Sponsored
Sponsored
This approach uses two pointers to simultaneously traverse both strings. The goal is to see if `str2` can be formed as a subsequence of `str1` with at most one cyclic increment operation. During the traversal, if `str1[i]` can be incremented to match `str2[j]` by a cyclic shift, move both pointers forward. If `str1[i]` cannot be shifted to match `str2[j]`, and only one operation is allowed, then return false. Otherwise, continue checking until either `str2` is fully traversed (return true) or `str1` is exhausted (return false).
Time Complexity: O(n + m), where n is the length of `str1` and m is the length of `str2` since we process each character once.
Space Complexity: O(1), since no additional space proportional to input size is used.
1function canFormSubsequence(str1, str2) {
2 let i = 0, j = 0, n = str1.length, m = str2.length;
3 whileThis JavaScript solution adeptly utilizes character iteration and ASCII manipulation to ascertain if `str2` can traceably emerge from cyclic manipulation of `str1`. The strategy is similar to two-pointers, adjusting based on condition validation.
This method determines the feasibility of forming `str2` as a subsequence by transforming `str1` using character frequency transitions, considering cyclic alphabet rules. Using a frequency map for `str1`, iterate through `str2`, checking if corresponding increments can map `str1` characters to those in `str2`. For each character needed in `str2`, see if it's obtainable via allowable operations on `str1` characters, optimizing through potential full rotations using the alphabet cycle property.
Time Complexity: O(n + m + 26) to cover all characters and frequency evaluations.
Space Complexity: O(1), preserving essential space outside of constant size arrays.
1def can_form_subsequence_by_char_freq(str1: str, str2: str) -> bool:
2 char_count = [0] * 26
3 for char in str1:
4 char_count[ord(char) - ord('a')] += 1
5 for char in str2:
6 idx = ord(char) - ord('a')
7 if char_count[idx] > 0:
8 char_count[idx] -= 1
9 else:
10 can_transform = False
11 for j in range(26):
12 if char_count[j] > 0 and (chr((j + 1) % 26 + ord('a')) == char):
13 can_transform = True
14 char_count[j] -= 1
15 break
16 if not can_transform:
17 return False
18 return TruePython's solution conducts character transformation checks by maintaining a frequency register for `str1`, testing potential transformations in seeking sub-sequential emergence within `str2`. Prompt checking ceases unsuccessful allocation.