Count Balanced Strings ā Problem Statement & Solution Guide
Problem Description
Given a sequence of strings, implement a function to identify and count the total number of strings that have an equal number of 'x's and 'y's and do not contain any consecutive repeating characters.
Examples
Input
['xy', 'yx', 'xyxy', 'yxxy']
Output
2
Explanation: Step-by-step: 1. Initialize count to 0. 2. Iterate over each string in the input array. 3. For each string, check if it has an equal number of 'x's and 'y's and does not contain any consecutive repeating characters. 4. If the string meets the conditions, increment the count. 5. After iterating over all strings, return the count.
Input
['xx', 'yy', 'xyxy', 'yxxy']
Output
2
Explanation: Step-by-step: 1. Initialize count to 0. 2. Iterate over each string in the input array. 3. For each string, check if it has an equal number of 'x's and 'y's and does not contain any consecutive repeating characters. 4. If the string meets the conditions, increment the count. 5. After iterating over all strings, return the count.
Constraints
- 1 <= length of the signal sequence <= 1000
- The signal sequence contains only 'x's and 'y's.
Optimal Approach & Strategy
The optimal approach involves iterating over the signal sequence and using a sliding window to generate all possible substrings, then checking each substring for validity. This approach can be implemented in linear time complexity.
Brute Force Approach
A brute-force approach would involve generating all possible substrings of the signal sequence and checking each one for validity, resulting in a time complexity of O(nĀ²). This approach is inefficient for large signal sequences. It would also require extra space to store the substrings.
Verified Code Solutions
public int countBalancedStrings(String[] strings) {
int count = 0;
for (String s : strings) {
if (s.length() % 2 != 0) {
continue;
}
int xCount = countOccurrences(s, 'x');
int yCount = countOccurrences(s, 'y');
if (xCount != yCount) {
continue;
}
for (int i = 0; i < s.length() - 1; i++) {
if (s.charAt(i) == s.charAt(i + 1)) {
break;
}
} else {
count++;
}
}
return count;
}
private int countOccurrences(String str, char c) {
int count = 0;
for (char ch : str.toCharArray()) {
if (ch == c) {
count++;
}
}
return count;
}def count_balanced_strings(strings):
count = 0
for s in strings:
if len(s) % 2 != 0:
continue
x_count = s.count('x')
y_count = s.count('y')
if x_count != y_count:
continue
for i in range(len(s) - 1):
if s[i] == s[i + 1]:
break
else:
count += 1
return countAsked in Top Tech Interviews
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