LeetCode #2824 — EASY

Count Pairs Whose Sum is Less than Target

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

Given a 0-indexed integer array nums of length n and an integer target, return the number of pairs (i, j) where 0 <= i < j < n and nums[i] + nums[j] < target.

Example 1:

Input: nums = [-1,1,2,3,1], target = 2
Output: 3
Explanation: There are 3 pairs of indices that satisfy the conditions in the statement:
- (0, 1) since 0 < 1 and nums[0] + nums[1] = 0 < target
- (0, 2) since 0 < 2 and nums[0] + nums[2] = 1 < target 
- (0, 4) since 0 < 4 and nums[0] + nums[4] = 0 < target
Note that (0, 3) is not counted since nums[0] + nums[3] is not strictly less than the target.

Example 2:

Input: nums = [-6,2,5,-2,-7,-1,3], target = -2
Output: 10
Explanation: There are 10 pairs of indices that satisfy the conditions in the statement:
- (0, 1) since 0 < 1 and nums[0] + nums[1] = -4 < target
- (0, 3) since 0 < 3 and nums[0] + nums[3] = -8 < target
- (0, 4) since 0 < 4 and nums[0] + nums[4] = -13 < target
- (0, 5) since 0 < 5 and nums[0] + nums[5] = -7 < target
- (0, 6) since 0 < 6 and nums[0] + nums[6] = -3 < target
- (1, 4) since 1 < 4 and nums[1] + nums[4] = -5 < target
- (3, 4) since 3 < 4 and nums[3] + nums[4] = -9 < target
- (3, 5) since 3 < 5 and nums[3] + nums[5] = -3 < target
- (4, 5) since 4 < 5 and nums[4] + nums[5] = -8 < target
- (4, 6) since 4 < 6 and nums[4] + nums[6] = -4 < target

Constraints:

1 <= nums.length == n <= 50
-50 <= nums[i], target <= 50

Step 01

Brute Force Baseline

Problem summary: Given a 0-indexed integer array nums of length n and an integer target, return the number of pairs (i, j) where 0 <= i < j < n and nums[i] + nums[j] < target.

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

[-1,1,2,3,1]
2

Example 2

[-6,2,5,-2,-7,-1,3]
-2

Core Insight

What unlocks the optimal approach

The constraints are small enough for a brute-force solution to pass

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 #2824: Count Pairs Whose Sum is Less than Target
class Solution {
    public int countPairs(List<Integer> nums, int target) {
        Collections.sort(nums);
        int ans = 0;
        for (int j = 0; j < nums.size(); ++j) {
            int x = nums.get(j);
            int i = search(nums, target - x, j);
            ans += i;
        }
        return ans;
    }

    private int search(List<Integer> nums, int x, int r) {
        int l = 0;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (nums.get(mid) >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}

// Accepted solution for LeetCode #2824: Count Pairs Whose Sum is Less than Target
func countPairs(nums []int, target int) (ans int) {
	sort.Ints(nums)
	for j, x := range nums {
		i := sort.SearchInts(nums[:j], target-x)
		ans += i
	}
	return
}

# Accepted solution for LeetCode #2824: Count Pairs Whose Sum is Less than Target
class Solution:
    def countPairs(self, nums: List[int], target: int) -> int:
        nums.sort()
        ans = 0
        for j, x in enumerate(nums):
            i = bisect_left(nums, target - x, hi=j)
            ans += i
        return ans

// Accepted solution for LeetCode #2824: Count Pairs Whose Sum is Less than Target
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2824: Count Pairs Whose Sum is Less than Target
// class Solution {
//     public int countPairs(List<Integer> nums, int target) {
//         Collections.sort(nums);
//         int ans = 0;
//         for (int j = 0; j < nums.size(); ++j) {
//             int x = nums.get(j);
//             int i = search(nums, target - x, j);
//             ans += i;
//         }
//         return ans;
//     }
// 
//     private int search(List<Integer> nums, int x, int r) {
//         int l = 0;
//         while (l < r) {
//             int mid = (l + r) >> 1;
//             if (nums.get(mid) >= x) {
//                 r = mid;
//             } else {
//                 l = mid + 1;
//             }
//         }
//         return l;
//     }
// }

// Accepted solution for LeetCode #2824: Count Pairs Whose Sum is Less than Target
function countPairs(nums: number[], target: number): number {
    nums.sort((a, b) => a - b);
    let ans = 0;
    const search = (x: number, r: number): number => {
        let l = 0;
        while (l < r) {
            const mid = (l + r) >> 1;
            if (nums[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    };
    for (let j = 0; j < nums.length; ++j) {
        const i = search(target - nums[j], j);
        ans += i;
    }
    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.