LeetCode #2109 — MEDIUM

Adding Spaces to a String

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

Solve on LeetCode
The Problem

Problem Statement

You are given a 0-indexed string s and a 0-indexed integer array spaces that describes the indices in the original string where spaces will be added. Each space should be inserted before the character at the given index.

  • For example, given s = "EnjoyYourCoffee" and spaces = [5, 9], we place spaces before 'Y' and 'C', which are at indices 5 and 9 respectively. Thus, we obtain "Enjoy Your Coffee".

Return the modified string after the spaces have been added.

Example 1:

Input: s = "LeetcodeHelpsMeLearn", spaces = [8,13,15]
Output: "Leetcode Helps Me Learn"
Explanation: 
The indices 8, 13, and 15 correspond to the underlined characters in "LeetcodeHelpsMeLearn".
We then place spaces before those characters.

Example 2:

Input: s = "icodeinpython", spaces = [1,5,7,9]
Output: "i code in py thon"
Explanation:
The indices 1, 5, 7, and 9 correspond to the underlined characters in "icodeinpython".
We then place spaces before those characters.

Example 3:

Input: s = "spacing", spaces = [0,1,2,3,4,5,6]
Output: " s p a c i n g"
Explanation:
We are also able to place spaces before the first character of the string.

Constraints:

  • 1 <= s.length <= 3 * 105
  • s consists only of lowercase and uppercase English letters.
  • 1 <= spaces.length <= 3 * 105
  • 0 <= spaces[i] <= s.length - 1
  • All the values of spaces are strictly increasing.
Patterns Used

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 a 0-indexed string s and a 0-indexed integer array spaces that describes the indices in the original string where spaces will be added. Each space should be inserted before the character at the given index. For example, given s = "EnjoyYourCoffee" and spaces = [5, 9], we place spaces before 'Y' and 'C', which are at indices 5 and 9 respectively. Thus, we obtain "Enjoy Your Coffee". Return the modified string after the spaces have been added.

Baseline thinking

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

Pattern signal: Array · Two Pointers

Example 1

"LeetcodeHelpsMeLearn"
[8,13,15]

Example 2

"icodeinpython"
[1,5,7,9]

Example 3

"spacing"
[0,1,2,3,4,5,6]
Step 02

Core Insight

What unlocks the optimal approach

  • Create a new string, initially empty, as the modified string. Iterate through the original string and append each character of the original string to the new string. However, each time you reach a character that requires a space before it, append a space before appending the character.
  • Since the array of indices for the space locations is sorted, use a pointer to keep track of the next index to place a space. Only increment the pointer once a space has been appended.
  • Ensure that your append operation can be done in O(1).
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 #2109: Adding Spaces to a String
class Solution {
    public String addSpaces(String s, int[] spaces) {
        StringBuilder ans = new StringBuilder();
        for (int i = 0, j = 0; i < s.length(); ++i) {
            if (j < spaces.length && i == spaces[j]) {
                ans.append(' ');
                ++j;
            }
            ans.append(s.charAt(i));
        }
        return ans.toString();
    }
}
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(n + m)
Space
O(n + m)

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.