Find the Difference

Find the Difference - Problem

You are given two strings s and t.

String t is generated by randomly shuffling string s and then adding one more letter at a random position.

Return the letter that was added to t.

Input & Output

Example 1 — Basic Case

$ Input: s = "abcd", t = "abcde"

› Output: "e"

💡 Note: The letter 'e' was added to the end of string s to create string t.

Example 2 — Added in Middle

$ Input: s = "abc", t = "abcd"

› Output: "d"

💡 Note: The letter 'd' was added to string s. Even though t is shuffled, we can detect the extra character.

Example 3 — Single Character

$ Input: s = "", t = "y"

› Output: "y"

💡 Note: Edge case: empty string s, and t contains only the added character 'y'.

Constraints

0 ≤ s.length ≤ 1000
t.length == s.length + 1
s and t consist of lowercase English letters only

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8 f Facebook 6

The key insight is that we need to find the one character that appears differently between two strings. The XOR approach is most elegant since identical characters cancel out, leaving only the added character. Best approach: XOR bit manipulation with Time: O(n), Space: O(1).

Common Approaches

✓ XOR Bit Manipulation

⏱️ Time: O(n + m) Space: O(1)

XOR has the property that x ⊕ x = 0 and x ⊕ 0 = x. By XORing all characters from both strings, identical characters cancel out, leaving only the added character.

Character Counting with Arrays

⏱️ Time: O(n + m) Space: O(1)

Use frequency arrays to count occurrences of each character in both strings. The character with different counts is the added one.

Sort and Compare

⏱️ Time: O(n log n) Space: O(n)

Sort both strings and iterate through them simultaneously. The first position where characters differ or when one string ends reveals the added character.

Hash Map Frequency Count

⏱️ Time: O(n + m) Space: O(1)

Build a frequency map for string s, then iterate through string t decrementing counts. The character that goes negative or remains positive is the added one.

XOR Bit Manipulation — Algorithm Steps

Initialize result as 0
XOR all characters from string s
XOR all characters from string t
Result contains the added character

Visualization

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Step-by-Step Walkthrough

XOR s

XOR all characters in s: a ⊕ b ⊕ c

XOR t

XOR all characters in t: a ⊕ b ⊕ c ⊕ d

Cancel

Identical characters cancel: only d remains

Code -

solution.c — C

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

char solution(char* s, char* t) {
    int result = 0;
    
    for (int i = 0; s[i]; i++) {
        result ^= s[i];
    }
    
    for (int i = 0; t[i]; i++) {
        result ^= t[i];
    }
    
    return (char)result;
}

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;
    
    char result = solution(s, t);
    printf("%c\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n + m)

Single pass through each string where n and m are string lengths

✓ Linear Growth

Space Complexity

O(1)

Only uses one variable to store XOR result

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

XOR Bit Manipulation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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