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Formatted question description: https://leetcode.ca/all/925.html
925. Long Pressed Name (Easy)
Your friend is typing his name
into a keyboard. Sometimes, when typing a character c
, the key might get long pressed, and the character will be typed 1 or more times.
You examine the typed
characters of the keyboard. Return True
if it is possible that it was your friends name, with some characters (possibly none) being long pressed.
Example 1:
Input: name = "alex", typed = "aaleex" Output: true Explanation: 'a' and 'e' in 'alex' were long pressed.
Example 2:
Input: name = "saeed", typed = "ssaaedd" Output: false Explanation: 'e' must have been pressed twice, but it wasn't in the typed output.
Example 3:
Input: name = "leelee", typed = "lleeelee" Output: true
Example 4:
Input: name = "laiden", typed = "laiden" Output: true Explanation: It's not necessary to long press any character.
Constraints:
1 <= name.length <= 1000
1 <= typed.length <= 1000
name
andtyped
contain only lowercase English letters.
Related Topics:
Two Pointers, String
Solution 1.
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class Solution { public boolean isLongPressedName(String name, String typed) { if (name == null || typed == null || name.length() == 0 || typed.length() == 0) return false; int nameLength = name.length(), typedLength = typed.length(); int index2 = 0; for (int i = 0; i < nameLength; i++) { char c1 = name.charAt(i); if (c1 != typed.charAt(index2)) return false; if (i == nameLength - 1 || c1 != name.charAt(i + 1)) { while (index2 < typedLength && typed.charAt(index2) == c1) index2++; } else index2++; if (index2 == typedLength && i < nameLength - 1) return false; } return true; } } ############ class Solution { public boolean isLongPressedName(String name, String typed) { int m = name.length(), n = typed.length(); int i = 0, j = 0; for (; i < m && j < n; ++i, ++j) { if (name.charAt(i) != typed.charAt(j)) { return false; } int cnt1 = 0, cnt2 = 0; char c = name.charAt(i); while (i + 1 < m && name.charAt(i + 1) == c) { ++i; ++cnt1; } while (j + 1 < n && typed.charAt(j + 1) == c) { ++j; ++cnt2; } if (cnt1 > cnt2) { return false; } } return i == m && j == n; } }
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// OJ: https://leetcode.com/problems/long-pressed-name/ // Time: O(M + N) // Space: O(1) class Solution { public: bool isLongPressedName(string name, string typed) { int i = 0, M = name.size(), j = 0, N = typed.size(); while (i < M && j < N) { char c = name[i]; int a = 0, b = 0; while (i < M && name[i] == c) ++i, ++a; while (j < N && typed[j] == c) ++j, ++b; if (a > b) return false; } return i == M && j == N; } };
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class Solution: def isLongPressedName(self, name: str, typed: str) -> bool: m, n = len(name), len(typed) i = j = 0 while i < m and j < n: if name[i] != typed[j]: return False cnt1 = cnt2 = 0 c = name[i] while i + 1 < m and name[i + 1] == c: i += 1 cnt1 += 1 while j + 1 < n and typed[j + 1] == c: j += 1 cnt2 += 1 if cnt1 > cnt2: return False i, j = i + 1, j + 1 return i == m and j == n ############ class Solution(object): def isLongPressedName(self, name, typed): """ :type name: str :type typed: str :rtype: bool """ M = len(name) N = len(typed) i, j = 0, 0 while i < M: c_i = name[i] count_i = 0 count_j = 0 while i < M and name[i] == c_i: i += 1 count_i += 1 while j < N and typed[j] == c_i: j += 1 count_j += 1 if count_j < count_i: return False return True
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func isLongPressedName(name string, typed string) bool { m, n := len(name), len(typed) i, j := 0, 0 for ; i < m && j < n; i, j = i+1, j+1 { if name[i] != typed[j] { return false } cnt1, cnt2 := 0, 0 c := name[i] for i+1 < m && name[i+1] == c { i++ cnt1++ } for j+1 < n && typed[j+1] == c { j++ cnt2++ } if cnt1 > cnt2 { return false } } return i == m && j == n }