without using additional space. The only reason the problem statement allows you to make modifications in place is that it hints at avoiding any additional memory.
The main problem with not using additional memory is that we might override elements due to the zero duplication requirement of the problem statement. How do we get around that?
If we had enough space available, we would be able to accommodate all the elements properly. The new length would be the original length of the array plus the number of zeros. Can we use this information somehow to solve the problem?

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 #1089: Duplicate Zeros
class Solution {
    public void duplicateZeros(int[] arr) {
        int n = arr.length;
        int i = -1, k = 0;
        while (k < n) {
            ++i;
            k += arr[i] > 0 ? 1 : 2;
        }
        int j = n - 1;
        if (k == n + 1) {
            arr[j--] = 0;
            --i;
        }
        while (j >= 0) {
            arr[j] = arr[i];
            if (arr[i] == 0) {
                arr[--j] = arr[i];
            }
            --i;
            --j;
        }
    }
}

// Accepted solution for LeetCode #1089: Duplicate Zeros
func duplicateZeros(arr []int) {
	n := len(arr)
	i, k := -1, 0
	for k < n {
		i, k = i+1, k+1
		if arr[i] == 0 {
			k++
		}
	}
	j := n - 1
	if k == n+1 {
		arr[j] = 0
		i, j = i-1, j-1
	}
	for j >= 0 {
		arr[j] = arr[i]
		if arr[i] == 0 {
			j--
			arr[j] = arr[i]
		}
		i, j = i-1, j-1
	}
}

# Accepted solution for LeetCode #1089: Duplicate Zeros
class Solution:
    def duplicateZeros(self, arr: List[int]) -> None:
        """
        Do not return anything, modify arr in-place instead.
        """
        n = len(arr)
        i, k = -1, 0
        while k < n:
            i += 1
            k += 1 if arr[i] else 2
        j = n - 1
        if k == n + 1:
            arr[j] = 0
            i, j = i - 1, j - 1
        while ~j:
            if arr[i] == 0:
                arr[j] = arr[j - 1] = arr[i]
                j -= 1
            else:
                arr[j] = arr[i]
            i, j = i - 1, j - 1

// Accepted solution for LeetCode #1089: Duplicate Zeros
impl Solution {
    pub fn duplicate_zeros(arr: &mut Vec<i32>) {
        let n = arr.len();
        let mut i = 0;
        let mut j = 0;
        while j < n {
            if arr[i] == 0 {
                j += 1;
            }
            j += 1;
            i += 1;
        }
        while i > 0 {
            if arr[i - 1] == 0 {
                if j <= n {
                    arr[j - 1] = arr[i - 1];
                }
                j -= 1;
            }
            arr[j - 1] = arr[i - 1];
            i -= 1;
            j -= 1;
        }
    }
}

// Accepted solution for LeetCode #1089: Duplicate Zeros
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1089: Duplicate Zeros
// class Solution {
//     public void duplicateZeros(int[] arr) {
//         int n = arr.length;
//         int i = -1, k = 0;
//         while (k < n) {
//             ++i;
//             k += arr[i] > 0 ? 1 : 2;
//         }
//         int j = n - 1;
//         if (k == n + 1) {
//             arr[j--] = 0;
//             --i;
//         }
//         while (j >= 0) {
//             arr[j] = arr[i];
//             if (arr[i] == 0) {
//                 arr[--j] = arr[i];
//             }
//             --i;
//             --j;
//         }
//     }
// }

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.