Question

1163. Last Substring in Lexicographical Order

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Description

Given a string s, return _the last substring of_ s _in lexicographical order_.

 

Example 1:

Input: s = "abab"
Output: "bab"
Explanation: The substrings are ["a", "ab", "aba", "abab", "b", "ba", "bab"]. The lexicographically maximum substring is "bab".

Example 2:

Input: s = "leetcode"
Output: "tcode"

 

Constraints:

• 1 <= s.length <= 4 * 105
• s contains only lowercase English letters.

Solutions

Solution 1: Two pointers

We notice that if a substring starts from position $i$, then the largest substring with the largest dictionary order must be $s[i,..n-1]$, which is the longest suffix starting from position $i$. Therefore, we only need to find the largest suffix substring.

We use two pointers $i$ and $j$, where pointer $i$ points to the starting position of the current largest substring with the largest dictionary order, and pointer $j$ points to the starting position of the current substring being considered. In addition, we use a variable $k$ to record the current position being compared. Initially, $i = 0$, $j=1$, $k=0$.

Each time, we compare $s[i+k]$ and $s[j+k]$:

If $s[i + k] = s[j + k]$, it means that $s[i,..i+k]$ and $s[j,..j+k]$ are the same, and we add $k$ by $1$ and continue to compare $s[i+k]$ and $s[j+k]$;

If $s[i + k] \lt s[j + k]$, it means that the dictionary order of $s[j,..j+k]$ is larger. At this time, we update $i = i + k + 1$, and reset $k$ to $0$. If $i \geq j$ at this time, we update pointer $j$ to $i + 1$, that is, $j = i + 1$. Here we skip all suffix substrings with $s[i,..,i+k]$ as the starting position, because their dictionary orders are smaller than the suffix substrings with $s[j,..,j+k]$ as the starting position.

Similarly, if $s[i + k] \gt s[j + k]$, it means that the dictionary order of $s[i,..,i+k]$ is larger. At this time, we update $j = j + k + 1$ and reset $k$ to $0$. Here we skip all suffix substrings with $s[j,..,j+k]$ as the starting position, because their dictionary orders are smaller than the suffix substrings with $s[i,..,i+k]$ as the starting position.

Finally, we return the suffix substring starting from $i$, that is, $s[i,..,n-1]$.

The time complexity is $O(n)$, where $n$ is the length of string $s$. The space complexity is $O(1)$.

class Solution:
  def lastSubstring(self, s: str) -> str:
    i, j, k = 0, 1, 0
    while j + k < len(s):
      if s[i + k] == s[j + k]:
        k += 1
      elif s[i + k] < s[j + k]:
        i += k + 1
        k = 0
        if i >= j:
          j = i + 1
      else:
        j += k + 1
        k = 0
    return s[i:]

class Solution {
  public String lastSubstring(String s) {
    int n = s.length();
    int i = 0;
    for (int j = 1, k = 0; j + k < n;) {
      int d = s.charAt(i + k) - s.charAt(j + k);
      if (d == 0) {
        ++k;
      } else if (d < 0) {
        i += k + 1;
        k = 0;
        if (i >= j) {
          j = i + 1;
        }
      } else {
        j += k + 1;
        k = 0;
      }
    }
    return s.substring(i);
  }
}

class Solution {
public:
  string lastSubstring(string s) {
    int n = s.size();
    int i = 0;
    for (int j = 1, k = 0; j + k < n;) {
      if (s[i + k] == s[j + k]) {
        ++k;
      } else if (s[i + k] < s[j + k]) {
        i += k + 1;
        k = 0;
        if (i >= j) {
          j = i + 1;
        }
      } else {
        j += k + 1;
        k = 0;
      }
    }
    return s.substr(i);
  }
};

func lastSubstring(s string) string {
  i, n := 0, len(s)
  for j, k := 1, 0; j+k < n; {
    if s[i+k] == s[j+k] {
      k++
    } else if s[i+k] < s[j+k] {
      i += k + 1
      k = 0
      if i >= j {
        j = i + 1
      }
    } else {
      j += k + 1
      k = 0
    }
  }
  return s[i:]
}

function lastSubstring(s: string): string {
  const n = s.length;
  let i = 0;
  for (let j = 1, k = 0; j + k < n; ) {
    if (s[i + k] === s[j + k]) {
      ++k;
    } else if (s[i + k] < s[j + k]) {
      i += k + 1;
      k = 0;
      if (i >= j) {
        j = i + 1;
      }
    } else {
      j += k + 1;
      k = 0;
    }
  }
  return s.slice(i);
}