Skip to content
Programming101
Programmingoneonone

Learn everything about programming

  • Home
  • CS Subjects
    • IoT – Internet of Things
    • Digital Communication
    • Human Values
  • Programming Tutorials
    • C Programming
    • Data structures and Algorithms
    • 100+ Java Programs
    • 100+ C Programs
  • HackerRank Solutions
    • HackerRank Algorithms Solutions
    • HackerRank C problems solutions
    • HackerRank C++ problems solutions
    • HackerRank Java problems solutions
    • HackerRank Python problems solutions
Programming101
Programmingoneonone

Learn everything about programming

Leetcode Reverse Pairs problem solution

YASH PAL, 31 July 2024

In this Leetcode Reverse Pairs problem solution 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].

Leetcode Reverse Pairs problem solution

Topics we are covering

Toggle
  • Problem solution in Python.
  • Problem solution in Java.
  • Problem solution in C++.

Problem solution in Python.

import bisect 
class Solution:
    def reversePairs(self, nums: List[int]) -> int:
        count,l=0,[]    
        for i in nums :
            a=bisect.bisect_right  (l,i*2)#finding the index 
            count+=len (l)-a#finding the number of greater than equal to i*2
            idx=bisect.bisect(l,i)#finding the index where the i need to be inserted
            l[idx:idx]=[i]#trick for inserting at idx much faster than insert()
        return count

Problem solution in Java.

class Solution {
    public int reversePairs(int[] nums) {
        return mergeSort(nums, 0, nums.length - 1);
    }
    private int mergeSort(int[] nums, int l, int r) {
        if (l >= r) return 0;
        int count = 0;
        int mid = l + (r - l) / 2;
        count += mergeSort(nums, l, mid);
        count += mergeSort(nums, mid + 1, r);
        count += merge(nums, l, mid, r);
        return count;
    }
    private int merge(int[] nums, int l, int mid, int r) {
        int count = 0, l1 = l, l2 = mid + 1, idx = 0;
        while (l1 <= mid && l2 <= r) {
            if ((long) nums[l1] > (long) 2 * nums[l2]) {
                count += mid - l1 + 1;
                l2++;
            } else l1++;
        }
        l1 = l;
        l2 = mid + 1;
        int[] copy = new int[r - l + 1];
        while (l1 <= mid && l2 <= r) {
            if (nums[l1] > nums[l2]) copy[idx++] = nums[l2++];
            else copy[idx++] = nums[l1++];
        }
        while (l1 <= mid) copy[idx++] = nums[l1++];
        while (l2 <= r) copy[idx++] = nums[l2++];
        System.arraycopy(copy, 0, nums, l, r - l + 1);
        return count;
    }
}

Problem solution in C++.

class Solution {
public:
    int reversePairs(vector<int>& nums) {
        multiset<long long> tb;
        int ret = 0;
        
        int n = nums.size();
        
        for(int i = n - 1; i >= 0; --i) {
            long long cur = nums[i];
            
            auto iter = lower_bound(tb.begin(), tb.end(), cur);
            
            ret +=  distance(tb.begin(), iter);
            tb.insert(2 * cur);
        }
        
        return(ret);
    }
};

coding problems solutions

Post navigation

Previous post
Next post
  • Automating Image Format Conversion with Python: A Complete Guide
  • HackerRank Separate the Numbers solution
  • How AI Is Revolutionizing Personalized Learning in Schools
  • GTA 5 is the Game of the Year for 2024 and 2025
  • Hackerrank Day 5 loops 30 days of code solution
How to download udemy paid courses for free

Pages

  • About US
  • Contact US
  • Privacy Policy

Programing Practice

  • C Programs
  • java Programs

HackerRank Solutions

  • C
  • C++
  • Java
  • Python
  • Algorithm

Other

  • Leetcode Solutions
  • Interview Preparation

Programming Tutorials

  • DSA
  • C

CS Subjects

  • Digital Communication
  • Human Values
  • Internet Of Things
  • YouTube
  • LinkedIn
  • Facebook
  • Pinterest
  • Instagram
©2025 Programmingoneonone | WordPress Theme by SuperbThemes