Payload Cipher Validator 41 — Problem Statement & Solution Guide

StringsHardFixed/Dynamic Window

TimeO(|s|)

SpaceO(1)

Problem Description

You are tasked with implementing a validation protocol for a stream of encrypted data packets. The input is a string s consisting of lowercase English letters and a positive integer k. The goal is to determine the maximum number of non-overlapping substrings of length k that satisfy a specific 'cipher balance' condition.

A substring of length k is considered 'balanced' if the frequency of every character within that substring is either 0 or exactly k. In other words, a valid window of length k must consist of a single repeated character (e.g., "aaa" or "bbbb").

Your task is to scan the string s and count how many such valid windows exist. The windows must be non-overlapping. Once a valid window is found starting at index i, the next potential window can only start at index i + k. If a window starting at index i is not valid, the next potential window starts at index i + 1. Return the total count of valid, non-overlapping windows found in the string.

DSA Pattern Breakdown

"Payload Cipher Validator 41"

hard

WHY DOES IT MATTER?

Detecting balanced fixed‑size windows is a core pattern in streaming validation, intrusion detection, and real‑time analytics where you must enforce constraints on sliding data without revisiting past elements.

OPTIMIZATION CHALLENGE

The breakthrough is realizing that the balance check can be maintained incrementally: when the window slides, only one character leaves and one enters, so the frequency map updates in constant time, eliminating the need for O(k) recomputation per window.

REAL-WORLD CONNECTION

Think of a network firewall that inspects every k‑byte packet and drops it if any byte repeats more than allowed – the algorithm mirrors how the firewall slides over a continuous byte stream, validates each packet in O(1), and decides instantly whether to accept or reject.

During an interview, implement the sliding window first, then add the greedy jump. If you get stuck, write a helper that returns true/false for a window – this isolates the balance logic and keeps the main loop clean.

COMPLEXITY AT A GLANCE

⏱ Time:O(|s|)

💾 Space:O(1)

Core Theory — Why This Approach?

The problem reduces to finding the maximum set of non‑overlapping length‑k windows whose internal character distribution satisfies a balance predicate (e.g., all characters appear at most once). A naïve solution would examine every possible window, verify the balance condition by counting frequencies from scratch, and then try all subsets of windows – an exponential blow‑up that fails for |s| up to 10^5. The optimal paradigm combines two classic techniques: (1) a sliding‑window frequency map that updates in O(1) per shift, allowing each window to be validated in constant amortized time, and (2) a greedy interval‑selection strategy that always picks the earliest valid window and then jumps k positions forward, guaranteeing the maximal count because all windows have identical length. This yields a linear‑time solution that scales to the hardest test cases.

Interview Questions on This Problem

Q1How would you modify the solution if the balance condition required each character to appear at most twice instead of once?

Replace the strict ‘count == 1’ check with ‘count <= 2’. The sliding window still updates in O(1); the greedy selection remains optimal because the window length is fixed, so picking the earliest valid window still maximizes the count.

Q2Can the greedy interval‑selection fail when windows have varying lengths? Explain why it works here.

Greedy fails for variable‑length intervals because a later longer interval might block multiple shorter ones. In this problem all windows are exactly length k, so picking the leftmost valid window never precludes a better solution; any alternative solution can be transformed into the greedy one without reducing the count.

Q3What is the time‑space trade‑off if you pre‑compute a prefix‑sum array of character frequencies for the whole string?

A prefix‑sum for each of the 26 letters lets you query any window’s frequencies in O(26) = O(1) time, but building the 26 × |s| table costs O(26·|s|) = O(|s|) space, which may be prohibitive for very large inputs. The sliding‑window map uses O(26) = O(1) extra space and the same O(|s|) time, making it preferable.

Examples

Example 1

Input

s = "aaabbbccc", k = 3

Output

Explanation: 1. Start at index 0: Substring "aaa" (indices 0-2). All characters are 'a'. Frequency of 'a' is 3, which equals k. This is a valid window. Count = 1. Next search starts at index 0 + 3 = 3. 2. Start at index 3: Substring "bbb" (indices 3-5). All characters are 'b'. Frequency of 'b' is 3, which equals k. This is a valid window. Count = 2. Next search starts at index 3 + 3 = 6. 3. Start at index 6: Substring "ccc" (indices 6-8). All characters are 'c'. Frequency of 'c' is 3, which equals k. This is a valid window. Count = 3. Next search starts at index 6 + 3 = 9. 4. Index 9 is out of bounds. Stop. Total count is 3.

Example 2

Input

s = "ababab", k = 2

Output

Explanation: 1. Start at index 0: Substring "ab" (indices 0-1). Characters are 'a' and 'b'. Frequencies are 1 and 1. Neither is 0 nor equal to k (2). Invalid. Next search starts at index 0 + 1 = 1. 2. Start at index 1: Substring "ba" (indices 1-2). Characters are 'b' and 'a'. Frequencies are 1 and 1. Invalid. Next search starts at index 1 + 1 = 2. 3. Start at index 2: Substring "ab" (indices 2-3). Invalid. Next search starts at index 3. 4. Start at index 3: Substring "ba" (indices 3-4). Invalid. Next search starts at index 4. 5. Start at index 4: Substring "ab" (indices 4-5). Invalid. Next search starts at index 5. 6. Index 5 + k = 7 > length 6. Stop. Total count is 0.

Example 3

Input

s = "aaabaaa", k = 3

Output

Explanation: 1. Start at index 0: Substring "aaa" (indices 0-2). Valid. Count = 1. Next search starts at index 3. 2. Start at index 3: Substring "baa" (indices 3-5). Characters 'b', 'a', 'a'. Frequencies: b=1, a=2. Invalid. Next search starts at index 4. 3. Start at index 4: Substring "aaa" (indices 4-6). Valid. Count = 2. Next search starts at index 7. 4. Index 7 is out of bounds. Stop. Total count is 2.

Example 4

Input

s = "aaaa", k = 2

Output

Explanation: 1. Start at index 0: Substring "aa" (indices 0-1). Valid. Count = 1. Next search starts at index 2. 2. Start at index 2: Substring "aa" (indices 2-3). Valid. Count = 2. Next search starts at index 4. 3. Index 4 is out of bounds. Stop. Total count is 2.

Constraints

1 <= s.length <= 10^5
1 <= k <= s.length
s consists of lowercase English letters only.

Optimal Approach & Strategy

Use a sliding window to update frequencies in O(1) per shift and greedily pick the earliest valid window, jumping k positions each time to ensure non‑overlap.

Brute Force Approach

Generate every possible length‑k substring, recompute its character frequencies from scratch, and then try all combinations of non‑overlapping substrings to find the maximum count.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(|s|)

/**
 * @param {string} s
 * @param {number} k
 * @return {number}
 */
var validatePayload = function(s, k) {
    let n = s.length;
    let count = 0;
    for (let i = 0; i + k <= n; i += k) {
        let isSorted = true;
        for (let j = i + 1; j < i + k; j++) {
            if (s[j] < s[j - 1]) {
                isSorted = false;
                break;
            }
        }
        if (isSorted) {
            count++;
        }
    }
    return count;
};

class Solution {
public:
    int validatePayload(const string& s, int k) {
        int n = s.size();
        int count = 0;
        for (int i = 0; i + k <= n; i += k) {
            bool isSorted = true;
            for (int j = i + 1; j < i + k; ++j) {
                if (s[j] < s[j - 1]) {
                    isSorted = false;
                    break;
                }
            }
            if (isSorted) {
                count++;
            }
        }
        return count;
    }
};

class Solution {
    public int validatePayload(String s, int k) {
        int n = s.length();
        int count = 0;
        for (int i = 0; i + k <= n; i += k) {
            boolean isSorted = true;
            for (int j = i + 1; j < i + k; j++) {
                if (s.charAt(j) < s.charAt(j - 1)) {
                    isSorted = false;
                    break;
                }
            }
            if (isSorted) {
                count++;
            }
        }
        return count;
    }
}

class Solution:
    def validatePayload(self, s: str, k: int) -> int:
        n = len(s)
        count = 0
        for i in range(0, n - k + 1, k):
            is_sorted = True
            for j in range(i + 1, i + k):
                if s[j] < s[j - 1]:
                    is_sorted = False
                    break
            if is_sorted:
                count += 1
        return count

/**
 * @param {string} s
 * @param {number} k
 * @return {number}
 */
var validatePayload = function(s, k) {
    let n = s.length;
    let count = 0;
    for (let i = 0; i + k <= n; i += k) {
        let isSorted = true;
        for (let j = i + 1; j < i + k; j++) {
            if (s[j] < s[j - 1]) {
                isSorted = false;
                break;
            }
        }
        if (isSorted) {
            count++;
        }
    }
    return count;
};

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Payload Cipher Validator 41 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Strings

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Payload Cipher Validator 41 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor