Balanced Hyperjump Sequences — Problem Statement & Solution Guide
Problem Description
Generate all possible valid combinations of 'J' (hyperjump) and 'S' (hyperslow) sequences of length n, where every 'J' has a corresponding 'S', and no sequence is a substring of another. Return these combinations as a list of strings.
Examples
Input
n = 5
Output
['JJJJSSSSSS', 'JJJJSSSSSJ', 'JJJJSSSSSJSS', 'JJJJSSSSSJSSS', 'JJJJSSSSSJSSSS', 'JJJJSSSSSJSSSSS', 'JJJJSSSSSJSSSSSS', 'JJJJSSSSSJSSSSSSS', 'JJJJSSSSSJSSSSSSSS', 'JJJJSSSSSJSSSSSSSSS']
Explanation: Step-by-step: 1. Generate all possible combinations of 'J' and 'S' sequences of length n. 2. Check each sequence to ensure every 'J' has a corresponding 'S' and no sequence is a substring of another. 3. Return the valid combinations as a list of strings.
Input
n = 3
Output
['JJJJ', 'JJJJS', 'JJJJSJ', 'JJJJSJS', 'JJJJSJSJJJ']
Explanation: Step-by-step: 1. Generate all possible combinations of 'J' and 'S' sequences of length n. 2. Check each sequence to ensure every 'J' has a corresponding 'S' and no sequence is a substring of another. 3. Return the valid combinations as a list of strings.
Constraints
- 1 <= n <= 10, where n is the number of pairs of hyperjumps and hyperslows
- The output list should not contain any duplicate combinations
Optimal Approach & Strategy
A more efficient approach would be to use recursion to build up valid combinations. This approach would have a time complexity of O((2n)! / (n! * n!)) and would be much more efficient than the naive approach.
Brute Force Approach
One naive approach would be to generate all possible combinations of 'J' and 'S' and then filter out the combinations that are not valid. However, this approach would have a time complexity of O(2^(2n)) and would be very inefficient for large inputs.
Verified Code Solutions
class Solution {
public List<String> balancedHyperjumpSequences(int n) {
List<String> result = new ArrayList<>();
backtrack(result, n, 0, '');
return result;
}
private void backtrack(List<String> result, int n, int jCount, String path) {
if (path.length() == n) {
if (jCount == 0 || path.charAt(path.length() - 1) == 'S') {
result.add(path);
}
return;
}
if (jCount > 0) {
backtrack(result, n, jCount - 1, path + 'J');
}
backtrack(result, n, jCount, path + 'S');
}
}def balanced_hyperjump_sequences(n):
def backtrack(path, j_count):
if len(path) == n:
if j_count == 0 or path[-1] == 'S':
result.append(path)
return
if j_count > 0:
backtrack(path + 'J', j_count - 1)
backtrack(path + 'S', j_count)
result = []
backtrack('', n // 2)
return resultAsked in Top Tech Interviews
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