Backspace String Compare

Backspace String Compare - Problem

Given two strings s and t, return true if they are equal when both are typed into empty text editors. '#' means a backspace character.

Note: After backspacing an empty text, the text will continue to remain empty.

Follow up: Can you solve it in O(1) space?

Input & Output

Example 1 — Basic Backspace

$ Input: s = "ab#c", t = "ad#c"

› Output: true

💡 Note: Both strings become "ac" after processing backspaces: s removes 'b', t removes 'd'

Example 2 — Multiple Backspaces

$ Input: s = "ab##", t = "c#d#"

› Output: true

💡 Note: Both strings become empty after processing: s removes 'b' then 'a', t removes 'd' then 'c'

Example 3 — Different Results

$ Input: s = "a#c", t = "b"

› Output: false

💡 Note: s becomes "c" after removing 'a', t remains "b", so they are not equal

Constraints

1 ≤ s.length, t.length ≤ 200
s and t only contain lowercase letters and '#' characters

Visualization

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Asked in

G Google 25 f Facebook 20 a Amazon 15

The key insight is to process strings from right to left to handle backspaces naturally. Best approach uses two pointers from the end for O(1) space. Time: O(n + m), Space: O(1).

Common Approaches

✓ Build Final Strings

⏱️ Time: O(n + m) Space: O(n + m)

Process each string character by character to build the final result after all backspaces are applied. Then compare the two final strings for equality.

Two Pointers from End

⏱️ Time: O(n + m) Space: O(1)

Start from the end of both strings and move backwards. When encountering backspaces, skip the appropriate number of characters. Compare valid characters directly without building intermediate strings.

Build Final Strings — Algorithm Steps

Process string s: add characters, remove on '#'
Process string t: add characters, remove on '#'
Compare the two final strings

Visualization

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Step-by-Step Walkthrough

Process Characters

Add characters to stack, pop on backspace

Build Final String

Convert stack to final string

Compare

Compare both final strings

Code -

solution.c — C

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

char* buildString(const char* str) {
    int len = strlen(str);
    char* stack = (char*)malloc((len + 1) * sizeof(char));
    int top = 0;
    
    for (int i = 0; i < len; i++) {
        if (str[i] == '#') {
            if (top > 0) {
                top--;
            }
        } else {
            stack[top++] = str[i];
        }
    }
    stack[top] = '\0';
    return stack;
}

bool solution(char* s, char* t) {
    char* s_built = buildString(s);
    char* t_built = buildString(t);
    bool result = strcmp(s_built, t_built) == 0;
    free(s_built);
    free(t_built);
    return result;
}

int main() {
    char s[201], t[201];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    fgets(t, sizeof(t), stdin);
    t[strcspn(t, "\n")] = 0;
    
    bool result = solution(s, t);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n + m)

Single pass through each string where n and m are lengths

✓ Linear Growth

Space Complexity

O(n + m)

Stack or string builder stores up to n + m characters

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Build Final Strings — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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