2903. Find Indices With Index and Value Difference I


Problem Description

In this problem, you're given an integer array called nums, with a length of n. You're also given two integers: indexDifference and valueDifference. Your task is to find two indices i and j such that both i and j are within the range from 0 to n - 1 and they meet the following criteria:

  1. The absolute difference between i and j must be at least indexDifference,
  2. The absolute difference between the values at nums[i] and nums[j] must be at least valueDifference.

In terms of the outcome, you need to return an array called answer. This array should consist of the indices [i, j] if such a pair of indices exists. If there are multiple valid pairs, you can return any one of them. If no valid pairs are found, then return [-1, -1]. An interesting point to note is that according to the problem statement, i and j can be the same index, which implies that indexDifference could potentially be 0.

Intuition

The primary intuition behind the solution is the usage of a sliding window technique, combined with the maintenance of the minimum and maximum values within the window. The sliding window is defined by two pointers, i and j, that maintain a distance apart specified by indexDifference. The pointers are used as markers to capture a subarray within nums to check against our two conditions.

At the outset, i starts at the position indexDifference, and j starts at 0. By doing this, the gap between i and j reflects the indexDifference requirement of our problem.

We maintain two variables, mi and mx, to keep track of the indices where the minimum and maximum values are found within our sliding window that ends at the current j index. While sliding i further along the array, we update mi and mx to account for the entry of new values into the window and the exit of old values.

When updating mi and mx, if nums[j] is less than nums[mi], we reassign mi to j, because we have found a new minimum. Conversely, if nums[j] is greater than nums[mx], we reassign mx to j due to identifying a new maximum.

After every movement of i and update of mi and mx, we check our two conditions against nums[i] (the current value at i). If the difference between nums[i] and the value at nums[mi] is greater than or equal to valueDifference, we have found a valid pair [mi, i]. Alternatively, if the difference between the maximum value (nums[mx]) and nums[i] is also greater than or equal to valueDifference, then [mx, i] is a valid pair. In this situation, we immediately return the pair as it meets our requirements.

If we reach the end of the array without finding a pair that satisfies both conditions, we conclude that no such pair exists, and we return the default output [-1, -1].

The approach's efficiency comes from the fact that it avoids checking every possible pair of indices, which would otherwise lead to a less efficient solution with a higher time complexity.

Solution Approach

The solution uses a sliding window approach, which involves moving a fixed-size window across the array to examine sub-sections one at a time. This allows for checking the conditions over smaller segments in a single pass through the array, making the solution more efficient than a brute force approach that would involve examining all possible index pairs.

To implement this technique, the algorithm maintains two pointers: i and j. These pointers define the bounds of the sliding window. The pointer i starts at the index equal to indexDifference while j starts at 0, thus immediately satisfying the condition abs(i - j) >= indexDifference because i - j is initialized to indexDifference.

As the algorithm iterates over the array, starting from i = indexDifference, it keeps track of the indices of the minimum and the maximum values found so far to the left of j. These indices are stored in variables mi and mx, respectively.

1for i in range(indexDifference, len(nums)):
2    j = i - indexDifference

Within the loop, we first check whether the current element at index j changes the minimum or maximum:

1if nums[j] < nums[mi]:
2    mi = j
3if nums[j] > nums[mx]:
4    mx = j

After updating mi and mx, we check if nums[i] differs enough from the minimum or maximum value to satisfy the valueDifference condition:

1if nums[i] - nums[mi] >= valueDifference:
2    return [mi, i]
3if nums[mx] - nums[i] >= valueDifference:
4    return [mx, i]

If either of these checks succeeds, the function immediately returns the corresponding pair of indices, as they meet both prescribed conditions. If the function reaches the end of the array without returning, this means no valid pairs were found, and thus it returns [-1, -1].

In terms of data structures, no additional structures are needed beyond the use of a few variables to keep track of the indices and values encountered. This algorithm is space-efficient because it operates directly on the input array without requiring extra space proportional to the input size.

The choice of a sliding window and keeping track of minimum and maximum values eliminates the need to compare every element with every other element, thereby reducing the time complexity from O(n^2) to O(n), where n is the length of the input array.

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Example Walkthrough

Let's walk through an example to illustrate the solution approach described above. Consider the integer array nums = [1, 2, 3, 4, 5], with indexDifference = 3 and valueDifference = 3.

Our task is to find indices i and j such that abs(i - j) >= indexDifference and abs(nums[i] - nums[j]) >= valueDifference.

According to the given solution approach, we initialize the sliding window by setting i to indexDifference and j to 0. This immediately satisfies the first condition as the difference between the indices i = 3 and j = 0 is 3, which is equal to indexDifference.

We then maintain mi and mx to keep track of the minimum and maximum values within the window.

Here's how we proceed step by step:

  1. On the first iteration where i = 3:
    • We have j = 0
    • We initiate mi to j since there are no previous values to compare with, and similarly, mx is also initiated to j
    • The elements under consideration are [nums[0], nums[3]] i.e., [1, 4]
    • We see that nums[3] - nums[0] = 4 - 1 = 3, which is equal to valueDifference
    • Since nums[3] - nums[0] fulfills the valueDifference condition, we return [0, 3] as the indices that satisfy both conditions.

In this example, we directly found a pair that met both conditions, and thus we would return [0, 3]. However, if we needed to continue the iteration, we would:

  1. Increment i to the next position and decrement j to keep the window size constant while satisfying the indexDifference. If nums[j] changes the minimum or maximum within the new window, update mi or mx accordingly.
  2. Check if nums[i] differs enough from nums[mi] or nums[mx] as explained previously.
  3. If we find a pair, we return it. If not, we continue iterating until i reaches the end of the array.

If no valid pairs are found by the end of the array, we return [-1, -1] as specified.

Solution Implementation

1from typing import List
2
3class Solution:
4    def findIndices(self, nums: List[int], idx_diff: int, val_diff: int) -> List[int]:
5        # Initialize min and max index pointers
6        min_idx = max_idx = 0
7      
8        # Traverse the array, starting from the index that enables the required index difference
9        for current_idx in range(idx_diff, len(nums)):
10          
11            # Compute the comparison index that matches the index difference
12            compare_idx = current_idx - idx_diff
13          
14            # Check and update the min and max indices based on the values at compare_idx
15            if nums[compare_idx] < nums[min_idx]:
16                min_idx = compare_idx
17            if nums[compare_idx] > nums[max_idx]:
18                max_idx = compare_idx
19          
20            # If the value difference requirement is met with the minimum, return the indices
21            if nums[current_idx] - nums[min_idx] >= val_diff:
22                return [min_idx, current_idx]
23          
24            # If the value difference requirement is met with the maximum, return the indices
25            if nums[max_idx] - nums[current_idx] >= val_diff:
26                return [max_idx, current_idx]
27      
28        # If the required value difference isn't found, return [-1, -1] as per problem statement
29        return [-1, -1]
30
1class Solution {
2  
3    // Method to find indices in an array such that the difference between their values is at least a given value and their positions are no less than a given index difference
4    public int[] findIndices(int[] nums, int indexDifference, int valueDifference) {
5        int minIndex = 0;   // Initializing the minimum value index
6        int maxIndex = 0;   // Initializing the maximum value index
7
8        // Loop through the array starting from the index equal to the indexDifference to the end of the array
9        for (int i = indexDifference; i < nums.length; ++i) {
10            int currentIndex = i - indexDifference;  // Calculate the index to compare with
11
12            // Update the minimum value index if a new minimum is found
13            if (nums[currentIndex] < nums[minIndex]) {
14                minIndex = currentIndex;
15            }
16
17            // Update the maximum value index if a new maximum is found
18            if (nums[currentIndex] > nums[maxIndex]) {
19                maxIndex = currentIndex;
20            }
21
22            // Check if the difference between the current value and the minimum value found so far is at least valueDifference
23            if (nums[i] - nums[minIndex] >= valueDifference) {
24                return new int[] {minIndex, i}; // Return the indices if condition is met
25            }
26
27            // Check if the difference between the maximum value found so far and the current value is at least valueDifference
28            if (nums[maxIndex] - nums[i] >= valueDifference) {
29                return new int[] {maxIndex, i}; // Return the indices if condition is met
30            }
31        }
32
33        // Return [-1, -1] if no such pair of indices is found
34        return new int[] {-1, -1};
35    }
36}
37
1#include <vector>
2
3class Solution {
4public:
5    // Method to find the two indices in the array nums such that the difference
6    // between their values is at least valueDifference and their index difference is exactly indexDifference
7    // Args:
8    //   nums: The input vector of integers
9    //   indexDiff: The required difference between the indices of the two elements
10    //   valueDiff: The minimum required value difference between the two elements
11    // Returns:
12    //   A vector with two elements: the indices of the elements in nums that satisfy the above criteria
13    //   If no such pair exists, returns {-1, -1}    
14    std::vector<int> findIndices(std::vector<int>& nums, int indexDiff, int valueDiff) {
15        int minIndex = 0, maxIndex = 0; // Initialized to store the index of minimum and maximum values seen so far
16        for (int i = indexDiff; i < nums.size(); ++i) {
17            int j = i - indexDiff; // Calculate the corresponding index
18            if (nums[j] < nums[minIndex]) {
19                minIndex = j; // Update minIndex if a new minimum is found
20            }
21            if (nums[j] > nums[maxIndex]) {
22                maxIndex = j; // Update maxIndex if a new maximum is found
23            }
24            // Check if the difference between the current value and the minimum value seen so far is at least valueDiff
25            if (nums[i] - nums[minIndex] >= valueDiff) {
26                return {minIndex, i}; // Pair found, return indices
27            }
28            // Check if the difference between the maximum value seen so far and the current value is at least valueDiff
29            if (nums[maxIndex] - nums[i] >= valueDiff) {
30                return {maxIndex, i}; // Pair found, return indices
31            }
32        }
33        return {-1, -1}; // If no pair found, return {-1, -1}
34    }
35};
36
1// This function finds two indices such that the difference of the elements at these indices 
2// is at least the given valueDifference and the indices are separated by at most the given 
3// indexDifference.
4// nums: The array of numbers to search within
5// indexDifference: The maximum allowed difference between the indices
6// valueDifference: The minimum required difference between the values at the indices
7// Returns an array with two numbers representing the indices, or [-1, -1] if no such pair exists
8function findIndices(nums: number[], indexDifference: number, valueDifference: number): number[] {
9    // Initialize the indices for the minimum value (minIndex) and maximum value (maxIndex) found.
10    let minIndex = 0;
11    let maxIndex = 0;
12  
13    // Iterate over the array, starting from the element at the indexDifference.
14    for (let currentIndex = indexDifference; currentIndex < nums.length; currentIndex++) {
15        // Calculate the index of the element we are comparing against,
16        // which is indexDifference behind the current index.
17        const compareIndex = currentIndex - indexDifference;
18      
19        // Update minIndex if the current compareIndex points to a new minimum value
20        if (nums[compareIndex] < nums[minIndex]) {
21            minIndex = compareIndex;
22        }
23      
24        // Update maxIndex if the current compareIndex points to a new maximum value
25        if (nums[compareIndex] > nums[maxIndex]) {
26            maxIndex = compareIndex;
27        }
28      
29        // Check if the difference between the current element and the minimum value is large enough.
30        if (nums[currentIndex] - nums[minIndex] >= valueDifference) {
31            return [minIndex, currentIndex]; // Return the indices if the condition is met.
32        }
33      
34        // Check if the difference between the maximum value and the current element is large enough.
35        if (nums[maxIndex] - nums[currentIndex] >= valueDifference) {
36            return [maxIndex, currentIndex]; // Return the indices if the condition is met.
37        }
38    }
39  
40    // If no suitable pair of indices is found, return [-1, -1].
41    return [-1, -1];
42}
43

Time and Space Complexity

Time Complexity

The time complexity of the provided code is O(n). This is achieved by iterating over the array once from indexDifference to the length of the array len(nums). Only constant time checks and updates are performed within the loop, leading to a linear time complexity relative to the array's size.

Space Complexity

The space complexity of the code is O(1). No additional space proportional to the input size is used. Only a fixed number of variables mi, mx, and j are used, which occupy constant space regardless of the input array size.

Learn more about how to find time and space complexity quickly using problem constraints.


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