How do you solve Number of Pairs Satisfying Inequality on LeetCode?

Number of Pairs Satisfying Inequality (LeetCode #2426) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Rearranging the inequality can simplify the problem significantly.

What is the time complexity of Number of Pairs Satisfying Inequality?

The optimal solution for Number of Pairs Satisfying Inequality runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to sorting the transformed array, and the space complexity is O(n) for storing the transformed values.

What is the brute force solution for Number of Pairs Satisfying Inequality?

The brute force approach for Number of Pairs Satisfying Inequality is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible pair of indices (i, j) to see if they satisfy the given inequality. This is straightforward but inefficient for large arrays. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for Number of Pairs Satisfying Inequality?

Always start with a brute-force solution to understand the problem better. Look for patterns in the problem that can lead to optimizations. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Number of Pairs Satisfying Inequality?

Not considering the constraints properly, leading to incorrect assumptions. Overlooking the need for sorting or efficient searching techniques.

#2426

Number of Pairs Satisfying Inequality

Hard

ArrayBinary SearchDivide and ConquerBinary Indexed TreeSegment TreeMerge SortOrdered SetBinary SearchSortingTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

By rearranging the inequality and using a data structure like a Fenwick Tree (Binary Indexed Tree), we can efficiently count valid pairs without checking every combination, reducing the time complexity significantly.

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Algorithm

4 steps

1Step 1: Create a new array 'transformed' where each element is calculated as nums1[i] - nums2[i] + diff.
2Step 2: Sort the 'transformed' array to facilitate efficient counting.
3Step 3: For each element in the original array, use binary search to find how many elements in 'transformed' are less than or equal to the current element.
4Step 4: Sum these counts to get the total number of valid pairs.

solution.py8 lines

1def countPairs(nums1, nums2, diff):
2    n = len(nums1)
3    transformed = [nums1[i] - nums2[i] + diff for i in range(n)]
4    transformed.sort()
5    count = 0
6    for i in range(n):
7        count += bisect.bisect_right(transformed, nums1[i]) - (i + 1)
8    return count

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Complexity note: The time complexity is O(n log n) due to sorting the transformed array, and the space complexity is O(n) for storing the transformed values.

1Rearranging the inequality can simplify the problem significantly.
2Using efficient data structures can reduce the time complexity from quadratic to logarithmic.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.