Strings Differ by One Character

Strings Differ by One Character - Problem

Given a list of strings dict where all the strings are of the same length.

Return true if there are 2 strings that only differ by 1 character in the same index, otherwise return false.

Note: Two strings differ by one character means exactly one position has different characters while all other positions have the same character.

Input & Output

Example 1 — Basic Case

$ Input: dict = ["abcd","acbd","aacd"]

› Output: true

💡 Note: Strings "abcd" and "aacd" differ by exactly 1 character at index 1 (b vs a)

Example 2 — No Match

$ Input: dict = ["ab","cd","yz"]

› Output: false

💡 Note: No two strings differ by exactly 1 character. Each pair differs by 2 characters

Example 3 — Multiple Differences

$ Input: dict = ["abcd","cccc","abcy","ghij"]

› Output: true

💡 Note: Strings "abcd" and "abcy" differ by exactly 1 character at index 3 (d vs y)

Constraints

1 ≤ dict.length ≤ 10⁵
1 ≤ dict[i].length ≤ 20
dict[i] consists of lowercase English letters only
All strings in dict have the same length

Visualization

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Asked in

G Google 35 f Facebook 28 a Amazon 22

The key insight is to use pattern matching by generating masks for each character position. The hash map approach is optimal with O(n×m²) time complexity by creating patterns like "ab*" and detecting duplicates.

Common Approaches

✓ Brute Force Comparison

⏱️ Time: O(n² × m) Space: O(1)

For each pair of strings, compare them character by character. If exactly one character differs, return true. This approach checks all possible combinations.

Hash Map with Patterns

⏱️ Time: O(n × m²) Space: O(n × m²)

For each string, generate patterns by replacing each character position with '*'. If two strings differ by one character, they will generate the same pattern. Use a hash map to detect duplicate patterns.

Brute Force Comparison — Algorithm Steps

Compare each string with every other string
Count differences between each pair
Return true if any pair differs by exactly 1 character

Visualization

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Step-by-Step Walkthrough

Compare Pairs

Check each pair of strings

Count Differences

Count character differences

Result

Return true if exactly 1 difference found

Code -

solution.c — C

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

bool solution(char dict[][100], int n, int len) {
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            int diffCount = 0;
            for (int k = 0; k < len; k++) {
                if (dict[i][k] != dict[j][k]) {
                    diffCount++;
                    if (diffCount > 1) break;
                }
            }
            if (diffCount == 1) return true;
        }
    }
    return false;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    char dict[100][100];
    int n = 0, len = 0;
    char *token = strtok(line, "[],\" \n");
    while (token != NULL && n < 100) {
        strcpy(dict[n], token);
        if (len == 0) len = strlen(token);
        n++;
        token = strtok(NULL, "[],\" \n");
    }
    printf(solution(dict, n, len) ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n² × m)

n² pairs to compare, each comparison takes m characters

✓ Linear Growth

Space Complexity

O(1)

Only using variables for comparison, no extra data structures

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force Comparison — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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