Find the lexicographically smallest sequence which can be formed by re-arranging elements of second array in C++

Suppose we have two arrays A and B with n numbers, we have to rearrange the elements of B in itself in a way such that the sequence formed by (A[i] + B[i]) % n after rearranging it is lexicographically smallest. Finally we will return the lexicographically smallest possible sequence.

So, if the input is like A = {1, 2, 3, 2}, B = {4, 3, 2, 2}, then the output will be [0, 0, 1, 2]

To solve this, we will follow these steps −

n := size of a
Define one map my_map
Define one set my_set
for initialize i := 0, when i < n, update (increase i by 1), do −
- (increase my_map[b[i]] by 1)
- insert b[i] into my_set
Define an array sequence
for initialize i := 0, when i < n, update (increase i by 1), do −
- if a[i] is same as 0, then −
  - it := elements of my_set which are not smaller than 0, initially point to first element
  - value := it
  - insert value mod n at the end of sequence
  - (decrease my_map[value] by 1)
  - if my_map[value] is zero, then −
    - delete value from my_set
Otherwise
- x := n - a[i]
- it := elements of my_set which are not smaller than x, initially point to first element
- if it is not in my_set, then −
  - it := elements of my_set which are not smaller than 0, initially point to first element
- value := it
- insert (a[i] + value) mod n at the end of sequence
- (decrease my_map[value] by 1)
- if my_map[value] is zero, then −
  - delete value from my_set
return sequence

Example

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;
}
vector<int> solve(vector<int>&a, vector<int> &b) {
   int n = a.size();
   unordered_map<int, int> my_map;
   set<int> my_set;
   for (int i = 0; i < n; i++) {
      my_map[b[i]]++;
      my_set.insert(b[i]);
   }
   vector<int> sequence;
   for (int i = 0; i < n; i++) {
      if (a[i] == 0) {
         auto it = my_set.lower_bound(0);
         int value = *it;
         sequence.push_back(value % n);
         my_map[value]--;
         if (!my_map[value])
            my_set.erase(value);
         }
         else {
            int x = n - a[i];
            auto it = my_set.lower_bound(x);
            if (it == my_set.end())
               it = my_set.lower_bound(0);
            int value = *it;
            sequence.push_back((a[i] + value) % n);
            my_map[value]--;
            if (!my_map[value])
               my_set.erase(value);
         }
      }
   return sequence;
}
int main() {
   vector<int> a = {1, 2, 3, 2};
   vector<int> b = {4, 3, 2, 2};
   vector<int> res = solve(a, b);
   print_vector(res);
}