Frequency Window Constraint Resolver 8 — Problem Statement & Solution Guide
Problem Description
You are given 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 the beginning, but the relative order of the remaining segments is preserved. Your task is to identify the index of the minimum element in this rotated sequence, which corresponds to the pivot point where the rotation occurred.
The input is a single array of integers. The output must be the zero-based index of the smallest value in the array. If the array is not rotated (i.e., the smallest element is already at index 0), return 0. The solution must operate in logarithmic time complexity, leveraging the binary search paradigm to efficiently narrow down the search space by comparing the mid-point element with the boundaries of the current search interval.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Frequency Window Constraint Resolver 8"
WHY DOES IT MATTER?
Binary search on a rotated array demonstrates how to adapt a classic algorithm to non-standard data layouts, a skill highly valued in interviews and production code. It shows mastery of invariant maintenance and boundary conditions.
OPTIMIZATION CHALLENGE
The core insight is that the pivot is the only element where the next element is smaller. By comparing the middle element to the high end, we can discard half the array each iteration, reducing time from linear to logarithmic.
REAL-WORLD CONNECTION
Consider a circular buffer in a network router that stores packet timestamps. When the buffer wraps around, the earliest packet is no longer at index 0; finding it quickly is analogous to locating the pivot in a rotated array.
Always verify that the array is not fully sorted (i.e., pivot at index 0) before starting the binary search; this edge case can be handled by a simple check of arr[0] < arr[n-1].
COMPLEXITY AT A GLANCE
O(log n)O(1)Core Theory — Why This Approach?
Finding the minimum element in a rotated sorted array is a classic application of binary search on a modified search space. The array is originally sorted in ascending order, but a rotation at an unknown pivot splits it into two ascending subarrays: the right part contains the smallest elements, and the left part contains the larger ones. A naive linear scan would examine each element, yielding O(n) time, which is inefficient for large inputs. By exploiting the fact that the array is mostly sorted, we can compare the middle element with the high end to determine which half contains the pivot. If the middle element is greater than the high element, the pivot lies to the right; otherwise, it lies to the left or at mid. Repeating this halving process reduces the search interval logarithmically, achieving O(log n) time while using only constant extra space.
The key insight is that the rotated array preserves the relative order within each segment, so the minimum is the only element that violates the ascending property with its predecessor. Binary search leverages this property by checking the monotonicity of the subarray boundaries rather than inspecting every element. This paradigm is widely used in problems involving rotated or partially sorted data structures, such as finding rotation counts, searching in circular buffers, or locating breakpoints in time-series data.
Because the array contains distinct integers, there is a unique pivot, and the algorithm can safely discard half of the search space at each step. If duplicates were allowed, additional checks would be required to handle equal values, but the distinctness guarantees that the comparison between mid and high is decisive. Thus, binary search provides a clean, optimal solution that scales gracefully with input size.
Interview Questions on This Problem
Q1How would you modify the algorithm if the array could contain duplicate values?
With duplicates, the comparison mid vs high may not be decisive when arr[mid]==arr[high]. In that case, we can safely decrement high by one and continue, because the duplicate does not affect the pivot location. This degrades the worst-case to O(n) but still works for most cases.
Q2In a distributed system, you have a sorted list of timestamps that has been rotated due to a clock skew. How would you locate the earliest timestamp efficiently?
Treat the timestamps as a rotated sorted array and apply the same binary search logic. Compare the middle timestamp with the last one; if the middle is greater, the earliest timestamp is to the right, otherwise to the left. This gives O(log n) time across the distributed log shards.
Q3What is the time complexity if you were to use a linear scan versus binary search for this problem, and why is binary search preferable in production?
Linear scan is O(n) and would become a bottleneck for large logs or real-time analytics. Binary search reduces the search to O(log n), which is essential for high-throughput services where latency must be minimized.
Examples
Input
nums = [15, 18, 22, 3, 6, 9, 12]
Output
3
Explanation: The array is rotated such that the segment [3, 6, 9, 12] follows [15, 18, 22]. The minimum value is 3, which is located at index 3. Binary search compares mid (index 3, value 3) with the right boundary (index 6, value 12). Since 3 < 12, the minimum must be in the left half including mid. The search converges to index 3.
Input
nums = [4, 5, 6, 7, 8, 1, 2, 3]
Output
5
Explanation: The pivot is at index 5 where the value 1 resides. The array is split into [4, 5, 6, 7, 8] and [1, 2, 3]. Comparing mid (index 3, value 7) with the right boundary (index 7, value 3), since 7 > 3, the minimum is in the right half. The search continues until index 5 is isolated as the minimum.
Input
nums = [1, 2, 3, 4, 5, 6, 7]
Output
0
Explanation: The array is not rotated; it remains in ascending order. The minimum value is 1 at index 0. Binary search will eventually compare the mid-point with the right boundary and determine that the left half contains the minimum, converging to index 0.
Input
nums = [50, 10, 20, 30, 40]
Output
1
Explanation: The array is rotated at index 1. The minimum value is 10. Comparing mid (index 2, value 20) with the right boundary (index 4, value 40), since 20 < 40, the minimum is in the left half. The next mid (index 0, value 50) is compared with the right boundary (index 1, value 10). Since 50 > 10, the minimum is in the right half, isolating index 1.
Constraints
- 1 <= nums.length <= 10^5
- All elements in nums are distinct integers.
- -10^9 <= nums[i] <= 10^9
- The array is guaranteed to be a rotated version of a strictly increasing sorted array.
Optimal Approach & Strategy
Apply binary search: maintain low and high indices; while low < high, compute mid; if arr[mid] > arr[high], set low = mid+1 else high = mid. Return low as the pivot index.
Brute Force Approach
Scan the array from left to right, keeping track of the smallest element seen so far. Return its index after one full pass.
Verified Code Solutions
function solution(nums) {
let pivot = findPivot(nums);
let maxFreq = 0;
let windowStart = pivot;
for (let windowEnd = pivot; windowEnd < nums.length; windowEnd++) {
let count = 0;
for (let i = windowStart; i <= windowEnd; i++) {
if (nums[i] === nums[windowEnd]) {
count++;
}
}
maxFreq = Math.max(maxFreq, count);
windowStart = windowEnd + 1;
}
return maxFreq;
function findPivot(nums) {
let left = 0;
let right = nums.length - 1;
while (left < right) {
let mid = Math.floor((left + right) / 2);
if (nums[mid] > nums[right]) {
left = mid + 1;
} else {
right = mid;
}
}
return left;
}
}class Solution {
public:
int solution(vector<int>& nums) {
int pivot = findPivot(nums);
int maxFreq = 0;
int windowStart = pivot;
for (int windowEnd = pivot; windowEnd < nums.size(); windowEnd++) {
int count = 0;
for (int i = windowStart; i <= windowEnd; i++) {
if (nums[i] == nums[windowEnd]) {
count++;
}
}
maxFreq = max(maxFreq, count);
windowStart = windowEnd + 1;
}
return maxFreq;
}
int findPivot(vector<int>& nums) {
int left = 0;
int right = nums.size() - 1;
while (left < right) {
int mid = left + (right - left) / 2;
if (nums[mid] > nums[right]) {
left = mid + 1;
} else {
right = mid;
}
}
return left;
}
}class Solution {
public int solution(int[] nums) {
int pivot = findPivot(nums);
int maxFreq = 0;
int windowStart = pivot;
for (int windowEnd = pivot; windowEnd < nums.length; windowEnd++) {
int count = 0;
for (int i = windowStart; i <= windowEnd; i++) {
if (nums[i] == nums[windowEnd]) {
count++;
}
}
maxFreq = Math.max(maxFreq, count);
windowStart = windowEnd + 1;
}
return maxFreq;
}
public int findPivot(int[] nums) {
int left = 0;
int right = nums.length - 1;
while (left < right) {
int mid = left + (right - left) / 2;
if (nums[mid] > nums[right]) {
left = mid + 1;
} else {
right = mid;
}
}
return left;
}
}def solution(nums):
pivot = find_pivot(nums)
max_freq = 0
window_start = pivot
for window_end in range(pivot, len(nums)):
count = 0
for i in range(window_start, window_end + 1):
if nums[i] == nums[window_end]:
count += 1
max_freq = max(max_freq, count)
window_start = window_end + 1
return max_freq
def find_pivot(nums):
left = 0
right = len(nums) - 1
while left < right:
mid = (left + right) // 2
if nums[mid] > nums[right]:
left = mid + 1
else:
right = mid
return leftfunction solution(nums) {
let pivot = findPivot(nums);
let maxFreq = 0;
let windowStart = pivot;
for (let windowEnd = pivot; windowEnd < nums.length; windowEnd++) {
let count = 0;
for (let i = windowStart; i <= windowEnd; i++) {
if (nums[i] === nums[windowEnd]) {
count++;
}
}
maxFreq = Math.max(maxFreq, count);
windowStart = windowEnd + 1;
}
return maxFreq;
function findPivot(nums) {
let left = 0;
let right = nums.length - 1;
while (left < right) {
let mid = Math.floor((left + right) / 2);
if (nums[mid] > nums[right]) {
left = mid + 1;
} else {
right = mid;
}
}
return left;
}
}Asked in Top Tech Interviews
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