LeetCode #493 — HARD

Reverse Pairs

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

Given an integer array nums, return the number of reverse pairs in the array.

A reverse pair is a pair (i, j) where:

0 <= i < j < nums.length and
nums[i] > 2 * nums[j].

Example 1:

Input: nums = [1,3,2,3,1]
Output: 2
Explanation: The reverse pairs are:
(1, 4) --> nums[1] = 3, nums[4] = 1, 3 > 2 * 1
(3, 4) --> nums[3] = 3, nums[4] = 1, 3 > 2 * 1

Example 2:

Input: nums = [2,4,3,5,1]
Output: 3
Explanation: The reverse pairs are:
(1, 4) --> nums[1] = 4, nums[4] = 1, 4 > 2 * 1
(2, 4) --> nums[2] = 3, nums[4] = 1, 3 > 2 * 1
(3, 4) --> nums[3] = 5, nums[4] = 1, 5 > 2 * 1

Constraints:

1 <= nums.length <= 5 * 10⁴
-2³¹ <= nums[i] <= 2³¹ - 1

Step 01

Brute Force Baseline

Problem summary: Given an integer array nums, return the number of reverse pairs in the array. A reverse pair is a pair (i, j) where: 0 <= i < j < nums.length and nums[i] > 2 * nums[j].

Baseline thinking

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

Pattern signal: Array · Binary Search · Segment Tree

Example 1

[1,3,2,3,1]

Example 2

[2,4,3,5,1]

Core Insight

What unlocks the optimal approach

Use the merge-sort technique.
Divide the array into two parts and sort them.
For each integer in the first part, count the number of integers that satisfy the condition from the second part. Use the pointer to help you in the counting process.

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

Largest constraint values

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 #493: Reverse Pairs
class Solution {
    private int[] nums;
    private int[] t;

    public int reversePairs(int[] nums) {
        this.nums = nums;
        int n = nums.length;
        this.t = new int[n];
        return mergeSort(0, n - 1);
    }

    private int mergeSort(int l, int r) {
        if (l >= r) {
            return 0;
        }
        int mid = (l + r) >> 1;
        int ans = mergeSort(l, mid) + mergeSort(mid + 1, r);
        int i = l, j = mid + 1, k = 0;
        while (i <= mid && j <= r) {
            if (nums[i] <= nums[j] * 2L) {
                ++i;
            } else {
                ans += mid - i + 1;
                ++j;
            }
        }
        i = l;
        j = mid + 1;
        while (i <= mid && j <= r) {
            if (nums[i] <= nums[j]) {
                t[k++] = nums[i++];
            } else {
                t[k++] = nums[j++];
            }
        }
        while (i <= mid) {
            t[k++] = nums[i++];
        }
        while (j <= r) {
            t[k++] = nums[j++];
        }
        for (i = l; i <= r; ++i) {
            nums[i] = t[i - l];
        }
        return ans;
    }
}

// Accepted solution for LeetCode #493: Reverse Pairs
func reversePairs(nums []int) int {
	n := len(nums)
	t := make([]int, n)
	var mergeSort func(l, r int) int
	mergeSort = func(l, r int) int {
		if l >= r {
			return 0
		}
		mid := (l + r) >> 1
		ans := mergeSort(l, mid) + mergeSort(mid+1, r)
		i, j, k := l, mid+1, 0
		for i <= mid && j <= r {
			if nums[i] <= nums[j]*2 {
				i++
			} else {
				ans += mid - i + 1
				j++
			}
		}
		i, j = l, mid+1
		for i <= mid && j <= r {
			if nums[i] <= nums[j] {
				t[k] = nums[i]
				k, i = k+1, i+1
			} else {
				t[k] = nums[j]
				k, j = k+1, j+1
			}
		}
		for ; i <= mid; i, k = i+1, k+1 {
			t[k] = nums[i]
		}
		for ; j <= r; j, k = j+1, k+1 {
			t[k] = nums[j]
		}
		for i = l; i <= r; i++ {
			nums[i] = t[i-l]
		}
		return ans
	}
	return mergeSort(0, n-1)
}

# Accepted solution for LeetCode #493: Reverse Pairs
class Solution:
    def reversePairs(self, nums: List[int]) -> int:
        def merge_sort(l, r):
            if l >= r:
                return 0
            mid = (l + r) >> 1
            ans = merge_sort(l, mid) + merge_sort(mid + 1, r)
            t = []
            i, j = l, mid + 1
            while i <= mid and j <= r:
                if nums[i] <= 2 * nums[j]:
                    i += 1
                else:
                    ans += mid - i + 1
                    j += 1
            i, j = l, mid + 1
            while i <= mid and j <= r:
                if nums[i] <= nums[j]:
                    t.append(nums[i])
                    i += 1
                else:
                    t.append(nums[j])
                    j += 1
            t.extend(nums[i : mid + 1])
            t.extend(nums[j : r + 1])
            nums[l : r + 1] = t
            return ans

        return merge_sort(0, len(nums) - 1)

// Accepted solution for LeetCode #493: Reverse Pairs
struct Solution;

impl Solution {
    fn reverse_pairs(mut nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let mut temp = vec![0; n];
        Self::merge_sort(0, n, &mut nums, &mut temp) as i32
    }

    fn merge_sort(start: usize, end: usize, nums: &mut Vec<i32>, temp: &mut Vec<i32>) -> usize {
        if start + 1 >= end {
            return 0;
        }
        let mid = start + (end - start) / 2;
        let mut res = 0;
        res += Self::merge_sort(start, mid, nums, temp);
        res += Self::merge_sort(mid, end, nums, temp);
        let mut i = start;
        let mut j = mid;
        while i < mid {
            while j < end && nums[i] as i64 > 2 * nums[j] as i64 {
                j += 1;
            }
            res += j - mid;
            i += 1;
        }
        let mut k = start;
        let mut i = start;
        let mut j = mid;
        while i < mid || j < end {
            if i == mid {
                temp[k] = nums[j];
                k += 1;
                j += 1;
                continue;
            }
            if j == end {
                temp[k] = nums[i];
                k += 1;
                i += 1;
                continue;
            }
            if nums[i] < nums[j] {
                temp[k] = nums[i];
                i += 1;
            } else {
                temp[k] = nums[j];
                j += 1;
            }
            k += 1;
        }
        nums[start..end].clone_from_slice(&temp[start..end]);
        res
    }
}

#[test]
fn test() {
    let nums = vec![1, 3, 2, 3, 1];
    let res = 2;
    assert_eq!(Solution::reverse_pairs(nums), res);
    let nums = vec![2, 4, 3, 5, 1];
    let res = 3;
    assert_eq!(Solution::reverse_pairs(nums), res);
    let nums = vec![
        2147483647, 2147483647, 2147483647, 2147483647, 2147483647, 2147483647,
    ];
    let res = 0;
    assert_eq!(Solution::reverse_pairs(nums), res);
}

// Accepted solution for LeetCode #493: Reverse Pairs
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #493: Reverse Pairs
// class Solution {
//     private int[] nums;
//     private int[] t;
// 
//     public int reversePairs(int[] nums) {
//         this.nums = nums;
//         int n = nums.length;
//         this.t = new int[n];
//         return mergeSort(0, n - 1);
//     }
// 
//     private int mergeSort(int l, int r) {
//         if (l >= r) {
//             return 0;
//         }
//         int mid = (l + r) >> 1;
//         int ans = mergeSort(l, mid) + mergeSort(mid + 1, r);
//         int i = l, j = mid + 1, k = 0;
//         while (i <= mid && j <= r) {
//             if (nums[i] <= nums[j] * 2L) {
//                 ++i;
//             } else {
//                 ans += mid - i + 1;
//                 ++j;
//             }
//         }
//         i = l;
//         j = mid + 1;
//         while (i <= mid && j <= r) {
//             if (nums[i] <= nums[j]) {
//                 t[k++] = nums[i++];
//             } else {
//                 t[k++] = nums[j++];
//             }
//         }
//         while (i <= mid) {
//             t[k++] = nums[i++];
//         }
//         while (j <= r) {
//             t[k++] = nums[j++];
//         }
//         for (i = l; i <= r; ++i) {
//             nums[i] = t[i - l];
//         }
//         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(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

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.

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.