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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.
1function canFormSubsequenceByCharFreq(str1, str2) {
2 const charCount = new Array(26).fill(0);
3 for (let char of str1) {
4 charCount[char.charCodeAt(0) - 'a'.charCodeAt(0)]++;
5 }
6 for (let char of str2) {
7 const idx = char.charCodeAt(0) - 'a'.charCodeAt(0);
8 if (charCount[idx] > 0) {
9 charCount[idx]--;
10 } else {
11 let canTransform = false;
12 for (let j = 0; j < 26; j++) {
13 if (charCount[j] > 0 && (String.fromCharCode((j + 1) % 26 + 'a'.charCodeAt(0)) === char)) {
14 canTransform = true;
15 charCount[j]--;
16 break;
17 }
18 }
19 if (!canTransform) return false;
20 }
21 }
22 return true;
23}In JavaScript, the frequency of each character is assessed and considered for transformation feasibility. This strategy checks potential creation of `str2` from cyclic transit adjustments via `str1`. Any unaligning components obstruct final success.