Min Length Alternating Subarray — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Master coding challenges related to Arrays and solve the Min Length Alternating Subarray problem optimally.
O(n)O(1)Problem Description
Given an integer array nums of length n and a non‑negative integer k, determine the smallest possible length of a contiguous subarray that contains exactly k positions j such that nums[j] > nums[j+1] (i.e., a descending adjacent pair). If no subarray satisfies the condition, return -1. The array is zero‑indexed; the subarray may start and end at any indices as long as it is contiguous. The algorithm must run efficiently for large inputs.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Min Length Alternating Subarray"
WHY DOES IT MATTER?
Exact‑count sliding‑window patterns appear frequently in performance‑critical code where you need the smallest window meeting a precise metric (e.g., minimum latency window with k errors). Mastering this pattern improves a candidate’s ability to design linear‑time solutions for a broad class of subarray problems.
OPTIMIZATION CHALLENGE
The breakthrough is realizing that the descent count changes only at the boundaries of the window, allowing constant‑time updates when moving pointers. This eliminates the need to recount the whole window each time, collapsing O(n^2) work to O(n).
REAL-WORLD CONNECTION
Think of a network packet monitor that must raise an alert as soon as exactly k packet drops occur within a contiguous time window. The monitor slides a time‑based window over the stream, updating the drop count in O(1) per event, mirroring the algorithm’s mechanics.
When coding, keep a separate boolean array or compute on‑the‑fly whether each adjacent pair is descending; then update the count only when the left or right pointer crosses a pair boundary. This avoids off‑by‑one bugs and keeps the code clean.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem asks for the minimum length of a contiguous subarray that contains exactly k descending adjacent pairs (i.e., indices j where nums[j]>nums[j+1]). A naive solution would enumerate every possible subarray, count the descents inside each, and keep the smallest length that matches k. This brute‑force method is O(n^2) time because there are O(n^2) subarrays and counting descents for each costs O(n) in the worst case, which quickly becomes infeasible for n up to 10^5. The optimal paradigm is a two‑pointer (sliding‑window) technique that treats the array as a stream and maintains a running count of descents inside the current window. By moving the right pointer to expand the window and the left pointer to shrink it only when the descent count exceeds k, we can examine each element a constant number of times, achieving linear time. The key insight is that the descent count is a monotonic property with respect to window expansion: adding a new element can increase the count by at most one, and removing the leftmost element can decrease it by at most one, which makes the sliding window viable for exact‑k constraints.
Interview Questions on This Problem
Q1How would you modify the sliding‑window solution if the requirement changed from exactly k descents to at most k descents?
Maintain the same window but only shrink when the count exceeds k; whenever the count is ≤k, update the answer with the current window length. This yields the shortest subarray with ≤k descents.
Q2Can this problem be solved using prefix sums? If so, outline the approach and its time complexity.
Yes. Pre‑compute an array desc[i] = 1 if nums[i]>nums[i+1] else 0, then build its prefix sum pref. For any subarray [l,r] the number of descents is pref[r-1]-pref[l-1]. Use a hashmap to store earliest index for each prefix value and look for pref[r-1]-k, giving O(n) time but O(n) extra space.
Q3Why does a binary‑search on answer length combined with a check‑function work for this problem, and what is its overall complexity?
Binary‑search on length L (1…n) tests whether any subarray of length L contains exactly k descents using a sliding window in O(n). The outer binary search adds a log n factor, so total O(n log n) time, which is slower than the optimal O(n) but still acceptable for large n.
Examples
Input
nums = [5,3,4,2,1], k = 2
Output
3
Explanation: All descending pairs in the whole array are (5,3), (4,2) and (2,1). The subarray [4,2,1] (indices 2‑4) has exactly two descending pairs: (4,2) and (2,1). Its length is 3, which is the minimum possible.
Input
nums = [1,2,3,4], k = 1
Output
-1
Explanation: No adjacent pair satisfies nums[j] > nums[j+1]; therefore no subarray can contain exactly one descending pair.
Input
nums = [9,7,5,6,4,2], k = 3
Output
5
Explanation: The subarray [9,7,5,6,4] (indices 0‑4) has descending pairs (9,7), (7,5) and (6,4) – exactly three. Its length is 5. All subarrays of length 4 contain at most two descending pairs, so 5 is minimal.
Constraints
- 1 <= nums.length <= 100000
- -1000000000 <= nums[i] <= 1000000000
- 0 <= k < nums.length
Optimal Approach & Strategy
Use a sliding window with two pointers, maintaining the current number of descents and shrinking the left side whenever the count exceeds k, recording lengths when it equals k.
Brute Force Approach
Enumerate all O(n^2) subarrays and count descents inside each, updating the minimum length when the count equals k.
Code Solutions
/**
* @param {number[]} nums
* @param {number} k
* @return {number}
*/
var minLengthAlternatingSubarray = function(nums, k) {
const n = nums.length;
if (n === 0) return -1;
if (k === 0) return 1;
// Create a binary array where 1 indicates a descending pair
const desc = new Array(n - 1).fill(0);
for (let i = 0; i < n - 1; i++) {
if (nums[i] > nums[i + 1]) {
desc[i] = 1;
}
}
// Use sliding window to find the minimum length subarray with exactly k descents
let left = 0;
let count = 0;
let minLen = n + 1;
for (let right = 0; right < n - 1; right++) {
count += desc[right];
while (count >= k) {
// The subarray in nums corresponds to indices [left, right+1]
// Length is (right + 1) - left + 1 = right - left + 2
const len = right - left + 2;
minLen = Math.min(minLen, len);
count -= desc[left];
left++;
}
}
return minLen === n + 1 ? -1 : minLen;
};
// Example usage
console.log(minLengthAlternatingSubarray([5, 3, 4, 2, 1], 2));#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
class Solution {
public:
int minLengthAlternatingSubarray(vector<int>& nums, int k) {
int n = nums.size();
if (n == 0) return -1;
if (k == 0) return 1;
// Create a binary array where 1 indicates a descending pair
vector<int> desc(n - 1, 0);
for (int i = 0; i < n - 1; i++) {
if (nums[i] > nums[i + 1]) {
desc[i] = 1;
}
}
// Use sliding window to find the minimum length subarray with exactly k descents
int left = 0;
int count = 0;
int minLen = n + 1;
for (int right = 0; right < n - 1; right++) {
count += desc[right];
while (count >= k) {
// The subarray in nums corresponds to indices [left, right+1]
// Length is (right + 1) - left + 1 = right - left + 2
int len = right - left + 2;
minLen = min(minLen, len);
count -= desc[left];
left++;
}
}
return minLen == n + 1 ? -1 : minLen;
}
};
int main() {
vector<int> nums = {5, 3, 4, 2, 1};
int k = 2;
Solution sol;
cout << sol.minLengthAlternatingSubarray(nums, k) << endl;
return 0;
}import java.util.*;
class Solution {
public int minLengthAlternatingSubarray(int[] nums, int k) {
int n = nums.length;
if (n == 0) return -1;
if (k == 0) return 1;
// Create a binary array where 1 indicates a descending pair
int[] desc = new int[n - 1];
for (int i = 0; i < n - 1; i++) {
if (nums[i] > nums[i + 1]) {
desc[i] = 1;
}
}
// Use sliding window to find the minimum length subarray with exactly k descents
int left = 0;
int count = 0;
int minLen = n + 1;
for (int right = 0; right < n - 1; right++) {
count += desc[right];
while (count >= k) {
// The subarray in nums corresponds to indices [left, right+1]
// Length is (right + 1) - left + 1 = right - left + 2
int len = right - left + 2;
minLen = Math.min(minLen, len);
count -= desc[left];
left++;
}
}
return minLen == n + 1 ? -1 : minLen;
}
}
public class Main {
public static void main(String[] args) {
int[] nums = {5, 3, 4, 2, 1};
int k = 2;
Solution sol = new Solution();
System.out.println(sol.minLengthAlternatingSubarray(nums, k));
}
}from typing import List
class Solution:
def min_length_alternating_subarray(self, nums: List[int], k: int) -> int:
n = len(nums)
if n == 0:
return -1
if k == 0:
return 1
# Create a binary array where 1 indicates a descending pair
desc = [0] * (n - 1)
for i in range(n - 1):
if nums[i] > nums[i + 1]:
desc[i] = 1
# Use sliding window to find the minimum length subarray with exactly k descents
left = 0
count = 0
min_len = n + 1
for right in range(n - 1):
count += desc[right]
while count >= k:
# The subarray in nums corresponds to indices [left, right+1]
# Length is (right + 1) - left + 1 = right - left + 2
length = right - left + 2
min_len = min(min_len, length)
count -= desc[left]
left += 1
return -1 if min_len == n + 1 else min_len
# Example usage
if __name__ == "__main__":
sol = Solution()
print(sol.min_length_alternating_subarray([5, 3, 4, 2, 1], 2))/**
* @param {number[]} nums
* @param {number} k
* @return {number}
*/
var minLengthAlternatingSubarray = function(nums, k) {
const n = nums.length;
if (n === 0) return -1;
if (k === 0) return 1;
// Create a binary array where 1 indicates a descending pair
const desc = new Array(n - 1).fill(0);
for (let i = 0; i < n - 1; i++) {
if (nums[i] > nums[i + 1]) {
desc[i] = 1;
}
}
// Use sliding window to find the minimum length subarray with exactly k descents
let left = 0;
let count = 0;
let minLen = n + 1;
for (let right = 0; right < n - 1; right++) {
count += desc[right];
while (count >= k) {
// The subarray in nums corresponds to indices [left, right+1]
// Length is (right + 1) - left + 1 = right - left + 2
const len = right - left + 2;
minLen = Math.min(minLen, len);
count -= desc[left];
left++;
}
}
return minLen === n + 1 ? -1 : minLen;
};
// Example usage
console.log(minLengthAlternatingSubarray([5, 3, 4, 2, 1], 2));Asked in Top Tech Interviews
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