LeetCode #2563 — MEDIUM

Count the Number of Fair Pairs

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

Given a 0-indexed integer array nums of size n and two integers lower and upper, return the number of fair pairs.

A pair (i, j) is fair if:

0 <= i < j < n, and
lower <= nums[i] + nums[j] <= upper

Example 1:

Input: nums = [0,1,7,4,4,5], lower = 3, upper = 6
Output: 6
Explanation: There are 6 fair pairs: (0,3), (0,4), (0,5), (1,3), (1,4), and (1,5).

Example 2:

Input: nums = [1,7,9,2,5], lower = 11, upper = 11
Output: 1
Explanation: There is a single fair pair: (2,3).

Constraints:

1 <= nums.length <= 10⁵
nums.length == n
-10⁹ <= nums[i] <= 10⁹
-10⁹ <= lower <= upper <= 10⁹

Step 01

Brute Force Baseline

Problem summary: Given a 0-indexed integer array nums of size n and two integers lower and upper, return the number of fair pairs. A pair (i, j) is fair if: 0 <= i < j < n, and lower <= nums[i] + nums[j] <= upper

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Two Pointers · Binary Search

Example 1

[0,1,7,4,4,5]
3
6

Example 2

[1,7,9,2,5]
11
11

Core Insight

What unlocks the optimal approach

Sort the array in ascending order.
For each number in the array, keep track of the smallest and largest numbers in the array that can form a fair pair with this number.
As you move to larger number, both boundaries move down.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2563: Count the Number of Fair Pairs
class Solution {
    public long countFairPairs(int[] nums, int lower, int upper) {
        Arrays.sort(nums);
        long ans = 0;
        int n = nums.length;
        for (int i = 0; i < n; ++i) {
            int j = search(nums, lower - nums[i], i + 1);
            int k = search(nums, upper - nums[i] + 1, i + 1);
            ans += k - j;
        }
        return ans;
    }

    private int search(int[] nums, int x, int left) {
        int right = nums.length;
        while (left < right) {
            int mid = (left + right) >> 1;
            if (nums[mid] >= x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

// Accepted solution for LeetCode #2563: Count the Number of Fair Pairs
func countFairPairs(nums []int, lower int, upper int) (ans int64) {
	sort.Ints(nums)
	for i, x := range nums {
		j := sort.Search(len(nums), func(h int) bool { return h > i && nums[h] >= lower-x })
		k := sort.Search(len(nums), func(h int) bool { return h > i && nums[h] >= upper-x+1 })
		ans += int64(k - j)
	}
	return
}

# Accepted solution for LeetCode #2563: Count the Number of Fair Pairs
class Solution:
    def countFairPairs(self, nums: List[int], lower: int, upper: int) -> int:
        nums.sort()
        ans = 0
        for i, x in enumerate(nums):
            j = bisect_left(nums, lower - x, lo=i + 1)
            k = bisect_left(nums, upper - x + 1, lo=i + 1)
            ans += k - j
        return ans

// Accepted solution for LeetCode #2563: Count the Number of Fair Pairs
struct Solution;

fn lower_bound(nums: &[i32], bound: i32) -> i64 {
    let mut l = 0;
    let mut r = nums.len() - 1;
    let mut res = 0;
    while l < r {
        let sum = nums[l] + nums[r];
        if sum < bound {
            res += r - l;
            l += 1;
        } else {
            r -= 1;
        }
    }
    res as i64
}

impl Solution {
    pub fn count_fair_pairs(mut nums: Vec<i32>, lower: i32, upper: i32) -> i64 {
        nums.sort();
        lower_bound(&nums, upper + 1) - lower_bound(&nums, lower)
    }
}

#[test]
fn test() {
    let nums = vec![0, 1, 7, 4, 4, 5];
    let lower = 3;
    let upper = 6;
    let res = 6;
    assert_eq!(Solution::count_fair_pairs(nums, lower, upper), res);
    let nums = vec![1, 7, 9, 2, 5];
    let lower = 11;
    let upper = 11;
    let res = 1;
    assert_eq!(Solution::count_fair_pairs(nums, lower, upper), res);
    let nums = vec![0, 0, 0, 0, 0, 0];
    let lower = 0;
    let upper = 0;
    let res = 15;
    assert_eq!(Solution::count_fair_pairs(nums, lower, upper), res);
}

// Accepted solution for LeetCode #2563: Count the Number of Fair Pairs
function countFairPairs(nums: number[], lower: number, upper: number): number {
    const search = (x: number, l: number): number => {
        let r = nums.length;
        while (l < r) {
            const mid = (l + r) >> 1;
            if (nums[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    };

    nums.sort((a, b) => a - b);
    let ans = 0;
    for (let i = 0; i < nums.length; ++i) {
        const j = search(lower - nums[i], i + 1);
        const k = search(upper - nums[i] + 1, i + 1);
        ans += k - j;
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(log n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.