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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.
1import java.util.HashMap;
2
3public class Solution {
4 public boolean canFormSubsequenceByCharFreq(String str1, String str2) {
5 int[] charCount = new int[26];
6 for (char c : str1.toCharArray()) {
7 charCount[c - 'a']++;
8 }
9 for (char c : str2.toCharArray()) {
10 int idx = c - 'a';
11 if (charCount[idx] > 0) {
12 charCount[idx]--;
13 } else {
14 boolean canTransform = false;
15 for (int j = 0; j < 26; j++) {
16 if (charCount[j] > 0 && (((j + 1) % 26 + 'a') == c)) {
17 canTransform = true;
18 charCount[j]--;
19 break;
20 }
21 }
22 if (!canTransform) return false;
23 }
24 }
25 return true;
26 }
27}Java's solution involves preparing an array to track `str1` character frequencies, iteratively checking the possibility of forming each `str2` character through available data post-cyclic application. Mismatches halting transformations obstruct outcomes.