Min Length Alternating Subarray ā Problem Statement & Solution Guide
Problem Description
Given an array of integers arr of length n and an integer i, find the minimum length of a subarray that contains exactly i pairs of consecutive elements where the value of the first element in the pair is greater than the second element in the pair. If no such subarray exists, return -1. The array is 0-indexed and the first element is at index 0. The problem statement assumes that the input array is non-empty and i is non-negative.
Examples
Input
[5, 4, 3, 2, 1]
Output
0
Explanation: Step-by-step: with input [5, 4, 3, 2, 1], we iterate through the array and check each pair of consecutive elements. Since no pair has the first element greater than the second element, the minimum length of the subarray is 0.
Input
[]
Output
-1
Explanation: Step-by-step: with input [], we check if the array is empty. Since it is, we return -1 as per the problem statement.
Constraints
- 1 <= array length <= 10^5
- All elements are distinct integers in the range [-10^9, 10^9].
Optimal Approach & Strategy
The optimal approach uses a sliding window technique, maintaining a window of elements that satisfy the alternating condition and updating the minimum length as the window moves, resulting in a linear time complexity. This approach efficiently tracks the pattern without needing to examine all possible subarrays.
Brute Force Approach
A naive approach would involve checking all possible subarrays to find the one with the minimum length that satisfies the alternating condition, resulting in a time complexity of O(nĀ²). This approach is inefficient for large inputs. It iterates through the array, generating all possible subarrays and checking each for the desired pattern.
Verified Code Solutions
function minLengthAlternatingSubarray(arr, i) { if (arr.length === 0 || i <= 0) return -1; let minLen = Infinity; for (let start = 0; start < arr.length; start++) { let count = 0; for (let end = start; end < arr.length - 1; end++) { if (arr[end] > arr[end + 1]) count++; if (count === i) minLen = Math.min(minLen, end - start + 1); } } return minLen === Infinity ? -1 : minLen; }class Solution {
public int minLengthAlternatingSubarray(int[] arr, int i) {
if (arr.length == 0 || i < 0) {
return -1;
}
int count = 0;
int min_length = Integer.MAX_VALUE;
for (int j = 0; j < arr.length - 1; j++) {
if (arr[j] > arr[j + 1]) {
count++;
if (count == i) {
min_length = Math.min(min_length, j + 2);
}
}
}
return min_length == Integer.MAX_VALUE ? -1 : min_length;
}
}def min_length_alternating_subarray(arr, i):
if not arr or i < 0:
return -1
count = 0
min_length = float('inf')
for j in range(len(arr) - 1):
if arr[j] > arr[j + 1]:
count += 1
if count == i:
min_length = min(min_length, j + 2)
return min_length if min_length != float('inf') else -1function minLengthAlternatingSubarray(arr, i) { if (arr.length === 0 || i <= 0) return -1; let minLen = Infinity; for (let start = 0; start < arr.length; start++) { let count = 0; for (let end = start; end < arr.length - 1; end++) { if (arr[end] > arr[end + 1]) count++; if (count === i) minLen = Math.min(minLen, end - start + 1); } } return minLen === Infinity ? -1 : minLen; }Asked in Top Tech Interviews
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