LeetCode 1055: Shortest Way to Form String Solution

Master LeetCode problem 1055 (Shortest Way to Form String), a medium 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.

1055. Shortest Way to Form String

Medium

Two Pointers String Binary Search Greedy

Problem Explanation

Explanation:

The problem asks to find the minimum number of subsequences of source that can be concatenated to form the target string. One subsequence of source can be formed by deleting some characters in source without changing the order of the remaining characters.

To solve this problem, we can iterate through the target string and for each character, we try to find the shortest subsequence in source that contains this character. We can use a pointer to keep track of our position in source while iterating through the target string.

Algorithm:

Initialize a variable result to keep track of the number of subsequences needed.
Initialize a pointer index to 0.
Iterate through the target string:
- For each character in the target string, find the shortest subsequence in source starting from the current position index that contains the current character.
- If we cannot find the character in source starting from index, increment result and reset the pointer index to 0.
- Otherwise, update the index to the next character position in source.
Return the result.

Time Complexity:

The time complexity of this algorithm is O(M*N), where M is the length of the target string and N is the length of the source string.

Space Complexity:

The space complexity of this algorithm is O(1) since we are using constant extra space.

Solution Code

class Solution {
    public int shortestWay(String source, String target) {
        int result = 0;
        int index = 0;
        
        while (index < target.length()) {
            int start = index;
            for (int i = 0; i < source.length(); i++) {
                if (index < target.length() && source.charAt(i) == target.charAt(index)) {
                    index++;
                }
            }
            if (index == start) {
                return -1;
            }
            result++;
        }
        
        return result;
    }
}

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What is the time complexity of LeetCode 1055 (Shortest Way to Form String)?

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