LeetCode 1989: Maximum Number of People That Can Be Caught in Tag Solution
Master LeetCode problem 1989 (Maximum Number of People That Can Be Caught in Tag), 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.
Problem Explanation
Explanation:
To solve this problem, we can use two pointers approach. We iterate through the array of tags and maintain two pointers, start and end, representing the start and end indices of the current valid sequence of tags. We also keep track of the number of tags caught so far. Whenever a 'X' tag is encountered, we update the number of tags caught by incrementing the difference between end and start + 1, and move the start pointer to the next position. If the number of tags caught exceeds the limit k, we move the end pointer to the next position until the number of tags caught is within the limit.
Time Complexity: O(n) where n is the number of tags in the input array. Space Complexity: O(1)
:
Solution Code
class Solution {
public int catchMaximumAmountofTags(char[] tags, int k) {
int n = tags.length;
int start = 0, end = 0;
int caught = 0;
while (end < n) {
if (tags[end] == 'X') {
caught += end - start + 1;
start = end + 1;
}
if (caught > k) {
caught -= end - start + 1;
start++;
}
end++;
}
return Math.min(caught, k);
}
}Try It Yourself
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