Adding Spaces to a String

Adding Spaces to a String - Problem

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.

Input & Output

Example 1 — Basic Case

$ Input: s = "EnjoyYourCoffee", spaces = [5,9]

› Output: "Enjoy Your Coffee"

💡 Note: Insert space before index 5 (character 'Y') and index 9 (character 'C') to get "Enjoy Your Coffee"

Example 2 — Multiple Spaces

$ Input: s = "icodeinpython", spaces = [1,5,7,9]

› Output: "i code in py thon"

💡 Note: Insert spaces at positions 1, 5, 7, and 9 to separate the string into readable segments

Example 3 — Single Space

$ Input: s = "spacing", spaces = [0]

› Output: " spacing"

💡 Note: Insert space at the beginning (index 0) to get " spacing"

Constraints

1 ≤ s.length ≤ 3 × 10⁴
s consists only of lowercase and uppercase English letters.
1 ≤ spaces.length ≤ 3 × 10⁴
0 ≤ spaces[i] ≤ s.length - 1
All the values of spaces are strictly increasing.

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to use two pointers to track the original string position and space array index simultaneously. The optimal approach builds the result in a single pass by checking if the current position matches a space position and inserting a space before adding the character. Time: O(n + m), Space: O(n + m).

Common Approaches

✓ String Concatenation

⏱️ Time: O(n²) Space: O(n)

Split the original string into segments based on space positions and concatenate them with spaces in between. This approach creates multiple string objects during concatenation.

Two Pointers with Pre-allocation

⏱️ Time: O(n + m) Space: O(n + m)

Calculate the final string length first, then use two pointers - one for the original string and one for the spaces array - to build the result character by character in a single pass.

String Concatenation — Algorithm Steps

Split string at each space position
Concatenate segments with spaces
Return final result

Visualization

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Step-by-Step Walkthrough

Extract Segments

Split original string at space positions

Add Spaces

Concatenate segments with spaces between them

Final Result

Complete string with spaces inserted

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char* solution(char* s, int* spaces, int spacesSize) {
    int len = strlen(s);
    char* result = malloc((len + spacesSize + 1) * sizeof(char));
    int resultIdx = 0;
    int prev = 0;
    
    for (int i = 0; i < spacesSize; i++) {
        int spacePos = spaces[i];
        for (int j = prev; j < spacePos; j++) {
            result[resultIdx++] = s[j];
        }
        result[resultIdx++] = ' ';
        prev = spacePos;
    }
    
    for (int j = prev; j < len; j++) {
        result[resultIdx++] = s[j];
    }
    
    result[resultIdx] = '\0';
    return result;
}

int main() {
    char s[30001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    
    char line[300001];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int spaces[30000];
    int spacesSize = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        spaces[spacesSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    char* result = solution(s, spaces, spacesSize);
    printf("%s\n", result);
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

String concatenation creates new strings each time, leading to quadratic complexity

⚠ Quadratic Growth

Space Complexity

O(n)

Multiple temporary strings are created during concatenation process

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

String Concatenation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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