Longest Homogeneous Segment — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Master coding challenges related to Sliding Window and solve the Longest Homogeneous Segment problem optimally.
O(N)O(1)Problem Description
Given an array chars containing only lowercase English letters and an integer K, you may replace at most K elements of the array with any lowercase letter of your choice. Determine the greatest possible length of a contiguous sub‑array that can be transformed into a sequence of identical letters after performing no more than K replacements. The solution must run in linear time relative to the array size and use only constant extra space beyond the input.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Longest Homogeneous Segment"
WHY DOES IT MATTER?
Sliding‑window patterns turn problems with a global constraint on a sub‑array into a local, incremental check, enabling O(N) solutions that scale to massive inputs—essential for real‑time analytics and streaming data pipelines.
OPTIMIZATION CHALLENGE
The key insight is that you never need to recompute the most frequent character from scratch when the window slides; you can keep a running frequency table and a single maxFreq variable, shrinking the window only when the replacement count surpasses K, which collapses the naive quadratic search to linear time.
REAL-WORLD CONNECTION
Think of a production line where you can rework at most K defective items in a batch; the algorithm finds the longest batch you can ship without exceeding the rework budget, mirroring quality‑control decisions in manufacturing or network packet sanitization.
During an interview, maintain a clear invariant: the current window always satisfies "requiredReplacements <= K". When it breaks, move the left pointer until the invariant is restored—this mental model prevents off‑by‑one errors and keeps the code concise.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem is a classic application of the sliding‑window (two‑pointer) technique for finding the longest sub‑array that satisfies a constraint on the number of “bad” elements. By keeping a window that expands to the right and only contracts when the number of characters that differ from the most frequent character inside the window exceeds K, we can guarantee that every window considered is feasible. Naïve solutions, such as checking every possible sub‑array and counting replacements, run in O(N^2) time because each window requires a full scan to compute the frequency distribution, which is prohibitive for N up to 10^5 or more. The optimal paradigm leverages the fact that the window’s feasibility depends solely on the count of the most common character; as the window grows, this count can be maintained incrementally, allowing us to adjust the left pointer only when the replacement budget is exceeded, yielding a linear‑time algorithm.
Interview Questions on This Problem
Q1How would you adapt the sliding‑window solution if the alphabet size were large (e.g., Unicode characters) and you could not afford an O(Alphabet) update per step?
Use a hash map to store frequencies only for characters that appear in the current window; the map’s size is bounded by the window length, and you still maintain the max frequency by updating it when the count of the added character exceeds the current max. This keeps amortized O(1) updates while avoiding a full alphabet scan.
Q2Explain why the condition "windowSize - maxFreq <= K" is sufficient to guarantee that the window can be transformed into a homogeneous segment.
windowSize is the total number of characters in the window, maxFreq is the count of the most frequent character. The difference is exactly the number of characters that need to be changed to match that majority character. If that difference does not exceed K, we can replace those outliers within the allowed budget, making the whole window homogeneous.
Q3In a distributed system where logs are sharded by time, how could the longest homogeneous segment algorithm help in detecting anomalies in a stream of status codes?
Treat each status code as a character and K as the tolerated noise level. Running the sliding‑window algorithm on each shard identifies the longest period where the system behaved uniformly except for at most K anomalies, highlighting stable intervals and pinpointing when the pattern breaks, which is useful for alerting and root‑cause analysis.
Examples
Input
chars = ["a","b","b","a","b","b","b"], K = 2
Output
7
Explanation: The sub‑array from index 0 to 6 contains the letters a,b,b,a,b,b,b. By changing the two a's (positions 0 and 3) to b, the entire segment becomes seven consecutive b's. No longer segment can be made homogeneous with only two changes, so the answer is 7.
Input
chars = ["c","c","c","a","c","c","b","c"], K = 1
Output
6
Explanation: Changing the single a at index 3 to c produces the segment indices 0 through 5: c,c,c,c,c,c, which has length 6. Any other choice of a single replacement yields a shorter homogeneous block, so the maximum length is 6.
Input
chars = ["z","y","x","w"], K = 0
Output
1
Explanation: With zero allowed replacements, the longest run of identical letters is any single element, giving a maximum length of 1.
Constraints
- 1 <= chars.length <= 100000
- 0 <= K <= chars.length
- chars[i] is a lowercase English letter ('a'‑'z')
Optimal Approach & Strategy
Maintain a sliding window with a frequency map and the current max frequency; expand right, shrink left only when replacements exceed K, updating the answer on the fly—O(N) time.
Brute Force Approach
Check every possible sub‑array, count the frequency of each letter inside it, compute the needed replacements, and keep the longest feasible length—O(N^2) time.
Code Solutions
const fs = require('fs');
const input = fs.readFileSync(0,'utf8').trim().split(/\s+/);
let idx=0;
function longestHomogeneousSegment(chars,K){
const n = chars.length;
if(n===0) return 0;
const cnt = new Array(26).fill(0);
let left=0, maxCount=0, best=0;
for(let right=0; right<n; ++right){
const cIdx = chars[right].charCodeAt(0)-97;
cnt[cIdx]++;
maxCount = Math.max(maxCount, cnt[cIdx]);
while((right-left+1)-maxCount > K){
const lIdx = chars[left].charCodeAt(0)-97;
cnt[lIdx]--;
left++;
// recompute maxCount
maxCount = 0;
for(let v of cnt) if(v>maxCount) maxCount=v;
}
best = Math.max(best, right-left+1);
}
return best;
}
const n = parseInt(input[idx++]);
let chars = [];
for(let i=0;i<n;i++) chars.push(input[idx++]);
const K = parseInt(input[idx++]);
console.log(longestHomogeneousSegment(chars,K).toString());#include <bits/stdc++.h>
using namespace std;
int longestHomogeneousSegment(const vector<char>& chars,int K){
int n=chars.size();
if(n==0) return 0;
array<int,26> cnt{}; cnt.fill(0);
int left=0, maxCount=0, best=0;
for(int right=0; right<n; ++right){
cnt[chars[right]-'a']++;
maxCount = max(maxCount, cnt[chars[right]-'a']);
while((right-left+1)-maxCount > K){
cnt[chars[left]-'a']--;
++left;
// recompute maxCount could be stale but works because window only shrinks
maxCount = 0;
for(int c:cnt) maxCount = max(maxCount,c);
}
best = max(best, right-left+1);
}
return best;
}
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n; if(!(cin>>n)) return 0;
vector<char> chars(n);
for(int i=0;i<n;++i) cin>>chars[i];
int K; cin>>K;
cout<<longestHomogeneousSegment(chars,K);
return 0;
}import java.io.*;
import java.util.*;
public class Main {
public static int longestHomogeneousSegment(char[] chars, int K) {
int n = chars.length;
if(n==0) return 0;
int[] cnt = new int[26];
int left = 0, maxCount = 0, best = 0;
for(int right=0; right<n; ++right){
int idx = chars[right]-'a';
cnt[idx]++;
maxCount = Math.max(maxCount, cnt[idx]);
while((right-left+1)-maxCount > K){
cnt[chars[left]-'a']--;
left++;
maxCount = 0;
for(int v:cnt) if(v>maxCount) maxCount=v;
}
best = Math.max(best, right-left+1);
}
return best;
}
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String line = br.readLine();
if(line==null||line.isEmpty()) return;
int n = Integer.parseInt(line.trim());
char[] chars = new char[n];
StringTokenizer st = new StringTokenizer(br.readLine());
for(int i=0;i<n;i++) chars[i] = st.nextToken().charAt(0);
int K = Integer.parseInt(br.readLine().trim());
System.out.println(longestHomogeneousSegment(chars, K));
}
}import sys
def longest_homogeneous_segment(chars, K):
n = len(chars)
if n == 0:
return 0
cnt = [0]*26
left = 0
max_count = 0
best = 0
for right in range(n):
idx = ord(chars[right]) - ord('a')
cnt[idx] += 1
max_count = max(max_count, cnt[idx])
while (right-left+1) - max_count > K:
cnt[ord(chars[left]) - ord('a')] -= 1
left += 1
max_count = max(cnt)
best = max(best, right-left+1)
return best
def main():
data = sys.stdin.read().strip().split()
if not data:
return
it = iter(data)
n = int(next(it))
chars = [next(it) for _ in range(n)]
K = int(next(it))
print(longest_homogeneous_segment(chars, K))
if __name__ == "__main__":
main()const fs = require('fs');
const input = fs.readFileSync(0,'utf8').trim().split(/\s+/);
let idx=0;
function longestHomogeneousSegment(chars,K){
const n = chars.length;
if(n===0) return 0;
const cnt = new Array(26).fill(0);
let left=0, maxCount=0, best=0;
for(let right=0; right<n; ++right){
const cIdx = chars[right].charCodeAt(0)-97;
cnt[cIdx]++;
maxCount = Math.max(maxCount, cnt[cIdx]);
while((right-left+1)-maxCount > K){
const lIdx = chars[left].charCodeAt(0)-97;
cnt[lIdx]--;
left++;
// recompute maxCount
maxCount = 0;
for(let v of cnt) if(v>maxCount) maxCount=v;
}
best = Math.max(best, right-left+1);
}
return best;
}
const n = parseInt(input[idx++]);
let chars = [];
for(let i=0;i<n;i++) chars.push(input[idx++]);
const K = parseInt(input[idx++]);
console.log(longestHomogeneousSegment(chars,K).toString());Asked in Top Tech Interviews
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