Java  

Remove Spaces from a String (Java)

Remove Spaces from a String in Java

The Remove Spaces problem is a simple string manipulation task where we are given a string s, and we need to remove all space characters ' ' while preserving the order of the remaining characters.

This is a common problem used to understand string traversal, filtering, and efficient string building techniques.

Problem Statement

Given a string s, remove all spaces and return the modified string.

  • You are not allowed to change the order of characters

  • Only space characters should be removed

Examples

Example 1

Input:
s = "g eeks for ge eks"

Output:
"geeksforgeeks"

Example 2

Input:
s = "abc d "

Output:
"abcd"

Key Idea

The approach is straightforward:

  • Iterate through the string character by character

  • Keep only non-space characters

  • Build a new result string

Since strings are immutable in Java, using StringBuilder is the most efficient approach.

Java Solution

class Solution {
    String removeSpaces(String s) {
        StringBuilder sb = new StringBuilder();

        for (int i = 0; i < s.length(); i++) {
            char ch = s.charAt(i);

            // keep only non-space characters
            if (ch != ' ') {
                sb.append(ch);
            }
        }

        return sb.toString();
    }
}

How the Code Works

Step 1: Create a Result Container

StringBuilder sb = new StringBuilder();

  • StringBuilder is used because it allows efficient modification of strings

  • Avoids creating multiple string objects

Step 2: Traverse the String

for (int i = 0; i < s.length(); i++)

  • Loop through each character in the string

  • Access characters using charAt(i)

Step 3: Filter Spaces

if (ch != ' ')

  • If the character is not a space → append it

  • If it is a space → skip it

Step 4: Build Final String

return sb.toString();

  • Convert the StringBuilder into a final string

  • Return the result

Execution Example

Input:

"g eeks for ge eks"

Step-by-step processing:

  • g → keep

  • → remove

  • e → keep

  • e → keep

  • k → keep

  • s → keep

  • → remove

  • f → keep

  • o → keep

  • r → keep

Final Output:

geeksforgeeks

Complexity Analysis

  • Time Complexity: O(n)

    • We traverse the string once

  • Auxiliary Space: O(n)

    • In the worst case, all characters are stored (no spaces present)

Conclusion

This problem demonstrates a fundamental string processing technique. It highlights how to efficiently filter characters while maintaining order. Using StringBuilder is crucial for performance, especially when dealing with large inputs (up to 10^5 characters).

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