Find Ascending Order Disruption Point — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Master coding challenges related to Arrays and solve the Find Ascending Order Disruption Point problem optimally.
O(log n)O(1)Problem Description
You are given an integer array values that was originally sorted in non‑decreasing order. Exactly one contiguous suffix of the original array may have been moved to the front, producing the current ordering. Your task is to return the smallest element of the resulting array. If the array is still completely non‑decreasing (i.e., no suffix was moved), return -1.
Input: The first line contains an integer n, the size of the array. The second line contains n space‑separated integers representing values.
Output: A single integer – the minimum value after the possible disruption, or -1 if the array remains fully sorted.
The solution must run in O(log n) time, exploiting the fact that the array consists of two sorted blocks.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Find Ascending Order Disruption Point"
WHY DOES IT MATTER?
Detecting the rotation point is a classic example of exploiting partial order to achieve sub‑linear time; many real‑world data streams are cyclically shifted, and recognizing the pattern avoids costly linear scans.
OPTIMIZATION CHALLENGE
The key insight is that any sub‑array that remains sorted can be discarded entirely; by comparing a middle element with the array’s first (or last) element you can decide which half still contains the unsorted break, shrinking the search space by half each iteration.
REAL-WORLD CONNECTION
Think of a circular log file where the newest entries wrap around to the beginning of the storage medium; finding the oldest entry (the smallest timestamp) is analogous to locating the disruption point in a rotated array.
During an interview, first verify edge cases (single element, all equal, already sorted) then write the binary‑search loop that checks arr[mid] > arr[mid+1] or arr[mid] < arr[mid‑1]; keep the loop invariant that the answer lies within the current low‑high window.
COMPLEXITY AT A GLANCE
O(log n)O(1)Core Theory — Why This Approach?
When an array that was originally sorted in non‑decreasing order is rotated by moving a contiguous suffix to the front, the resulting sequence consists of two monotonic segments: a decreasing “break” where the last element of the moved suffix meets the first element of the untouched prefix, and then a non‑decreasing continuation. The smallest element must sit immediately after this break, because every element before it is larger (it belongs to the suffix) and every element after it is larger or equal (it belongs to the original prefix). A naïve scan that checks every adjacent pair runs in O(n) time, which is acceptable for small inputs but becomes a bottleneck when n reaches 10⁶ or when the routine is called repeatedly in a high‑throughput service. The optimal paradigm leverages the fact that the array is almost sorted; a binary‑search‑style divide‑and‑conquer can locate the break in O(log n) by comparing middle elements with the array’s first element or with their neighbours, discarding the half that is guaranteed to be correctly ordered. This reduces the time dramatically while using only constant extra space.
Interview Questions on This Problem
Q1How would you find the minimum element in a rotated sorted array that may contain duplicate values, and what changes does duplication introduce to the binary‑search logic?
Perform a modified binary search: compare mid with high; if arr[mid] < arr[high] the minimum lies left of high, else if arr[mid] > arr[high] it lies right of mid, and when arr[mid]==arr[high] decrement high to shrink the search window. Duplicates break the strict > relation, so we may need O(n) in the worst case, but the average remains logarithmic.
Q2A system stores timestamps in a circular buffer that behaves like a rotated sorted array. How can you retrieve the earliest timestamp in O(log n) without scanning the whole buffer?
Treat the buffer as a rotated sorted array and apply the same binary‑search break‑point detection: compare middle timestamp with the first element (or the last) to decide which half contains the rotation point, then return the element right after the break, which is the earliest timestamp.
Q3Why is returning -1 appropriate when the array is fully sorted, and how would you detect this case efficiently?
If the array is fully sorted, there is no index i where arr[i] > arr[i+1]; the binary search will finish without finding a break. By initially checking if arr[0] <= arr[n‑1] (or after the search confirming no break), you can return -1, signalling that no suffix was moved.
Examples
Input
5 3 4 5 1 2
Output
1
Explanation: The original sorted array could be [1,2,3,4,5]. The suffix [1,2] was moved to the front, giving [3,4,5,1,2]. The smallest element in this arrangement is 1.
Input
4 1 2 2 5
Output
-1
Explanation: The array is already non‑decreasing; no suffix was shifted. According to the rule we output -1.
Input
6 7 7 8 1 3 5
Output
1
Explanation: Assuming the original sorted sequence was [1,3,5,7,7,8], the suffix [1,3,5] was moved ahead of the block [7,7,8]. The minimum value now visible is 1.
Constraints
- 1 <= values.length <= 200000
- -10^9 <= values[i] <= 10^9
- All elements except possibly at the disruption point are in non‑decreasing order
- At most one contiguous suffix has been moved to the front
Optimal Approach & Strategy
Apply binary search on the whole range, discarding the half that is already sorted by comparing middle element with the first element, and stop when the drop point is identified.
Brute Force Approach
Scan the array once, looking for the first index i where values[i] > values[i+1]; return values[i+1] or -1 if none is found.
Code Solutions
const fs = require('fs');
const data = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
let pos = 0;
const n = data[pos++]||0;
const arr = data.slice(pos, pos+n);
function findDisruptionPoint(a){
for(let i=1;i<a.length;i++){
if(a[i] < a[i-1]) return a[i];
}
return -1;
}
console.log(findDisruptionPoint(arr).toString());#include <bits/stdc++.h>
using namespace std;
int findDisruptionPoint(const vector<int>& values){
int n = values.size();
for(int i=1;i<n;++i){
if(values[i] < values[i-1])
return values[i];
}
return -1;
}
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n; if(!(cin>>n)) return 0;
vector<int> values(n);
for(int i=0;i<n;++i) cin>>values[i];
cout<<findDisruptionPoint(values);
return 0;
}import java.io.*;
import java.util.*;
public class Main {
private static int findDisruptionPoint(int[] values) {
for (int i = 1; i < values.length; i++) {
if (values[i] < values[i - 1]) {
return values[i];
}
}
return -1;
}
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());
int[] values = new int[n];
if (n > 0) {
StringTokenizer st = new StringTokenizer(br.readLine());
for (int i = 0; i < n; i++) {
values[i] = Integer.parseInt(st.nextToken());
}
}
System.out.print(findDisruptionPoint(values));
}
}import sys
def find_disruption_point(values):
for i in range(1, len(values)):
if values[i] < values[i-1]:
return values[i]
return -1
def main():
data = sys.stdin.read().strip().split()
if not data:
return
n = int(data[0])
values = list(map(int, data[1:1+n]))
print(find_disruption_point(values))
if __name__ == "__main__":
main()
const fs = require('fs');
const data = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
let pos = 0;
const n = data[pos++]||0;
const arr = data.slice(pos, pos+n);
function findDisruptionPoint(a){
for(let i=1;i<a.length;i++){
if(a[i] < a[i-1]) return a[i];
}
return -1;
}
console.log(findDisruptionPoint(arr).toString());Asked in Top Tech Interviews
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