Count of Smaller Numbers After Self in C++

Suppose we have an array nums, we have to find another array called count, in this count array, the count[i] stores the number of smaller elements to the right of nums[i].
So if the input is like: [5,2,7,1], then the result will be [2,1,1,0].
To solve this, we will follow these steps −
Define one method called update(), this will take index, array bit and n
Define one method called update(), this will take index, array bit and n
while index <= n, do −increase bit[index]index = index + (index) AND ( - index)
while index <= n, do −
increase bit[index]
increase bit[index]
index = index + (index) AND ( - index)
index = index + (index) AND ( - index)
Define one array called query(), this will take index and bits arrayans := 0while index > 0, do −ans = ans + bit[index]index = index – (index AND - index)return ans
Define one array called query(), this will take index and bits array
ans := 0
ans := 0
while index > 0, do −ans = ans + bit[index]index = index – (index AND - index)
while index > 0, do −
ans = ans + bit[index]
ans = ans + bit[index]
index = index – (index AND - index)
index = index – (index AND - index)
return ans
return ans
From the main method, do the following −
From the main method, do the following −
n := size of nums
n := size of nums
Define an array res of size n
Define an array res of size n
if (n is non-zero) is false, then, return res
if (n is non-zero) is false, then, return res
maxx := 0, minn = nums [0]
maxx := 0, minn = nums [0]
for initializing i := 1, when i < n, increase i by 1 do −minn := min of nums[i] and minn
for initializing i := 1, when i < n, increase i by 1 do −
minn := min of nums[i] and minn
minn := min of nums[i] and minn
for initializing i := 0, when i < n, increase i by 1 do −nums[i] := nums[i] - minn + 1maxx := max of nums[i] and maxx
for initializing i := 0, when i < n, increase i by 1 do −
nums[i] := nums[i] - minn + 1
nums[i] := nums[i] - minn + 1
maxx := max of nums[i] and maxx
maxx := max of nums[i] and maxx
Define an array bit of size maxx + 1
Define an array bit of size maxx + 1
for initializing i := n - 1, when i >= 0, decrease i by 1 do −num = nums[i] – 1x := Call the function query(num,bit)res[i] := xCall the function update(num + 1, bit, maxx)
for initializing i := n - 1, when i >= 0, decrease i by 1 do −
num = nums[i] – 1
num = nums[i] – 1
x := Call the function query(num,bit)
x := Call the function query(num,bit)
res[i] := x
res[i] := x
Call the function update(num + 1, bit, maxx)
Call the function update(num + 1, bit, maxx)
return res
return res
Let us see the following implementation to get better understanding −
 Live Demo
#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<auto> v){
   cout << "[";
   for(int i = 0; i<v.size(); i++){
      cout << v[i] << ", ";
   }
   cout << "]"<<endl;
}
class Solution {
   public:
   void update(int index,vector <int>& bit, int n){
      while(index<=n){
         bit[index]++;
         index += (index)&(-index);
      }
   }
   int query(int index, vector <int> bit){
      int ans = 0;
      while(index>0){
         ans+=bit[index];
         index -= index&(-index);
      }
      return ans;
   }
   vector<int> countSmaller(vector<int>& nums) {
      int n = nums.size();
      vector <int> res(n);
      if(!n)return res;
      int maxx = 0;
      int minn = nums[0];
      for(int i =1;i<n;i++)minn = min(nums[i],minn);
      for(int i =0;i<n;i++){
         nums[i] = nums[i]-minn+1;
         maxx = max(nums[i],maxx);
      }
      vector <int> bit(maxx+1);
      for(int i =n-1;i>=0;i--){
         int num = nums[i]-1;
         int x = query(num,bit);
         res[i] = x;
         update(num+1,bit,maxx);
      }
      return res;
   }
};
main(){
   Solution ob;
   vector<int> v = {5,2,7,1};
   print_vector(ob.countSmaller(v));
}
[5,2,7,1]
[2, 1, 1, 0, ]