How do you solve Count the Number of Fair Pairs on LeetCode?

Count the Number of Fair Pairs (LeetCode #2563) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Sorting the array allows us to efficiently find pairs that meet the criteria using binary search.

What algorithm or data structure is used to solve Count the Number of Fair Pairs?

Count the Number of Fair Pairs uses the following concepts: Array, Two Pointers, Binary Search, Sorting. The recommended patterns to study are: Two Pointers, Binary Search, Sorting.

What are the interview tips for Count the Number of Fair Pairs?

Always explain your thought process clearly, especially when transitioning from brute force to optimal solutions. Practice identifying patterns in problems, such as when to use sorting or two-pointer techniques. Be mindful of edge cases, such as very small or very large values for lower and upper.

What are common mistakes when solving Count the Number of Fair Pairs?

Not considering the constraints of indices (i < j) when counting pairs. Overlooking the importance of sorting the array before applying binary search.

Count the Number of Fair Pairs — LeetCode #2563 (Medium)

Tags: Array, Two Pointers, Binary Search, Sorting

Related patterns: Two Pointers, Binary Search, Sorting

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

The brute force approach checks every possible pair of indices in the array to see if they meet the fair pair criteria. This is straightforward but inefficient, especially for large arrays.

This complexity arises because we are checking every possible pair of elements in the array, leading to n(n-1)/2 comparisons in the worst case.

Step 1: Initialize a counter to zero for counting fair pairs.
Step 2: Use two nested loops to iterate through all pairs (i, j) where i < j.
Step 3: For each pair, check if the sum of nums[i] + nums[j] is within the range [lower, upper]. If it is, increment the counter.
Step 4: Return the counter as the result.

Let's dry run the first example with nums = [0,1,7,4,4,5], lower = 3, upper = 6: 1. i=0, j=1: 0+1=1 (not fair) 2. i=0, j=2: 0+7=7 (not fair) 3. i=0, j=3: 0+4=4 (fair) 4. i=0, j=4: 0+4=4 (fair) 5. i=0, j=5: 0+5=5 (fair) 6. i=1, j=2: 1+7=8 (not fair) 7. i=1, j=3: 1+4=5 (fair) 8. i=1, j=4: 1+4=5 (fair) 9. i=1, j=5: 1+5=6 (fair) 10. i=2, j=3: 7+4=11 (not fair) 11. i=2, j=4: 7+4=11 (not fair) 12. i=2, j=5: 7+5=12 (not fair) 13. i=3, j=4: 4+4=8 (not fair) 14. i=3, j=5: 4+5=9 (not fair) 15. i=4, j=5: 4+5=9 (not fair) Total fair pairs found: 6.

Optimal Solution approach

Time complexity: O(n log n). Space complexity: O(1).

By sorting the array and using a two-pointer technique, we can efficiently find the number of fair pairs without checking every possible combination. This reduces the time complexity significantly.

The sorting step takes O(n log n), and the two-pointer technique runs in O(n), making this approach much more efficient than the brute force method.

Step 1: Sort the array nums.
Step 2: Initialize a counter for fair pairs.
Step 3: For each element nums[i], use binary search to find the range of indices that can form fair pairs with nums[i].
Step 4: Count the valid pairs using the indices found in the previous step and update the counter.
Step 5: Return the counter as the result.

Let's dry run the first example with nums = [0,1,7,4,4,5], lower = 3, upper = 6: 1. Sort nums: [0, 1, 4, 4, 5, 7] 2. For i=0 (nums[0]=0): - left = binarySearch(3-0=3) -> index 2 - right = binarySearch(6-0=6) -> index 5 - count += 5 - 2 = 3 3. For i=1 (nums[1]=1): - left = binarySearch(3-1=2) -> index 2 - right = binarySearch(6-1=5) -> index 5 - count += 5 - 2 = 3 4. For i=2 (nums[2]=4): - left = binarySearch(3-4=-1) -> index 3 - right = binarySearch(6-4=2) -> index 3 - count += 3 - 3 = 0 5. For i=3 (nums[3]=4): - left = binarySearch(3-4=-1) -> index 4 - right = binarySearch(6-4=2) -> index 4 - count += 4 - 4 = 0 6. For i=4 (nums[4]=5): - left = binarySearch(3-5=-2) -> index 5 - right = binarySearch(6-5=1) -> index 5 - count += 5 - 5 = 0 Final count = 6.

Key Insights

Sorting the array allows us to efficiently find pairs that meet the criteria using binary search.
Using two pointers or binary search reduces the number of comparisons significantly.

Common Mistakes

Not considering the constraints of indices (i < j) when counting pairs.
Overlooking the importance of sorting the array before applying binary search.

Interview Tips

Always explain your thought process clearly, especially when transitioning from brute force to optimal solutions.
Practice identifying patterns in problems, such as when to use sorting or two-pointer techniques.
Be mindful of edge cases, such as very small or very large values for lower and upper.

#2563

Count the Number of Fair Pairs

Medium

Array↗Two Pointers↗Binary Search↗Sorting↗Two Pointers↗Binary Search↗Sorting↗

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(1)

By sorting the array and using a two-pointer technique, we can efficiently find the number of fair pairs without checking every possible combination. This reduces the time complexity significantly.

⚙️

Algorithm

5 steps

1Step 1: Sort the array nums.
2Step 2: Initialize a counter for fair pairs.
3Step 3: For each element nums[i], use binary search to find the range of indices that can form fair pairs with nums[i].
4Step 4: Count the valid pairs using the indices found in the previous step and update the counter.
5Step 5: Return the counter as the result.

solution.py15 lines

1# Full working Python code
2import bisect
3
4def countFairPairs(nums, lower, upper):
5    nums.sort()
6    count = 0
7    n = len(nums)
8    for i in range(n):
9        left = bisect.bisect_left(nums, lower - nums[i], i + 1)
10        right = bisect.bisect_right(nums, upper - nums[i], i + 1)
11        count += right - left
12    return count
13
14# Example usage
15print(countFairPairs([0,1,7,4,4,5], 3, 6))  # Output: 6

ℹ

Complexity note: The sorting step takes O(n log n), and the two-pointer technique runs in O(n), making this approach much more efficient than the brute force method.

1Sorting the array allows us to efficiently find pairs that meet the criteria using binary search.
2Using two pointers or binary search reduces the number of comparisons significantly.

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