Find All Good Indices In The Given Array In C++

Overview:

In the realm of troubleshooting and coding, efficiently exploring an array's attributes for specific indices is a common challenge. Locating optimal indices within an array poses a similar challenge. An ideal index typically meets specific criteria such as maintaining non-decreasing or non-increasing subarrays of a defined length in its vicinity. This tutorial will demonstrate the process of identifying all favorable indices within a provided array using C++ programming language.

Problem Statement:

For an array arr of length n and an integer k, an index i is good if:

• ✅ The k elements up to index i constitute a non-increasing subarray.
• ✅ The k elements at or after index i is a non-decreasing subarray.
• ✅ The index i must be valid, i.e., it must have at least k elements on both sides of it.

The task is to identify and retrieve all such valid indexes within the array.

Approach: Two-Pass Approach

In order to solve this problem efficiently, we use a two-pass approach:

• ✅ Precompute non-increasing counts: Create an array left where left[i] stores the number of consecutive elements before i that are non-increasing.
• ✅ Precompute non-decreasing counts: Create an array right where right[i] stores the number of consecutive elements after i that are non-decreasing.
• ✅ Find valid indices: Traverse the array and check if both left[i] >= k and right[i] >= k. If both conditions are satisfied, i is a good index.

This method guarantees that we only have to iterate through the array a limited number of times, leading to an effective solution with a time complexity of O(n).

Example:

Let's consider an example to locate all valid indices within the provided array in C++.

#include <iostream>
#include <vector>
using namespace std;
vector<int> goodIndices(vector<int>& nums, int k) 
{
    int n = nums.size();
    vector<int> left(n, 0), right(n, 0), result;    
    // Compute non-increasing sequence lengths
    for (int i = 1; i < n; i++) 
{
        if (nums[i] <= nums[i - 1]) 
{
            left[i] = left[i - 1] + 1;
        }
    }    
    // Compute non-decreasing sequence lengths
    for (int i = n - 2; i >= 0; i--) 
{
        if (nums[i] <= nums[i + 1]) 
{
            right[i] = right[i + 1] + 1;
        }
    }    
    // Find good indices
    for (int i = k; i < n - k; i++) 
{
        if (left[i - 1] >= k - 1 && right[i + 1] >= k - 1) 
{
            result.push_back(i);
        }
    }  
    return result;
}
int main() 
{
    vector<int> nums = {2, 1, 1, 1, 3, 4, 1, 2, 5, 6};
    int k = 2;
    vector<int> indices = goodIndices(nums, k);
    cout << "Good Indices: ";
    for (int idx : indices) 
{
        cout << idx << " ";
    }
    return 0;
}

Output:

Good Indices: 2 3 7

Explanation of Code:

• ✅ Initialization: The left keeps track of the length of the previous consecutive non-increasing elements. The right keeps track of the length of the next consecutive non-decreasing elements. The result keeps track of the good indices.
• ✅ Populating the left array: If nums[i] <= nums[i-1], we increase left[i].
• ✅ Populating the right array: If nums[i] <= nums[i+1], we increase right[i].
• ✅ Valid index checking: If left[i-1] >= k-1 and right[i+1] >= k-1, index i is a good index and added to the result.
• ✅ The left keeps track of the length of the previous consecutive non-increasing elements.
• ✅ The right keeps track of the length of the next consecutive non-decreasing elements.
• ✅ The result keeps track of the good indices.
• ✅ If nums[i] <= nums[i-1], we increase left[i].
• ✅ If nums[i] <= nums[i+1], we increase right[i].
• ✅ If left[i-1] >= k-1 and right[i+1] >= k-1, index i is a good index and added to the result.

Complexity Analysis:

• ✅ Building left and right: O(n)
• ✅ Checking valid indices: O(n)
• ✅ Overall complexity: O(n)

It enhances the efficiency and viability for processing extensive inputs.

Example Walkthrough:

Consider nums = {2, 1, 1, 1, 3, 4, 1, 2, 5, 6} with k = 2:

• ✅ Compute left: left = {0, 1, 2, 3, 0, 0, 1, 0, 0, 0}
• ✅ Compute right: right = {0, 0, 0, 0, 1, 2, 0, 1, 2, 3}
• ✅ Find good indices: Check if left[i-1] >= 1 and right[i+1] >= 1 for valid indices. The result is: {3, 4, 5}.
• ✅ Check if left[i-1] >= 1 and right[i+1] >= 1 for valid indices.
• ✅ The result is: {3, 4, 5}.

Edge Cases Considered:

• ✅ Small array size (n < 2k): No valid indices exist, so return an empty list.
• ✅ Already sorted increasing/decreasing array: The middle elements can be good indices.
• ✅ All elements are the same: Every index (except boundary conditions) can be good.
• ✅ No valid indices: If no index satisfies the condition, return an empty list.
• ✅ No valid indices exist, so return an empty list.
• ✅ The middle elements can be good indices.
• ✅ Every index (except boundary conditions) can be good.
• ✅ If no index satisfies the condition, return an empty list.

Alternative Approach (Sliding Window):

Instead of computing left and right beforehand, we can opt for a sliding window approach to dynamically verify the non-increasing and non-decreasing characteristics. This method maintains an O(n) time complexity but could pose challenges in achieving an elegant implementation.

Conclusion:

In summary, effective indexes in an array represent a common challenge addressed efficiently by precalculated arrays. The subsequent O(n) resolution delivers peak efficiency while retaining simplicity. This serves as a prime illustration of utilizing prefix calculations and dynamic criteria to enhance array-centric algorithms.

By employing the dual-pass technique, we can efficiently identify favorable indexes within an array, which can then be utilized to address similar challenges in competitive programming and technical interview scenarios.