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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.
1#include <vector>
2#include <string>
3using namespace std;
4
5bool canFormSubsequenceByCharFreq(string str1, string str2) {
6 vector<int> charCount(26, 0);
7 for (char c : str1) {
8 charCount[c - 'a']++;
9 }
10 for (char c : str2) {
11 int idx = c - 'a';
12 if (charCount[idx] > 0) {
13 charCount[idx]--;
14 } else {
15 bool canTransform = false;
16 for (int j = 0; j < 26; j++) {
17 if (charCount[j] > 0 && (((j + 1) % 26 + 'a') == c)) {
18 canTransform = true;
19 charCount[j]--;
20 break;
21 }
22 }
23 if (!canTransform) return false;
24 }
25 }
26 return true;
27}The C++ solution applies a counting array to maintain character frequencies within `str1`. Each character in `str2` verifies the feasibility of direct matching or transformation from within `str1`. Iterating through necessary transformations aims to ensure complete emergence.