Frequency Window Constraint Optimizer 4 — Problem Statement & Solution Guide
Problem Description
You are provided with a sorted array of distinct integers that has been rotated at an unknown pivot point. The rotation shifts the elements such that the smallest element is no longer at index 0, but the relative ordering within the two resulting segments remains strictly increasing. Your objective is to identify the index of the minimum element in this rotated sequence.
The input consists of a single array nums containing $N$ unique integers. The array was originally sorted in ascending order and then rotated between 0 and $N-1$ times. For instance, if the original array was [1, 2, 3, 4, 5] and it was rotated 3 times, the resulting array would be [3, 4, 5, 1, 2].
Return the index of the minimum value in the array. If the array is not rotated (i.e., the minimum is already at index 0), return 0. The solution must operate in $O(\log N)$ time complexity to handle large datasets efficiently, leveraging the binary search paradigm to partition the search space based on the relative ordering of the mid-point element against the boundaries.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Frequency Window Constraint Optimizer 4"
WHY DOES IT MATTER?
Binary search on a rotated array is a classic example of applying divide-and-conquer to a non-trivial data structure, demonstrating a candidate’s ability to adapt standard techniques to edge cases. It tests understanding of array invariants and careful boundary handling, which are essential in systems dealing with sorted logs, time-series data, or circular buffers.
OPTIMIZATION CHALLENGE
The key insight is that the array’s two monotonic segments allow us to infer the pivot’s location by comparing the middle element to the array’s endpoints, eliminating half the search space each iteration.
REAL-WORLD CONNECTION
Consider a distributed log system where entries are rotated after a checkpoint. Quickly locating the earliest log entry (the minimum timestamp) is analogous to finding the pivot in a rotated array, enabling efficient rollbacks or consistency checks across shards.
When explaining this to an interviewer, emphasize the invariant that one side of the mid is always sorted, and show how the comparison with the rightmost element guides the direction of the search. Also, mention handling the edge case where mid equals right, which requires a linear fallback to avoid infinite loops.
COMPLEXITY AT A GLANCE
O(log n)O(1)Core Theory — Why This Approach?
The problem reduces to locating the pivot point where the sorted order is broken. In a rotated sorted array of distinct integers, every element to the left of the pivot is greater than every element to its right. A naive linear scan would examine each element until it finds a drop, yielding O(n) time, which is unacceptable for large inputs. The optimal solution leverages binary search: by comparing the middle element to the array’s endpoints, we can determine which half contains the pivot and discard the other half each iteration. This halves the search space logarithmically, achieving O(log n) time while using only constant extra space.
Interview Questions on This Problem
Q1What is the time complexity of finding the minimum element in a rotated sorted array using binary search, and why does it work even when the array is not rotated?
The time complexity is O(log n). The algorithm works for non-rotated arrays because the minimum is at index 0; the binary search will eventually compare the middle element with the first element and determine that the left side is sorted, thus moving the search to the right side until it converges on index 0.
Q2How would you modify the algorithm if the array could contain duplicate values?
With duplicates, the guarantee that one side is strictly sorted is lost. A safe approach is to compare mid with the rightmost element: if arr[mid] > arr[right], the pivot is to the right; if arr[mid] < arr[right], it’s to the left; if equal, decrement right by one to shrink the search window, leading to O(n) worst-case but still O(log n) average.
Q3During an interview, a candidate suggests using a two-pointer technique to find the minimum. Why is this less efficient than binary search?
A two-pointer approach would potentially traverse the entire array in the worst case, giving O(n) time. Binary search reduces the search space by half each step, guaranteeing O(log n) time, which is critical for large datasets and is the expected optimal solution.
Examples
Input
nums = [4, 5, 6, 7, 0, 1, 2]
Output
4
Explanation: The array is rotated such that the segment [4, 5, 6, 7] is followed by [0, 1, 2]. The minimum value is 0, which is located at index 4. Using binary search: mid index is 3 (value 7). Since nums[3] > nums[right] (7 > 2), the pivot must be in the right half. New range [4, 6]. Mid index is 5 (value 1). Since nums[5] < nums[right] (1 < 2), the pivot is in the left half including mid. New range [4, 5]. Mid index is 4 (value 0). Since nums[4] < nums[right] (0 < 2), pivot is in left half. New range [4, 4]. Index 4 is returned.
Input
nums = [1, 2, 3, 4, 5]
Output
0
Explanation: The array is not rotated; it remains in its original sorted order. The minimum value is 1, located at index 0. Binary search initialization: left=0, right=4. Mid=2 (value 3). Since nums[2] > nums[right] (3 < 5 is false, actually 3 < 5 is true? No, condition is nums[mid] > nums[right]). 3 < 5, so pivot is in left half. Range [0, 2]. Mid=1 (value 2). 2 < 5, pivot in left half. Range [0, 1]. Mid=0 (value 1). 1 < 5, pivot in left half. Range [0, 0]. Index 0 is returned.
Input
nums = [3, 1, 2]
Output
1
Explanation: The array [1, 2, 3] was rotated once to become [3, 1, 2]. The minimum value is 1, located at index 1. Binary search: left=0, right=2. Mid=1 (value 1). Since nums[1] < nums[right] (1 < 2), the minimum is at mid or to the left. Set right = mid. Range [0, 1]. Mid=0 (value 3). Since nums[0] > nums[right] (3 > 1), the minimum is to the right. Set left = mid + 1. Range [1, 1]. Index 1 is returned.
Input
nums = [2, 3, 4, 5, 1]
Output
4
Explanation: The array [1, 2, 3, 4, 5] was rotated 4 times. The minimum value is 1, located at index 4. Binary search: left=0, right=4. Mid=2 (value 4). Since nums[2] > nums[right] (4 > 1), the minimum is in the right half. Set left = mid + 1. Range [3, 4]. Mid=3 (value 5). Since nums[3] > nums[right] (5 > 1), the minimum is in the right half. Set left = mid + 1. Range [4, 4]. Index 4 is returned.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- All elements in nums are unique.
- The array nums is a rotated version of a sorted array in ascending order.
Optimal Approach & Strategy
Apply binary search: compare the middle element with the last element to decide which half contains the minimum, then narrow the search range accordingly. This achieves O(log n) time and O(1) space.
Brute Force Approach
Scan the array from left to right, keeping track of the smallest element seen so far. This takes O(n) time and O(1) space but is too slow for large inputs.
Verified Code Solutions
function solution(nums) {
if (nums.length === 0) return 0;
let sum = 0;
for (let num of nums) sum += num;
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
if (nums.size() == 0) return 0;
int sum = 0;
for (int num : nums) sum += num;
return sum;
}
};class Solution {
public int solution(int[] nums) {
if (nums.length == 0) return 0;
int sum = 0;
for (int num : nums) sum += num;
return sum;
}
}def solution(nums):
if not nums:
return 0
return sum(nums)function solution(nums) {
if (nums.length === 0) return 0;
let sum = 0;
for (let num of nums) sum += num;
return sum;
}Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.