Leetcode 2903: Find Indices With Index and Value Difference I
You are given a 0-indexed array of integers
nums of length n, along with two integers, indexDifference and valueDifference. Your task is to find two indices i and j such that the difference between the indices is greater than or equal to indexDifference, and the absolute difference between the values at the indices is greater than or equal to valueDifference. Return the indices [i, j] if such a pair exists, otherwise return [-1, -1].Problem
Approach
Steps
Complexity
Input: The input consists of three parts: the array `nums` of integers, the integer `indexDifference`, and the integer `valueDifference`. All are provided in a 0-indexed format.
Example: nums = [7, 2, 5, 1], indexDifference = 2, valueDifference = 3
Constraints:
• 1 <= n == nums.length <= 100
• 0 <= nums[i] <= 50
• 0 <= indexDifference <= 100
• 0 <= valueDifference <= 50
Output: Return an array containing two indices `[i, j]` that satisfy both conditions or `[-1, -1]` if no such indices exist.
Example: [0, 3]
Constraints:
• If multiple pairs satisfy the conditions, return any valid pair.
Goal: To find two indices such that their difference in indices is at least `indexDifference` and their values have a difference of at least `valueDifference`.
Steps:
• 1. Iterate through the array `nums` while keeping track of possible pairs of indices.
• 2. For each pair of indices, check the difference in indices and the difference in their values.
• 3. If both conditions are satisfied, return the pair of indices.
• 4. If no such pair is found after iterating through the array, return `[-1, -1]`.
Goal: The constraints for the problem specify the length of the array and the ranges for the values of `indexDifference` and `valueDifference`.
Steps:
• The array `nums` has a length of at least 1 and at most 100.
• The values in `nums` are between 0 and 50.
• Both `indexDifference` and `valueDifference` are non-negative and bounded by 100 and 50 respectively.
Assumptions:
• The input array `nums` is non-empty.
• The values of `indexDifference` and `valueDifference` are always valid.
• Input: Input: [7, 2, 5, 1], indexDifference = 2, valueDifference = 3
• Explanation: In this example, the pair of indices `[0, 3]` satisfies both conditions: the index difference is `3`, which is greater than or equal to `2`, and the absolute value difference is `6`, which is greater than or equal to `3`.
• Input: Input: [9, 3], indexDifference = 1, valueDifference = 6
• Explanation: In this case, the pair `[0, 1]` satisfies both conditions. The index difference is `1` (equal to `1`) and the absolute value difference is `6` (equal to `6`).
Approach: We need to iterate through the array and check each possible pair of indices. For each pair, check if the conditions are met. If they are, return the pair of indices. If no such pair is found, return `[-1, -1]`.
Observations:
• The problem requires checking pairs of indices with a certain index difference and value difference.
• We will iterate through the array using two pointers, ensuring that the required conditions are satisfied for each pair of indices.
Steps:
• 1. Start with two pointers, one iterating through the array `nums`.
• 2. For each element, check all previous elements to find a valid pair of indices.
• 3. If a valid pair is found, return the indices.
• 4. If no valid pair is found by the end of the iteration, return `[-1, -1]`.
Empty Inputs:
• If `nums` is empty, return `[-1, -1]`.
Large Inputs:
• Handle arrays of length up to 100 efficiently.
Special Values:
• Consider cases where the values of `indexDifference` and `valueDifference` are zero.
Constraints:
• Ensure that the indices and values are within the given constraints.
vector<int> findIndices(vector<int>& nums, int idif, int vdif) {
int mn = 0, mx = 0;
for (int i = idif, j = 0; i < nums.size(); ++i, ++j) {
mn = nums[mn] < nums[j] ? mn : j;
mx = nums[mx] > nums[j] ? mx : j;
if (abs(nums[i] - nums[mn]) >= vdif)
return {mn, i};
if (abs(nums[i] - nums[mx]) >= vdif)
return {mx, i};
}
return {-1, -1};
}
1 : Function Definition
vector<int> findIndices(vector<int>& nums, int idif, int vdif) {
Defines the function `findIndices` which takes in an array `nums`, and two integers `idif` and `vdif`. The function aims to find two indices where the absolute difference between the elements satisfies certain conditions.
2 : Variable Initialization
int mn = 0, mx = 0;
Initializes two integer variables `mn` and `mx` to track the indices of the minimum and maximum values encountered during the iteration over the array.
3 : Loop Initialization
for (int i = idif, j = 0; i < nums.size(); ++i, ++j) {
Starts a loop from index `idif` to the end of the array, using two indices `i` and `j`. The variable `j` runs from the beginning of the array while `i` runs from `idif`.
4 : Min Index Update
mn = nums[mn] < nums[j] ? mn : j;
Updates the minimum index `mn` to be the index of the smaller of the two values at `mn` and `j`.
5 : Max Index Update
mx = nums[mx] > nums[j] ? mx : j;
Updates the maximum index `mx` to be the index of the larger of the two values at `mx` and `j`.
6 : Condition Check for Min
if (abs(nums[i] - nums[mn]) >= vdif)
Checks if the absolute difference between the current element `nums[i]` and the minimum element `nums[mn]` is greater than or equal to `vdif`.
7 : Return Statement for Min
return {mn, i};
If the condition is satisfied, returns the indices of the minimum element and the current element `i`.
8 : Condition Check for Max
if (abs(nums[i] - nums[mx]) >= vdif)
Checks if the absolute difference between the current element `nums[i]` and the maximum element `nums[mx]` is greater than or equal to `vdif`.
9 : Return Statement for Max
return {mx, i};
If the condition is satisfied, returns the indices of the maximum element and the current element `i`.
10 : Return Statement for No Result
return {-1, -1};
If no indices satisfy the conditions, returns {-1, -1}, indicating no valid pair of indices was found.
Best Case: O(n)
Average Case: O(n^2)
Worst Case: O(n^2)
Description: The time complexity is quadratic in the worst case, as we check each pair of indices.
Best Case: O(1)
Worst Case: O(1)
Description: We only need constant extra space, as we only store the current indices and perform comparisons.
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