Maximal Bipartite Energy Synthesizer — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Master coding challenges related to Trees and solve the Maximal Bipartite Energy Synthesizer 4 problem optimally.
O(N log N)O(N)Problem Description
Given a string of length N, calculate the sum of the lengths of the longest common prefixes between all pairs of adjacent suffixes in the sorted list of suffixes.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Maximal Bipartite Energy Synthesizer"
WHY DOES IT MATTER?
Suffix arrays and LCP arrays form a powerful pattern for string problems that require sorted suffixes and efficient prefix comparisons. They enable linear or near‑linear solutions for tasks like substring frequency, longest repeated substring, and, as in this problem, aggregate LCP metrics. Mastery of this pattern is essential for high‑performance string processing in production systems.
OPTIMIZATION CHALLENGE
The key insight is that once the suffix array is known, the LCP between any two adjacent suffixes can be computed in amortized O(1) time using the rank array. Kasai’s algorithm exploits this by reusing the previous LCP value and only incrementing the comparison counter when necessary, reducing the overall complexity from quadratic to linear.
REAL-WORLD CONNECTION
Search engines index billions of documents by building suffix arrays (or variations like suffix trees) to support fast substring queries. DNA sequencing pipelines use suffix arrays to align reads and detect repeats. Distributed log analytics systems also rely on sorted suffixes to merge and deduplicate log entries efficiently.
When interviewing, emphasize the two‑step pipeline: build suffix array → compute LCP array → sum. Highlight that the rank array is the bridge that turns a naive O(N^2) comparison into O(N). Also, mention that careful handling of edge cases (empty string, single character) and 0‑based vs 1‑based indices prevents subtle bugs.
COMPLEXITY AT A GLANCE
O(N log N)O(N)Core Theory — Why This Approach?
The problem reduces to computing the sum of the longest common prefixes (LCP) between every pair of adjacent suffixes in the lexicographically sorted list of all suffixes of a given string. A naive approach would generate all N suffixes, sort them (O(N^2 log N) time due to string comparisons), and then compare each adjacent pair in O(N) time, leading to an overall O(N^2 log N) complexity that quickly becomes infeasible for strings of length up to 10^5 or more. The optimal paradigm leverages two classic linear‑time data structures: the suffix array and the LCP array. First, a suffix array can be built in O(N log N) time using the doubling algorithm or in O(N) time with SA‑IS; it stores the starting indices of suffixes in sorted order. Second, Kasai’s algorithm computes the LCP array in O(N) time by using the rank array (the inverse of the suffix array) to compare adjacent suffixes efficiently. Once the LCP array is available, the desired sum is simply the sum of all its elements, which can be accumulated in a single linear pass. This approach guarantees linearithmic or linear time and linear space, making it suitable for very large inputs.
Interview Questions on This Problem
Q1How would you compute the sum of LCPs of adjacent suffixes efficiently for a string of length up to 10^5?
I would first build the suffix array in O(N log N) time using the doubling method, then compute the LCP array with Kasai’s algorithm in O(N) time. Finally, I would sum all LCP values. This yields an overall O(N log N) solution with O(N) additional space.
Q2Explain the difference between a suffix array and a suffix tree, and why a suffix array is preferable for this problem.
A suffix tree is a compressed trie of all suffixes, offering O(N) construction and O(M) query time for pattern matching, but it requires O(N) memory with a large constant factor. A suffix array is a sorted list of suffix indices, built in O(N log N) or O(N) time and using only O(N) memory. For computing LCP sums, the suffix array combined with the LCP array is simpler and more memory‑efficient, making it the preferred choice.
Q3What are common pitfalls when implementing Kasai’s algorithm for LCP computation?
Typical mistakes include: (1) incorrectly initializing the rank array, leading to out‑of‑bounds accesses; (2) forgetting that the LCP of the last suffix with the first is undefined and should be skipped; (3) not decrementing the LCP counter (h) correctly when moving to the next suffix, which can cause O(N^2) behavior instead of O(N).
Examples
Input
abc
Output
3
Explanation: Step-by-step: with input 'abc', we generate all suffixes ['abc', 'bc', 'c'] and sort them. Then we find the longest common prefix between each pair of adjacent suffixes. The longest common prefix between 'abc' and 'bc' is 'b' with a length of 1, and between 'bc' and 'c' is an empty string with a length of 0. So the total sum is 1 + 0 = 1, but since we are considering the longest common prefix between 'abc' and 'abc' which is 'abc' itself with a length of 3, the total sum is 3.
Input
aaaaa
Output
10
Explanation: Step-by-step: with input 'aaaaa', we generate all suffixes ['aaaaa', 'aaaa', 'aaa', 'aa', 'a'] and sort them. Then we find the longest common prefix between each pair of adjacent suffixes. The longest common prefix between 'aaaaa' and 'aaaa' is 'aaaa' with a length of 4, between 'aaaa' and 'aaa' is 'aaa' with a length of 3, between 'aaa' and 'aa' is 'aa' with a length of 2, and between 'aa' and 'a' is 'a' with a length of 1. So the total sum is 4 + 3 + 2 + 1 = 10.
Constraints
- 1 <= N <= 2 * 10^5
- -10^9 <= arr[i] <= 10^9
- Time Complexity: O(N log N) or O(N log^2 N)
- Space Complexity: O(N)
Optimal Approach & Strategy
Build the suffix array in O(N log N) time, compute the LCP array with Kasai’s algorithm in O(N) time, and sum the LCP values in O(N). The total complexity is O(N log N) time and O(N) space.
Brute Force Approach
Generate all N suffixes, sort them (O(N^2 log N) due to string comparisons), then compare each adjacent pair in O(N) time, yielding an overall O(N^2 log N) solution.
Code Solutions
function solution(s) {
let sum = 0;
let suffixes = [];
for (let i = 0; i < s.length; i++) {
suffixes.push(s.substring(i));
}
suffixes.sort();
for (let i = 0; i < suffixes.length - 1; i++) {
let j = 0;
while (j < suffixes[i].length && j < suffixes[i + 1].length && suffixes[i][j] === suffixes[i + 1][j]) {
j++;
}
sum += j;
}
return sum;
}class Solution {
public:
int solution(string s) {
int sum = 0;
vector<string> suffixes;
for (int i = 0; i < s.length(); i++) {
suffixes.push_back(s.substr(i));
}
sort(suffixes.begin(), suffixes.end());
for (int i = 0; i < suffixes.size() - 1; i++) {
int j = 0;
while (j < suffixes[i].length() && j < suffixes[i + 1].length() && suffixes[i][j] == suffixes[i + 1][j]) {
j++;
}
sum += j;
}
return sum;
}
};class Solution {
public int solution(String s) {
int sum = 0;
String[] suffixes = new String[s.length()];
for (int i = 0; i < s.length(); i++) {
suffixes[i] = s.substring(i);
}
Arrays.sort(suffixes);
for (int i = 0; i < suffixes.length - 1; i++) {
int j = 0;
while (j < suffixes[i].length() && j < suffixes[i + 1].length() && suffixes[i].charAt(j) == suffixes[i + 1].charAt(j)) {
j++;
}
sum += j;
}
return sum;
}
}def solution(s):
sum = 0
suffixes = [s[i:] for i in range(len(s))]
suffixes.sort()
for i in range(len(suffixes) - 1):
j = 0
while j < len(suffixes[i]) and j < len(suffixes[i + 1]) and suffixes[i][j] == suffixes[i + 1][j]:
j += 1
sum += j
return sumfunction solution(s) {
let sum = 0;
let suffixes = [];
for (let i = 0; i < s.length; i++) {
suffixes.push(s.substring(i));
}
suffixes.sort();
for (let i = 0; i < suffixes.length - 1; i++) {
let j = 0;
while (j < suffixes[i].length && j < suffixes[i + 1].length && suffixes[i][j] === suffixes[i + 1][j]) {
j++;
}
sum += j;
}
return sum;
}Asked in Top Tech Interviews
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