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:
CPP
// C++ implementation of the// above approach#include <bits/stdc++.h>using namespace std;// Function to get the smallest// sequence possiblevoid 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 appears, then erase it from set if (!mpp[el]) st.erase(el); } } for (auto it : sequence) cout << it << " ";}// Driver Codeint 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;} |
Java
// Java implementation of the// above approachimport java.util.*;class GFG { // Function to get the smallest // sequence possible static void solve(int[] a, int[] b, int n) { // Hash-table to count the // number of occurrences of b[i] HashMap<Integer, Integer> mpp = new HashMap<Integer, Integer>(); // Store the element in sorted order // for using binary search HashSet<Integer> st = new HashSet<Integer>(); // Iterate in the B array // and count the occurrences and // store in the set for (int i = 0; i < n; i++) { if (!mpp.containsKey(b[i])) mpp.put(b[i], 0); mpp.put(b[i], 1 + mpp.get(b[i])); st.add(b[i]); } ArrayList<Integer> sequence = new ArrayList<Integer>(); // Iterate for N elements for (int y = 0; y < n; y++) { int it = st.size(); ArrayList<Integer> st1 = new ArrayList<Integer>(); for (int elem : st) st1.add(elem); Collections.sort(st1); Collections.reverse(st1); // If the element is 0 if (a[y] == 0) { // Find the nearest number to 0 it = st.size(); for (var i = 0; i < st1.size(); i++) { if (st1.get(i) >= 0) it = i; } int el = st1.get(it); sequence.add(el % n); // Decrease the count if (!mpp.containsKey(el)) mpp.put(el, 0); mpp.put(el, mpp.get(el) - 1); // Erase if no more are there if (mpp.get(el) == 0) st.remove(el); } // If the element is other than 0 else { // Find the difference int x = n - a[y]; // Find the nearest number which can give us // 0 on modulo it = st.size(); for (int i = 0; i < st1.size(); i++) { if (st1.get(i) >= x) it = i; } // If no such number occurs then // find the number closest to 0 if (it == st.size()) { for (int i = 0; i < st1.size(); i++) { if (st1.get(i) >= 0) it = i; } } // Get the number int el = st1.get(it); // store the number sequence.add((a[y] + el) % n); // Decrease the count if (!mpp.containsKey(el)) mpp.put(el, 0); mpp.put(el, mpp.get(el) - 1); // If no more appears, then erase it from // set if (mpp.get(el) == 0) st.remove(el); } } for (int elem : sequence) System.out.print(elem + " "); } // Driver Code public static void main(String[] args) { int[] a = { 0, 1, 2, 1 }; int[] b = { 3, 2, 1, 1 }; int n = a.length; solve(a, b, n); }}// This code is contributed by phasing17 |
Python3
# Python3 implementation of the# above approach# Function to get the smallest# sequence possibledef solve(a, b, n): # Hash-table to count the # number of occurrences of b[i] mpp = dict(); # Store the element in sorted order # for using binary search st = set() # Iterate in the B array # and count the occurrences and # store in the set for i in range(len(b)): if b[i] not in mpp: mpp[b[i]] = 0 mpp[b[i]] += 1; st.add(b[i]); sequence = []; # Iterate for N elements for y in range(n): it = len(st); st1 = sorted(st) st1 = st1[::-1] # If the element is 0 if (a[y] == 0) : # Find the nearest number to 0 it = len(st1) for i in range(len(st1)): if (st1[i] >= 0): it = i; el = st1[it] sequence.append(el % n); # Decrease the count if el not in mpp: mpp[el] = 0 mpp[el] -= 1; # Erase if no more are there if (mpp[el] == 0): st.discard(el); # If the element is other than 0 else : # Find the difference x = n - a[y]; # Find the nearest number which can give us # 0 on modulo it = len(st) for i in range(len(st1)): if (st1[i] >= x): it = i; # If no such number occurs then # find the number closest to 0 if (it == len(st)): for i in range(len(st1)): if (st1[i] >= 0): it = i; # Get the number el = st1[it]; # store the number sequence.append((a[y] + el) % n); # Decrease the count if el not in mpp: mpp[el] = 0; mpp[el]-=1; # If no more appears, then erase it from set if (mpp[el] == 0): st.remove(el); print(*sequence)# Driver Codea = [ 0, 1, 2, 1 ];b = [ 3, 2, 1, 1 ];n = len(a);solve(a, b, n)# This code is contributed by phasing17 |
C#
// C# implementation of the// above approachusing System;using System.Collections.Generic;class GFG { // Function to get the smallest // sequence possible static void solve(int[] a, int[] b, int n) { // Hash-table to count the // number of occurrences of b[i] Dictionary<int, int> mpp = new Dictionary<int, int>(); // Store the element in sorted order // for using binary search HashSet<int> st = new HashSet<int>(); // Iterate in the B array // and count the occurrences and // store in the set for (int i = 0; i < n; i++) { if (!mpp.ContainsKey(b[i])) mpp[b[i]] = 0; mpp[b[i]]++; st.Add(b[i]); } List<int> sequence = new List<int>(); // Iterate for N elements for (int y = 0; y < n; y++) { int it = st.Count; List<int> st1 = new List<int>(); foreach(int elem in st) st1.Add(elem); st1.Sort(); st1.Reverse(); // If the element is 0 if (a[y] == 0) { // Find the nearest number to 0 it = st.Count; for (var i = 0; i < st1.Count; i++) { if (st1[i] >= 0) it = i; } int el = st1[it]; sequence.Add(el % n); // Decrease the count if (!mpp.ContainsKey(el)) mpp[el] = 0; mpp[el]--; // Erase if no more are there if (mpp[el] == 0) st.Remove(el); } // If the element is other than 0 else { // Find the difference int x = n - a[y]; // Find the nearest number which can give us // 0 on modulo it = st.Count; for (var i = 0; i < st1.Count; i++) { if (st1[i] >= x) it = i; } // If no such number occurs then // find the number closest to 0 if (it == st.Count) { for (int i = 0; i < st1.Count; i++) { if (st1[i] >= 0) it = i; } } // Get the number int el = st1[it]; // store the number sequence.Add((a[y] + el) % n); // Decrease the count if (!mpp.ContainsKey(el)) mpp[el] = 0; mpp[el]--; // If no more appears, then erase it from // set if (mpp[el] == 0) st.Remove(el); } } foreach(int elem in sequence) Console.Write(elem + " "); } // Driver Code public static void Main(string[] args) { int[] a = { 0, 1, 2, 1 }; int[] b = { 3, 2, 1, 1 }; int n = a.Length; solve(a, b, n); }}// This code is contributed by phasing17 |
Javascript
// JS implementation of the// above approach// Function to get the smallest// sequence possiblefunction solve(a, b, n){ // Hash-table to count the // number of occurrences of b[i] let mpp = {}; // Store the element in sorted order // for using binary search let st = new Set(); // Iterate in the B array // and count the occurrences and // store in the set for (var i = 0; i < n; i++) { if (!mpp.hasOwnProperty(b[i])) mpp[b[i]] = 0; mpp[b[i]]++; st.add(b[i]); } let sequence = []; // Iterate for N elements for (var y = 0; y < n; y++) { let it = st.size; let st1 = Array.from(st); st1.sort(function(a, b) { return a < b}) // If the element is 0 if (a[y] == 0) { // Find the nearest number to 0 it = st.size for (var i = 0; i < st1.length; i++) { if (st1[i] >= 0) it = i; } let el = st1[it] sequence.push(el % n); // Decrease the count if (!mpp.hasOwnProperty(el)) mpp[el] = 0 mpp[el]--; // Erase if no more are there if (mpp[el] == 0) st.delete(el); } // If the element is other than 0 else { // Find the difference let x = n - a[y]; // Find the nearest number which can give us // 0 on modulo it = st.size for (var i = 0; i < st1.length; i++) { if (st1[i] >= x) it = i; } // If no such number occurs then // find the number closest to 0 if (it == st.size) { for (var i = 0; i < st1.length; i++) { if (st1[i] >= 0) it = i; } } // Get the number let el = st1[it]; // store the number sequence.push((a[y] + el) % n); // Decrease the count if (!mpp.hasOwnProperty(el)) mpp[el] = 0; mpp[el]--; // If no more appears, then erase it from set if (mpp[el] == 0) st.delete(el); } } console.log(sequence)}// Driver Codelet a = [ 0, 1, 2, 1 ];let b = [ 3, 2, 1, 1 ];let n = a.length;solve(a, b, n)// This code is contributed by phasing17 |
Output:
1 0 0 2
Time Complexity: O(n)
Auxiliary Space: O(n)
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