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14. Longest Common Prefix

Description

Write a function to find the longest common prefix string amongst an array of strings.

If there is no common prefix, return an empty string "".

 

Example 1:

Input: strs = ["flower","flow","flight"]
Output: "fl"

Example 2:

Input: strs = ["dog","racecar","car"]
Output: ""
Explanation: There is no common prefix among the input strings.

 

Constraints:

  • 1 <= strs.length <= 200
  • 0 <= strs[i].length <= 200
  • strs[i] consists of only lowercase English letters.

Solutions

Solution 1: Character Comparison

We use the first string $strs[0]$ as a benchmark, and compare whether the $i$-th character of the subsequent strings is the same as the $i$-th character of $strs[0]$. If they are the same, we continue to compare the next character. Otherwise, we return the first $i$ characters of $strs[0]$.

If the traversal ends, it means that the first $i$ characters of all strings are the same, and we return $strs[0]$.

The time complexity is $O(n \times m)$, where $n$ and $m$ are the length of the string array and the minimum length of the strings, respectively. The space complexity is $O(1)$.

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class Solution:
    def longestCommonPrefix(self, strs: List[str]) -> str:
        for i in range(len(strs[0])):
            for s in strs[1:]:
                if len(s) <= i or s[i] != strs[0][i]:
                    return s[:i]
        return strs[0]

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class Solution {
    public String longestCommonPrefix(String[] strs) {
        int n = strs.length;
        for (int i = 0; i < strs[0].length(); ++i) {
            for (int j = 1; j < n; ++j) {
                if (strs[j].length() <= i || strs[j].charAt(i) != strs[0].charAt(i)) {
                    return strs[0].substring(0, i);
                }
            }
        }
        return strs[0];
    }
}

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class Solution {
public:
    string longestCommonPrefix(vector<string>& strs) {
        int n = strs.size();
        for (int i = 0; i < strs[0].size(); ++i) {
            for (int j = 1; j < n; ++j) {
                if (strs[j].size() <= i || strs[j][i] != strs[0][i]) {
                    return strs[0].substr(0, i);
                }
            }
        }
        return strs[0];
    }
};

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func longestCommonPrefix(strs []string) string {
    n := len(strs)
    for i := range strs[0] {
        for j := 1; j < n; j++ {
            if len(strs[j]) <= i || strs[j][i] != strs[0][i] {
                return strs[0][:i]
            }
        }
    }
    return strs[0]
}

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function longestCommonPrefix(strs: string[]): string {
    const len = strs.reduce((r, s) => Math.min(r, s.length), Infinity);
    for (let i = len; i > 0; i--) {
        const target = strs[0].slice(0, i);
        if (strs.every(s => s.slice(0, i) === target)) {
            return target;
        }
    }
    return '';
}

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impl Solution {
    pub fn longest_common_prefix(strs: Vec<String>) -> String {
        let mut len =