Maximum Length of Semi-Decreasing Subarrays

Maximum Length of Semi-Decreasing Subarrays - Problem

You are given an integer array nums. Return the length of the longest semi-decreasing subarray of nums, and 0 if there are no such subarrays.

A subarray is a contiguous non-empty sequence of elements within an array. A non-empty array is semi-decreasing if its first element is strictly greater than its last element.

Input & Output

Example 1 — Basic Case

$ Input: nums = [7,6,8,1,2]

› Output: 3

💡 Note: The subarray [8,1,2] has first element 8 > last element 2, with length 3. This is the longest such subarray.

Example 2 — No Semi-Decreasing

$ Input: nums = [1,2,3,4]

› Output: 0

💡 Note: All elements are in increasing order, so no subarray has first > last. Return 0.

Example 3 — Single Decreasing Pair

$ Input: nums = [5,1]

› Output: 2

💡 Note: The entire array [5,1] is semi-decreasing since 5 > 1, length is 2.

Constraints

1 ≤ nums.length ≤ 10⁵
-10⁹ ≤ nums[i] ≤ 10⁹

Visualization

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Understanding the Visualization

Input Array

Given array [7,6,8,1,2]

Find Semi-Decreasing

Check subarrays where first > last

Return Max Length

Longest valid subarray has length 3

Key Takeaway

🎯 Key Insight: We need to find the longest contiguous subarray where the first element is strictly greater than the last element

Asked in

G Google 25 M Microsoft 18 a Amazon 15

The key insight is that a semi-decreasing subarray must have its first element strictly greater than its last element. The brute force approach checks all possible subarrays in O(n²) time. The optimal approach uses the observation that for each ending position, we want to find the furthest left position with a larger value. Time: O(n²), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Check All Subarrays	O(n²)	O(1)	Check every possible subarray to find the longest semi-decreasing one
Monotonic Stack Optimization	O(n)	O(n)	Use a monotonic stack to efficiently find optimal starting and ending positions

Brute Force - Check All Subarrays — Algorithm Steps

Iterate through each starting index i
For each i, iterate through each ending index j
Check if nums[i] > nums[j] and update max length

Visualization

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Step-by-Step Walkthrough

Check All Pairs

Compare first and last element of each subarray

Track Maximum

Keep track of longest valid subarray found

Return Result

Return the maximum length found

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int* nums, int n) {
    int maxLength = 0;
    
    for (int i = 0; i < n; i++) {
        for (int j = i; j < n; j++) {
            if (nums[i] > nums[j]) {
                int length = j - i + 1;
                if (length > maxLength) {
                    maxLength = length;
                }
            }
        }
    }
    
    return maxLength;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int n = 0;
    
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        nums[n++] = atoi(token);
        token = strtok(NULL, "[,]");
    }
    
    int result = solution(nums, n);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops check all pairs of indices

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check All Subarrays — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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