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3216. Lexicographically Smallest String After a Swap

Description

Given a string s containing only digits, return the lexicographically smallest string that can be obtained after swapping adjacent digits in s with the same parity at most once.

Digits have the same parity if both are odd or both are even. For example, 5 and 9, as well as 2 and 4, have the same parity, while 6 and 9 do not.

 

Example 1:

Input: s = "45320"

Output: "43520"

Explanation:

s[1] == '5' and s[2] == '3' both have the same parity, and swapping them results in the lexicographically smallest string.

Example 2:

Input: s = "001"

Output: "001"

Explanation:

There is no need to perform a swap because s is already the lexicographically smallest.

 

Constraints:

  • 2 <= s.length <= 100
  • s consists only of digits.

Solutions

Solution 1: Greedy + Simulation

We can traverse the string $\textit{s}$ from left to right. For each pair of adjacent digits, if they have the same parity and the previous digit is greater than the next digit, then we swap these two digits to make the lexicographical order of the string $\textit{s}$ smaller, and then return the swapped string.

After the traversal, if no swappable pair of digits is found, it means the string $\textit{s}$ is already in its smallest lexicographical order, and we can return it directly.

The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the string $\textit{s}$.

  • class Solution {
    public String getSmallestString(String s) {
        char[] cs = s.toCharArray();
        int n = cs.length;
        for (int i = 1; i < n; ++i) {
            char a = cs[i - 1], b = cs[i];
            if (a > b && a % 2 == b % 2) {
                cs[i] = a;
                cs[i - 1] = b;
                return new String(cs);
            }
        }
        return s;
    }
}

  • class Solution {
public:
    string getSmallestString(string s) {
        int n = s.length();
        for (int i = 1; i < n; ++i) {
            char a = s[i - 1], b = s[i];
            if (a > b && a % 2 == b % 2) {
                s[i - 1] = b;
                s[i] = a;
                break;
            }
        }
        return s;
    }
};

  • class Solution:
    def getSmallestString(self, s: str) -> str:
        for i, (a, b) in enumerate(pairwise(map(ord, s))):
            if (a + b) % 2 == 0 and a > b:
                return s[:i] + s[i + 1] + s[i] + s[i + 2 :]
        return s


  • func getSmallestString(s string) string {
	cs := []byte(s)
	n := len(cs)
	for i := 1; i < n; i++ {
		a, b := cs[i-1], cs[i]
		if a > b && a%2 == b%2 {
			cs[i-1], cs[i] = b, a
			return string(cs)
		}
	}
	return s
}

  • function getSmallestString(s: string): string {
    const n = s.length;
    const cs: string[] = s.split('');
    for (let i = 1; i < n; ++i) {
        const a = cs[i - 1];
        const b = cs[i];
        if (a > b && +a % 2 === +b % 2) {
            cs[i - 1] = b;
            cs[i] = a;
            return cs.join('');
        }
    }
    return s;
}

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