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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.
1public class Solution {
2 public boolean canFormSubsequence(String str1, String str2) {
3 int i = 0, j = 0, n =The Java implementation uses two pointers to ensure that characters in `str1` can be cyclically incremented to match characters in `str2`. Using Java's `charAt` method for indexing, we check if each character in `str1` can help form a subsequence matching `str2`, considering cyclic increments.
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.
public bool CanFormSubsequenceByCharFreq(string str1, string str2) {
int[] charCount = new int[26];
foreach (char c in str1) {
charCount[c - 'a']++;
}
foreach (char c in str2) {
int idx = c - 'a';
if (charCount[idx] > 0) {
charCount[idx]--;
} else {
bool canTransform = false;
for (int j = 0; j < 26; j++) {
if (charCount[j] > 0 && (((j + 1) % 26 + 'a') == c)) {
canTransform = true;
charCount[j]--;
break;
}
}
if (!canTransform) return false;
}
}
return true;
}
}The C# method leverages a simple array to track `str1` frequency figures, examining character transitions across operations to determine the feasibility of reconstructing `str2`. If all purposeful transformations fail or are unauthorized, false results ensue.