2565. Subsequence With the Minimum Score
Explanation
To solve this problem, we can iterate through string s and keep track of the position of characters in string t that we encounter in s. We can then calculate the score based on the positions of the characters in t that we have found in s. The minimum score would be the difference between the maximum and minimum positions of the characters in t in s.
Algorithm
- Create a map to store the positions of characters in string
t. - Iterate through string
sand update the map with the positions of characters intfound ins. - Calculate the minimum score as the difference between the maximum and minimum positions of characters in
tins.
Time Complexity
The time complexity of this solution is O(n), where n is the length of string s or t.
Space Complexity
The space complexity of this solution is O(n), where n is the length of string t.
class Solution {
public int minScore(String s, String t) {
Map<Character, List<Integer>> positions = new HashMap<>();
for (int i = 0; i < t.length(); i++) {
positions.computeIfAbsent(t.charAt(i), k -> new ArrayList<>()).add(i);
}
int minScore = Integer.MAX_VALUE;
int left = -1, right = -1;
for (char c : s.toCharArray()) {
if (positions.containsKey(c)) {
for (int pos : positions.get(c)) {
if (pos > right) {
right = pos;
}
if (left == -1 || pos < left) {
left = pos;
}
}
}
if (left != -1 && right != -1) {
minScore = Math.min(minScore, right - left + 1);
}
}
return minScore == Integer.MAX_VALUE ? 0 : minScore;
}
}Code Editor (Testing phase)
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