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: "23"

💡 Note: Adjacent digits '2' and '3': digit 2 appears 2 times (matches its value), digit 3 appears 2 times (doesn't match its value 3). Continue checking. Next pair '3' and '3' are equal, skip. Pair '3' and '2': 3 appears 2 times ≠ 3, invalid. But wait - let me recheck: in "12332", pair "23" at positions 1,2 has 2 appearing 2 times = 2 ✓, and 3 appearing 2 times ≠ 3 ✗. Actually, looking for the first valid pair correctly.

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: "12"

💡 Note: First pair '1','2': digit 1 appears 2 times ≠ 1 ✗. Second pair '1','2': same issue. Third pair '2','2': digits are equal, skip. Need to find where both digits have correct frequencies.

Constraints

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

Visualization

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Understanding the Visualization

Input Analysis

String with digit frequency requirements

Frequency Check

Each digit must appear exactly as many times as its value

Pair Validation

Find first adjacent pair where both satisfy frequency rule

Key Takeaway

🎯 Key Insight: Pre-compute digit frequencies, then scan adjacent pairs for simultaneous validation

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

Approach	Time	Space	Notes
✓ Single-Pass Optimized	O(n)	O(1)	Count frequencies and check pairs in a single pass with early termination
Two-Pass Hash Map	O(n)	O(1)	First pass counts all digits, second pass finds first valid pair
Brute Force - Check Each Pair	O(n²)	O(1)	For each adjacent pair, count all digit frequencies to validate

Single-Pass Optimized — Algorithm Steps

Count all digit frequencies in one pass
Iterate through adjacent pairs
Use pre-computed frequencies for instant validation
Return first valid pair or empty string

Visualization

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Step-by-Step Walkthrough

Pre-count

Build frequency array for all digits

Smart Check

Skip invalid pairs early, validate efficiently

First Match

Return immediately when valid pair found

Code -

solution.c — C

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

char* solution(char* s, char* result) {
    int len = strlen(s);
    int freq[10] = {0}; // digits 0-9
    
    // Count frequencies
    for (int i = 0; i < len; i++) {
        freq[s[i] - '0']++;
    }
    
    // Check adjacent pairs with optimizations
    for (int i = 0; i < len - 1; i++) {
        char digit1 = s[i];
        char digit2 = s[i + 1];
        
        // Skip if digits are same
        if (digit1 == digit2) continue;
        
        // Quick validation
        int val1 = digit1 - '0';
        int val2 = digit2 - '0';
        if (freq[val1] == val1 && freq[val2] == val2) {
            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(n)

One pass to count + one pass to check pairs = O(n)

✓ Linear Growth

Space Complexity

O(1)

Hash map stores at most 10 digit frequencies (0-9)

✓ Linear Space

12.4K Views

Medium Frequency

~15 min Avg. Time

567 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Single-Pass Optimized — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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