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reverse pairs.cpp
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reverse pairs.cpp
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/*
///////////////////////////////////////////
//Question/Info
Reverse Pairs
Hard
1437
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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
Example 2:
Input: nums = [2,4,3,5,1]
Output: 3
Constraints:
1 <= nums.length <= 5 * 104
-231 <= nums[i] <= 231 - 1
author: srj_v
///////////////////////////////////////////
*/
#include <bits/stdc++.h>
using namespace std;
#define _IOS ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
typedef long double ld;
typedef long long int lli;
#pragma GCC optimize("Ofast")
void c_p_c()
{
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
}
int32_t main() {
///////////
c_p_c();
///////////
_IOS
//////////
// code
/*
int t ; cin >> t; while(t--){}
*/
class Solution {
public:
int count;
void merge(vector<int> &nums, int low , int mid , int high) {
int x = low; int y = mid + 1;
while (x <= mid and y <= high) {
if ((long)nums[x] > (long)2 * nums[y])
{
count += (mid - x + 1);
y++;
}
else {
x++;
}
}
sort(nums.begin() + low , nums.begin() + high + 1);
return;
}
int mergeSort(vector<int> &nums, int low, int high ) {
// int count = 0;
if (low < high) {
int mid = low + (high - low) / 2;
//count+=
mergeSort(nums, low, mid);
mergeSort(nums, mid + 1, high);
merge(nums, low, mid, high);
}
// return count;
}
int reversePairs(vector<int>& nums) {
count = 0;
int size = nums.size() - 1;
(mergeSort(nums, 0, size));
return count;
}
/*
int merge(vector<int> &nums, int l, int m, int h) {
int count=0;
int j=m+1;
for(int i=l;i<=m;++i) {
while(j<=h && nums[i] > 2LL*nums[j]) ++j;
count += (j - (m+1));
}
// merge of two sorted array
vector<int> temp;
int left=l,right=m+1;
// loop till on of them execuested
while(left<=m && right<=h) {
if(nums[left]<=nums[right]) temp.push_back(nums[left++]);
else temp.push_back(nums[right++]);
}
// remaining elements of left side
while(left<=m) temp.push_back(nums[left++]);
// remaining elements of right side
while(right<=h) temp.push_back(nums[right++]);
// assingn in main vector after merging
for(int i=l;i<=h;++i) {
nums[i]=temp[i-l];
}
return count;
}
int mergeSort(vector<int> &nums, int l, int h) {
if(l>=h) return 0; // A single element does not contain any pair
int m=(l+h)/2;
int count=0;
count += mergeSort(nums,l,m); // count from left side of vector
count += mergeSort(nums,m+1,h); // count from right side of vector
count += merge(nums,l,m,h); // count merge
return count;
}
int reversePairs(vector<int>& nums) {
return mergeSort(nums, 0, nums.size()-1);
}
*/
};
// cerr << "time: " << clock() << " ms" << '\n';
return 0;
}