# 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);
}

## Input

{1, 2, 3, 2}, {4, 3, 2, 2}

## Output

[0, 0, 1, 2, ]