Given two arrays **A** and** B** of **N** integers. Reorder the elements of B in itself in such a way that the sequence formed by **(A[i] + B[i]) % N** after re-ordering is the smallest lexicographically. The task is to print the **lexicographically** *smallest* sequence possible.

**Note**: The array elements are in range [0, n).

**Examples:**

Input:a[] = {0, 1, 2, 1}, b[] = {3, 2, 1, 1}Output:1 0 0 2

Reorder B to {1, 3, 2, 1} to get the smallest sequence possible.

Input:a[] = {2, 0, 0}, b[] = {1, 0, 2}Output:0 0 2

**Approach:** The problem can be solved *greedily*. Initially keep a count of all the numbers of array B using hashing, and store them in the set in C++, so that lower_bound() [* To check for an element* ] and erase() [ *To erase an element* ] operations can be done in logarithmic time.

For every element in the array, check for a number equal or greater than** n-a[i]** using *lower_bound* function. If there are no such elements then take the smallest element in the set. Decrease the value by 1 in the hash table for the number used, if the hash table’s value is 0, then erase the element from the set also.

However there is an exception if the array element is 0, then check for 0 at the first then for N, if both of them are not there, then take the smallest element.

Below is the implementation of the above approach:

`// C++ implementation of the ` `// above approach ` ` ` `#include <bits/stdc++.h> ` `using` `namespace` `std; `
` ` `// Function to get the smallest ` `// sequence possible ` `void` `solve(` `int` `a[], ` `int` `b[], ` `int` `n) `
`{ ` ` ` ` ` `// Hash-table to count the `
` ` `// number of occurrences of b[i] `
` ` `unordered_map<` `int` `, ` `int` `> mpp; `
` ` ` ` `// Store the element in sorted order `
` ` `// for using binary search `
` ` `set<` `int` `> st; `
` ` ` ` `// Iterate in the B array `
` ` `// and count the occurrences and `
` ` `// store in the set `
` ` `for` `(` `int` `i = 0; i < n; i++) { `
` ` `mpp[b[i]]++; `
` ` `st.insert(b[i]); `
` ` `} `
` ` ` ` `vector<` `int` `> sequence; `
` ` ` ` `// Iterate for N elements `
` ` `for` `(` `int` `i = 0; i < n; i++) { `
` ` ` ` `// If the element is 0 `
` ` `if` `(a[i] == 0) { `
` ` ` ` `// Find the nearest number to 0 `
` ` `auto` `it = st.lower_bound(0); `
` ` `int` `el = *it; `
` ` `sequence.push_back(el % n); `
` ` ` ` `// Decrease the count `
` ` `mpp[el]--; `
` ` ` ` `// Erase if no more are there `
` ` `if` `(!mpp[el]) `
` ` `st.erase(el); `
` ` `} `
` ` ` ` `// If the element is other than 0 `
` ` `else` `{ `
` ` ` ` `// Find the difference `
` ` `int` `x = n - a[i]; `
` ` ` ` `// Find the nearest number which can give us `
` ` `// 0 on modulo `
` ` `auto` `it = st.lower_bound(x); `
` ` ` ` `// If no such number occurs then `
` ` `// find the number closest to 0 `
` ` `if` `(it == st.end()) `
` ` `it = st.lower_bound(0); `
` ` ` ` `// Get the number `
` ` `int` `el = *it; `
` ` ` ` `// store the number `
` ` `sequence.push_back((a[i] + el) % n); `
` ` ` ` `// Decrease the count `
` ` `mpp[el]--; `
` ` ` ` `// If no more appers, then erase it from set `
` ` `if` `(!mpp[el]) `
` ` `st.erase(el); `
` ` `} `
` ` `} `
` ` ` ` `for` `(` `auto` `it : sequence) `
` ` `cout << it << ` `" "` `; `
`} ` ` ` `// Driver Code ` `int` `main() `
`{ ` ` ` `int` `a[] = { 0, 1, 2, 1 }; `
` ` `int` `b[] = { 3, 2, 1, 1 }; `
` ` `int` `n = ` `sizeof` `(a) / ` `sizeof` `(a[0]); `
` ` `solve(a, b, n); `
` ` `return` `0; `
`} ` |

*chevron_right*

*filter_none*

**Output:**

1 0 0 2

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