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3438. Find Valid Pair of Adjacent Digits in String

Description

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.
  • 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.

 

Example 1:

Input: s = "2523533"

Output: "23"

Explanation:

Digit '2' appears 2 times and digit '3' appears 3 times. Each digit in the pair "23" appears in s exactly as many times as its numeric value. Hence, the output is "23".

Example 2:

Input: s = "221"

Output: "21"

Explanation:

Digit '2' appears 2 times and digit '1' appears 1 time. Hence, the output is "21".

Example 3:

Input: s = "22"

Output: ""

Explanation:

There are no valid adjacent pairs.

 

Constraints:

  • 2 <= s.length <= 100
  • s only consists of digits from '1' to '9'.

Solutions

Solution 1: Counting

We can use an array $\textit{cnt}$ of length $10$ to record the occurrences of each digit in the string $\textit{s}$.

Then, we traverse the adjacent digit pairs in the string $\textit{s}$. If the two digits are not equal and the occurrences of these two digits are equal to the digits themselves, we have found a valid pair of adjacent digits and return it.

After traversing, if no valid pair of adjacent digits is found, we return an empty string.

The time complexity is $O(n)$, where $n$ is the length of the string $\textit{s}$. The space complexity is $O(|\Sigma|)$, where $\Sigma$ is the character set of the string $\textit{s}$. In this problem, $\Sigma = {1, 2, \ldots, 9}$.

  • class Solution {
    public String findValidPair(String s) {
        int[] cnt = new int[10];
        for (char c : s.toCharArray()) {
            ++cnt[c - '0'];
        }
        for (int i = 1; i < s.length(); ++i) {
            int x = s.charAt(i - 1) - '0';
            int y = s.charAt(i) - '0';
            if (x != y && cnt[x] == x && cnt[y] == y) {
                return s.substring(i - 1, i + 1);
            }
        }
        return "";
    }
}


  • class Solution {
public:
    string findValidPair(string s) {
        int cnt[10]{};
        for (char c : s) {
            ++cnt[c - '0'];
        }
        for (int i = 1; i < s.size(); ++i) {
            int x = s[i - 1] - '0';
            int y = s[i] - '0';
            if (x != y && cnt[x] == x && cnt[y] == y) {
                return s.substr(i - 1, 2);
            }
        }
        return "";
    }
};


  • class Solution:
    def findValidPair(self, s: str) -> str:
        cnt = [0] * 10
        for x in map(int, s):
            cnt[x] += 1
        for x, y in pairwise(map(int, s)):
            if x != y and cnt[x] == x and cnt[y] == y:
                return f"{x}{y}"
        return ""


  • func findValidPair(s string) string {
	cnt := [10]int{}
	for _, c := range s {
		cnt[c-'0']++
	}
	for i := 1; i < len(s); i++ {
		x, y := int(s[i-1]-'0'), int(s[i]-'0')
		if x != y && cnt[x] == x && cnt[y] == y {
			return s[i-1 : i+1]
		}
	}
	return ""
}


  • function findValidPair(s: string): string {
    const cnt: number[] = Array(10).fill(0);
    for (const c of s) {
        ++cnt[+c];
    }
    for (let i = 1; i < s.length; ++i) {
        const x = +s[i - 1];
        const y = +s[i];
        if (x !== y && cnt[x] === x && cnt[y] === y) {
            return `${x}${y}`;
        }
    }
    return '';
}

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