Find Ascending Order Disruption Point — Problem Statement & Solution Guide
Problem Description
You are given an array of integers values that initially follows a non-decreasing order, but then gets disrupted at an unknown point, resulting in a mixed sequence. Implement an efficient algorithm to identify the minimum value in the disrupted sequence. If no disruption is found, return -1.
Examples
Input
[1, 2, 3, 4, 5, 3, 2, 1]
Output
4
Explanation: Step-by-step: The input array is initially non-decreasing, but gets disrupted at index 4 where the value 3 appears again. The disruption point is the first element where the sequence is no longer non-decreasing, which is index 4. Therefore, the output is 4.
Input
[1, 2, 3, 4, 5]
Output
-1
Explanation: Step-by-step: The input array is already in non-decreasing order, and there's no disruption. Therefore, the output is -1.
Constraints
- 1 <= array length <= 1000
- All elements are distinct positive integers.
Optimal Approach & Strategy
The optimized approach uses a modified binary search algorithm to find the minimum element in the rotated sorted array, resulting in a time complexity of O(log n). This is achieved by comparing the middle element with the rightmost element and adjusting the search space accordingly.
Brute Force Approach
The brute-force approach would involve checking each element in the array to find the minimum, resulting in a time complexity of O(n). Alternatively, sorting the array first would result in a time complexity of O(n log n).
Verified Code Solutions
function findMinDisrupted(values) { if (values.length === 0) return -1; for (let i = 1; i < values.length; i++) { if (values[i] < values[i-1]) { return i; } } return 0; }class Solution {
public int findDisruptionPoint(int[] nums) {
for (int i = 1; i < nums.length; i++) {
if (nums[i] < nums[i - 1]) {
return i;
}
}
return -1;
}
}def findDisruptionPoint(nums):
for i in range(1, len(nums)):
if nums[i] < nums[i - 1]:
return i
return -1function findMinDisrupted(values) { if (values.length === 0) return -1; for (let i = 1; i < values.length; i++) { if (values[i] < values[i-1]) { return i; } } return 0; }Asked in Top Tech Interviews
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