925. Long Pressed Name
Explanation
To solve this problem, we can iterate through both the name and typed strings simultaneously. We keep track of the current characters being compared in both strings. If the characters match, we move to the next characters in both strings. If they don't match, we check if the current character in typed is the same as the previous character in typed. If it is, this means the current character in typed might have been long pressed, so we move to the next character in typed and continue comparing. At the end, if we reach the end of both strings, it means the typed string can be formed by long-pressing the name string.
Time complexity: O(n), where n is the length of the longer of the two input strings.
Space complexity: O(1)
class Solution {
public boolean isLongPressedName(String name, String typed) {
int i = 0, j = 0;
while (j < typed.length()) {
if (i < name.length() && name.charAt(i) == typed.charAt(j)) {
i++;
j++;
} else if (j > 0 && typed.charAt(j) == typed.charAt(j - 1)) {
j++;
} else {
return false;
}
}
return i == name.length();
}
}Code Editor (Testing phase)
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