763. Partition Labels LeetCode Solution

Last updated: February 20, 2025

In this guide, you will get 763. Partition Labels LeetCode Solution with the best time and space complexity. The solution to Partition Labels problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Partition Labels solution in C++
Partition Labels solution in Java
Partition Labels solution in Python
Additional Resources

763. Partition Labels LeetCode Solution image

Problem Statement of Partition Labels

You are given a string s. We want to partition the string into as many parts as possible so that each letter appears in at most one part. For example, the string “ababcc” can be partitioned into [“abab”, “cc”], but partitions such as [“aba”, “bcc”] or [“ab”, “ab”, “cc”] are invalid.
Note that the partition is done so that after concatenating all the parts in order, the resultant string should be s.
Return a list of integers representing the size of these parts.

Example 1:

Input: s = “ababcbacadefegdehijhklij”
Output: [9,7,8]
Explanation:
The partition is “ababcbaca”, “defegde”, “hijhklij”.
This is a partition so that each letter appears in at most one part.
A partition like “ababcbacadefegde”, “hijhklij” is incorrect, because it splits s into less parts.

Example 2:

Input: s = “eccbbbbdec”
Output: [10]

Constraints:

1 <= s.length <= 500
s consists of lowercase English letters.

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(128) = O(1)

763. Partition Labels LeetCode Solution in C++

class Solution {
 public:
  vector<int> partitionLabels(string s) {
    vector<int> ans;
    vector<int> rightmost(26);

    for (int i = 0; i < s.length(); ++i)
      rightmost[s[i] - 'a'] = i;

    int l = 0;  // the leftmost index of the current running string
    int r = 0;  // the rightmost index of the current running string

    for (int i = 0; i < s.length(); ++i) {
      r = max(r, rightmost[s[i] - 'a']);
      if (r == i) {
        ans.push_back(i - l + 1);
        l = i + 1;
      }
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

763. Partition Labels LeetCode Solution in Java

class Solution {
  public List<Integer> partitionLabels(String s) {
    List<Integer> ans = new ArrayList<>();
    int[] rightmost = new int[26];

    for (int i = 0; i < s.length(); ++i)
      rightmost[s.charAt(i) - 'a'] = i;

    int l = 0; // the leftmost index of the current running string
    int r = 0; // the rightmost index of the current running string

    for (int i = 0; i < s.length(); ++i) {
      r = Math.max(r, rightmost[s.charAt(i) - 'a']);
      if (r == i) {
        ans.add(i - l + 1);
        l = i + 1;
      }
    }

    return ans;
  }
}
// code provided by PROGIEZ

763. Partition Labels LeetCode Solution in Python

class Solution:
  def partitionLabels(self, s: str) -> list[int]:
    ans = []
    letterToRightmostIndex = {c: i for i, c in enumerate(s)}

    l = 0  # the leftmost index of the current running string
    r = 0  # the rightmost index of the current running string

    for i, c in enumerate(s):
      r = max(r, letterToRightmostIndex[c])
      if i == r:
        ans.append(r - l + 1)
        l = r + 1

    return ans
 # code by PROGIEZ

Additional Resources

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