LeetCode 1638: Count Substrings That Differ by One Character
LeetCode 1638 Solution Explanation
Explanation
To solve this problem, we can iterate through all possible substrings of string s and string t. For each pair of substrings, we check if they differ by exactly one character. To efficiently check this, we compare the substrings character by character and keep track of the number of differences encountered. If the number of differences is exactly one, we increment the count of valid substrings. The time complexity of this approach is O(n^4) where n is the length of the strings s and t.
LeetCode 1638 Solutions in Java, C++, Python
class Solution {
public int countSubstrings(String s, String t) {
int count = 0;
for (int i = 1; i <= s.length(); i++) {
for (int j = 0; j < s.length() - i + 1; j++) {
String subS = s.substring(j, j + i);
for (int k = 1; k <= t.length(); k++) {
for (int l = 0; l < t.length() - k + 1; l++) {
String subT = t.substring(l, l + k);
if (checkDifferByOne(subS, subT)) {
count++;
}
}
}
}
}
return count;
}
private boolean checkDifferByOne(String s, String t) {
int diffCount = 0;
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) != t.charAt(i)) {
diffCount++;
if (diffCount > 1) {
return false;
}
}
}
return diffCount == 1;
}
}Interactive Code Editor for LeetCode 1638
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