One Edit Distance - Practice Coding Problems

One Edit Distance - Problem

Given two strings s and t, return true if they are both one edit distance apart, otherwise return false.

A string s is said to be one distance apart from a string t if you can:

Insert exactly one character into s to get t
Delete exactly one character from s to get t
Replace exactly one character of s with a different character to get t

Input & Output

Example 1 — Insert Operation

$ Input: s = 'ab', t = 'abc'

› Output: true

💡 Note: We can insert 'c' at the end of 'ab' to get 'abc', which requires exactly one edit operation.

Example 2 — Replace Operation

$ Input: s = 'cab', t = 'ad'

› Output: false

💡 Note: Length difference is 2, which exceeds the maximum of 1 edit distance. No single edit can transform 'cab' to 'ad'.

Example 3 — Already Equal

$ Input: s = 'abc', t = 'abc'

› Output: false

💡 Note: The strings are already equal, so zero edits are needed, not exactly one edit.

Constraints

0 ≤ s.length, t.length ≤ 10⁴
s and t consist of lowercase English letters

Visualization

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Understanding the Visualization

Input

Two strings s and t to compare

Check Edit Types

Try insert, delete, or replace operations

Output

Return true if exactly one edit transforms s to t

Key Takeaway

🎯 Key Insight: Use two pointers to find the first mismatch, then check if the remaining parts are identical after applying the appropriate edit operation

Asked in

G Google 15 f Facebook 12 M Microsoft 8

The key insight is that we can use two pointers to scan both strings simultaneously, and when we find the first mismatch, we try the appropriate edit operation based on string lengths. The optimal two-pointer approach runs in O(n) time with O(1) space by handling each case (insert/delete/replace) directly without generating intermediate strings.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Generate All Single Edits	O(n × 26)	O(n)	Generate all possible strings with one edit and check if target matches any
Two Pointers - Direct Comparison	O(n)	O(1)	Use two pointers to compare strings character by character, tracking differences

Brute Force - Generate All Single Edits — Algorithm Steps

Step 1: Try deleting each character from s
Step 2: Try replacing each character in s with all possible characters
Step 3: Try inserting each character at all positions in s
Step 4: Check if any generated string equals t

Visualization

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Step-by-Step Walkthrough

Input

Two strings s='ab' and t='abc'

Generate Edits

Create all possible single edits of s

Check Match

See if any generated string equals t

Code -

solution.c — C

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

bool solution(char* s, char* t) {
    int sLen = strlen(s);
    int tLen = strlen(t);
    
    if (abs(sLen - tLen) > 1) {
        return false;
    }
    
    if (sLen == tLen) {
        return strcmp(s, t) == 0;
    }
    
    // For simplicity, check the three edit operations directly
    // This is a simplified version that checks basic cases
    
    if (sLen < tLen) {
        // Insert operation needed
        int i = 0, j = 0;
        bool inserted = false;
        while (i < sLen && j < tLen) {
            if (s[i] == t[j]) {
                i++; j++;
            } else {
                if (inserted) return false;
                inserted = true;
                j++; // skip character in t
            }
        }
        return true;
    } else {
        // Delete operation needed
        int i = 0, j = 0;
        bool deleted = false;
        while (i < sLen && j < tLen) {
            if (s[i] == t[j]) {
                i++; j++;
            } else {
                if (deleted) return false;
                deleted = true;
                i++; // skip character in s
            }
        }
        return true;
    }
}

int main() {
    char s[1001], t[1001];
    fgets(s, sizeof(s), stdin);
    fgets(t, sizeof(t), stdin);
    s[strcspn(s, "\n")] = 0;
    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 × 26)

For each position (n), try all 26 characters for insert/replace operations

✓ Linear Growth

Space Complexity

O(n)

Creating new strings for each edit operation

⚡ Linearithmic Space

23.4K Views

Medium Frequency

~15 min Avg. Time

890 Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Generate All Single Edits — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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