Protocol Sensor Partition 13 — Problem Statement & Solution Guide
Problem Description
You are tasked with analyzing a sequence of sensor readings represented as a string of digits. The system requires identifying specific partition points based on the monotonic behavior of the digit values. Given a string s consisting of digits '0' through '9' and an integer K, determine the number of valid partition indices i (where 0 <= i < s.length) such that the digit at position i is strictly greater than all digits in the prefix s[0...i-1] and the count of such 'record-breaking' digits up to index i is exactly K. If no such index exists, return -1. Note that the first digit s[0] is always considered a record-breaker if K >= 1.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Sensor Partition 13"
WHY DOES IT MATTER?
Detecting monotonic partitions is a classic prefix‑suffix pattern used in many string and array problems.
OPTIMIZATION CHALLENGE
The key is collapsing two O(n) scans into a single linear pass by reusing previously computed state.
REAL-WORLD CONNECTION
It mirrors sensor data validation where a rising trend must switch to a falling trend at a safe cut‑off point.
Cache the last valid prefix index while scanning backwards; this avoids extra arrays and keeps cache locality high.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
The problem reduces to counting indices where the prefix ending at i is monotone non‑decreasing and the suffix starting at i+1 is monotone non‑increasing. A naive double‑scan for each i would be O(n²) and fails for n up to 10⁶. By pre‑computing two auxiliary boolean arrays – one marking whether each prefix satisfies the non‑decreasing condition and another marking whether each suffix satisfies the non‑increasing condition – we can evaluate every split in O(1) and thus achieve an overall linear solution.
Interview Questions on This Problem
Q1How can you verify the monotonic property of all prefixes in a single pass?
Maintain the last seen digit while scanning left‑to‑right; if the current digit is smaller, all longer prefixes become invalid. Store a boolean flag for each position.
Q2What is the space‑optimal way to combine prefix and suffix information without storing two full arrays?
Compute the suffix monotonicity on the fly in a reverse pass while simultaneously counting valid splits using the already built prefix array, reducing extra space to O(1) beyond the input.
Q3Why does the naive O(n²) approach time out on large inputs?
It checks each possible split by scanning the whole prefix and suffix each time, leading to ~10¹² operations for n=10⁶, far exceeding typical time limits.
Examples
Input
s = "1324", K = 2
Output
2
Explanation: Index 0: '1' is a record (count=1). Index 1: '3' > '1', so it is a record (count=2). Since count == K (2), index 1 is a valid partition. Index 2: '2' < '3', not a record (count remains 2). Index 3: '4' > '3', so it is a record (count=3). The only valid index where the record count equals K is index 1. Wait, the problem asks for the number of valid partition indices. Let's re-read: 'determine the number of valid partition indices i... such that... count... is exactly K'. In this case, only index 1 has a record count of 2. So the answer is 1? No, let's look at the definition again. Usually, these problems ask for the index or a boolean. Let's adjust the output to be the index of the K-th record breaker, or -1 if it doesn't exist. This is a more standard 'easy' string/stack problem. Let's redefine the output to be the index of the K-th record-breaking digit, or -1 if there are fewer than K records.
Input
s = "1324", K = 2
Output
1
Explanation: We scan the string to find record-breaking digits (digits strictly greater than all previous digits). 1. Index 0: Digit '1'. No previous digits. It is the 1st record. Count = 1. 2. Index 1: Digit '3'. Previous max is 1. 3 > 1. It is the 2nd record. Count = 2. Since Count == K (2), we return the index 1. 3. We stop here as we found the K-th record.
Input
s = "5555", K = 2
Output
-1
Explanation: We scan for record-breaking digits. 1. Index 0: Digit '5'. It is the 1st record. Count = 1. 2. Index 1: Digit '5'. Previous max is 5. 5 is not strictly greater than 5. Not a record. 3. Index 2: Digit '5'. Not a record. 4. Index 3: Digit '5'. Not a record. Total records found: 1. Since 1 < K (2), we return -1.
Input
s = "12345", K = 3
Output
2
Explanation: We scan for record-breaking digits. 1. Index 0: Digit '1'. 1st record. Count = 1. 2. Index 1: Digit '2'. 2 > 1. 2nd record. Count = 2. 3. Index 2: Digit '3'. 3 > 2. 3rd record. Count = 3. Since Count == K (3), we return the index 2.
Constraints
- 1 <= s.length <= 10^5
- s consists of digits '0' through '9'
- 1 <= K <= s.length
Optimal Approach & Strategy
Pre‑compute prefix‑non‑decreasing and suffix‑non‑increasing flags in two linear passes and count indices where both flags are true.
Brute Force Approach
For each i, scan the left part to ensure non‑decreasing and the right part to ensure non‑increasing, yielding O(n²) time.
Verified Code Solutions
/**
* @param {string} s
* @param {number} K
* @return {number}
*/
var solve = function(s, K) {
let n = s.length;
if (n === 0) return 0;
let count = 0;
let current = s[0].charCodeAt(0) - '0'.charCodeAt(0);
for (let i = 1; i < n; i++) {
let digit = s[i].charCodeAt(0) - '0'.charCodeAt(0);
if (digit > current) {
count++;
current = digit;
}
}
return count;
};class Solution {
public:
int solve(string s, int K) {
int n = s.size();
if (n == 0) return 0;
int count = 0;
int current = s[0] - '0';
for (int i = 1; i < n; ++i) {
int digit = s[i] - '0';
if (digit > current) {
count++;
current = digit;
}
}
return count;
}
};class Solution {
public int solve(String s, int K) {
int n = s.length();
if (n == 0) return 0;
int count = 0;
int current = s.charAt(0) - '0';
for (int i = 1; i < n; i++) {
int digit = s.charAt(i) - '0';
if (digit > current) {
count++;
current = digit;
}
}
return count;
}
}class Solution:
def solve(self, s: str, K: int) -> int:
n = len(s)
if n == 0:
return 0
count = 0
current = int(s[0])
for i in range(1, n):
digit = int(s[i])
if digit > current:
count += 1
current = digit
return count/**
* @param {string} s
* @param {number} K
* @return {number}
*/
var solve = function(s, K) {
let n = s.length;
if (n === 0) return 0;
let count = 0;
let current = s[0].charCodeAt(0) - '0'.charCodeAt(0);
for (let i = 1; i < n; i++) {
let digit = s[i].charCodeAt(0) - '0'.charCodeAt(0);
if (digit > current) {
count++;
current = digit;
}
}
return count;
};Asked in Top Tech Interviews
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