Find Valid Pair of Adjacent Digits in String

Find Valid Pair of Adjacent Digits in String - Problem

You are given a string s consisting only of digits. A valid pair is defined as two adjacent digits in s such that:

The first digit is not equal to the second digit
Each digit in the pair appears in s exactly as many times as its numeric value

Return the first valid pair found in the string s when traversing from left to right. If no valid pair exists, return an empty string.

Input & Output

Example 1 — Basic Valid Pair

$ Input: s = "12332"

› Output: "12"

💡 Note: Adjacent digits '1' and '2': digit 1 appears 1 time (matches its value 1), digit 2 appears 2 times (matches its value 2). Both conditions satisfied, so return "12".

Example 2 — No Valid Pair

$ Input: s = "1234"

› Output: ""

💡 Note: Check each adjacent pair: '1' appears 1 time = 1 ✓, but '2' appears 1 time ≠ 2 ✗. Continue: '2' appears 1 time ≠ 2 ✗. No pair satisfies both conditions.

Example 3 — Early Valid Pair

$ Input: s = "1122"

› Output: ""

💡 Note: Check pairs: '1' appears 2 times ≠ 1, so no pair starting with '1' can be valid. '2' appears 2 times = 2, but needs to be paired with a valid digit. No valid pairs exist.

Constraints

1 ≤ s.length ≤ 1000
s consists only of digits '0' to '9'

Visualization

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Asked in

G Google 25 M Microsoft 18 a Amazon 15

The key insight is to count digit frequencies first, then scan for adjacent pairs where both digits appear exactly as many times as their numeric values. Best approach uses hash map for O(1) frequency lookups. Time: O(n), Space: O(1).

Common Approaches

✓ Brute Force - Check Each Pair

⏱️ Time: O(n²) Space: O(1)

Check every adjacent pair by counting the frequency of each digit in the entire string. For each pair, verify both digits appear exactly as many times as their numeric values and are different.

Single-Pass Optimized

⏱️ Time: O(n) Space: O(1)

Optimize by counting frequencies first, then immediately checking each adjacent pair. We can also optimize by skipping pairs where we already know one digit doesn't satisfy the frequency condition.

Two-Pass Hash Map

⏱️ Time: O(n) Space: O(1)

Use a hash map to count the frequency of each digit in the string during the first pass. Then scan through adjacent pairs and check if both digits satisfy the frequency condition using the pre-computed counts.

Brute Force - Check Each Pair — Algorithm Steps

Iterate through each adjacent pair
For each pair, count frequencies of all digits
Check if both digits satisfy the frequency condition
Return first valid pair or empty string

Visualization

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Step-by-Step Walkthrough

Check Pair

Find adjacent digits that are different

Count Frequencies

Count how many times each digit appears

Validate

Check if count equals digit value

Code -

solution.c — C

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

int countDigit(char digit, const char* s) {
    int count = 0;
    for (int i = 0; s[i]; i++) {
        if (s[i] == digit) count++;
    }
    return count;
}

char* solution(char* s, char* result) {
    int len = strlen(s);
    
    for (int i = 0; i < len - 1; i++) {
        char digit1 = s[i];
        char digit2 = s[i + 1];
        
        if (digit1 != digit2) {
            int count1 = countDigit(digit1, s);
            int count2 = countDigit(digit2, s);
            
            if (count1 == (digit1 - '0') && count2 == (digit2 - '0')) {
                result[0] = digit1;
                result[1] = digit2;
                result[2] = '\0';
                return result;
            }
        }
    }
    
    result[0] = '\0';
    return result;
}

int main() {
    char s[1001];
    char result[3];
    
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    
    solution(s, result);
    printf("%s\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n pairs, we count frequencies in O(n) time

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables for counting, no extra data structures

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Check Each Pair — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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