LeetCode 1520: Maximum Number of Non-Overlapping Substrings Solution

Master LeetCode problem 1520 (Maximum Number of Non-Overlapping Substrings), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1520. Maximum Number of Non-Overlapping Substrings

Hard

String Greedy

Problem Explanation

Explanation:

To solve this problem, we can use a greedy approach combined with interval merging. We will iterate over the string to find the valid substrings by keeping track of the first and last occurrence of each character in the string. Then, we will merge overlapping intervals to form non-overlapping substrings.

Iterate over the string to find the first and last occurrence of each character.
Expand the intervals to cover all characters within the range.
Merge overlapping intervals to create non-overlapping substrings.
Choose the substrings with the minimum total length.

Time Complexity: O(N), where N is the length of the input string. Space Complexity: O(N)

Solution Code

class Solution {
    public List<String> maxNumOfSubstrings(String s) {
        List<String> result = new ArrayList<>();
        Map<Character, Integer> first = new HashMap<>();
        Map<Character, Integer> last = new HashMap();
        
        // Find first and last occurrence of each character
        for (int i = 0; i < s.length(); i++) {
            last.put(s.charAt(i), i);
        }
        for (int i = s.length() - 1; i >= 0; i--) {
            first.put(s.charAt(i), i);
        }
        
        int end = -1, start = 0;
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (i == first.get(c)) {
                int newEnd = findEnd(s, i, first, last);
                if (newEnd != -1 && i > end) {
                    end = newEnd;
                    result.add("");
                }
            }
            if (i == last.get(c) && i == end) {
                result.set(result.size() - 1, s.substring(start, end + 1));
                start = i + 1;
            }
        }
        
        return result;
    }
    
    private int findEnd(String s, int i, Map<Character, Integer> first, Map<Character, Integer> last) {
        int end = last.get(s.charAt(i));
        for (int j = i; j <= end; j++) {
            if (first.get(s.charAt(j)) < i) {
                return -1;
            }
            end = Math.max(end, last.get(s.charAt(j)));
        }
        return end;
    }
}

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