Given an array arr[] and an 2D array query[][] consisting of queries. For each query q, replace all occurrences of query[i][0] in arr[], with query[i][1].
Examples:
Input: arr[] = {2, 2, 5, 1} query = {{2, 4}, {5, 2}}
Output: {4, 4, 2, 1}
Explanation: Following are the operations performed in the given array according to the queries given.
For first query {2, 4}, replace all occurrences of 2 in arr[] with 4. arr[] will be updated as arr[] = {4, 4, 5, 1}.
For second query {5, 2}, replace all occurrences of 5 with 2. arr[] will be updated as arr[] = {4, 4, 2, 1}.Input: arr[] ={2, 2, 5}, query = {{4, 5}, {2, 5}, {1, 3}, {2, 4}}
Output: {5, 5, 5}
Naive Approach: (Brute-Force solution) Naive approach would be to iterate through all the queries of query, and for each query[i][0], find all of its occurrences in arr[], and replace it with query[i][1].
Time Complexity: O(N*Q), where N is the size of arr[] and Q is size of query[][].
Auxiliary Space: O(1)
Efficient Approach: A better solution would be to use a Hashmap, that stores indexes of the element in the array. Follow the steps below to solve the given problem.
- Initialize a hashmap = {}, and fill it up with array element as key, and list indicating its position in the array.
- Iterate through each query q of query[][].
- If q[0] is present in the hashmap,
- If q[1] is present in the hashmap, then extend the value of q[0] to the value of q[1] key.
- Else, add the value to q[1] key with the value of q[0] key.
- Delete the key-value pair for q[0] from the hashmap.
- If q[0] is present in the hashmap,
- Now, create a new variable map = {}, from the values of the hashmap.
- Interchange the key-value pairs, so that map will contain key-value pairs as index, and key value of hashmap.
- Using this map, we can now update the original array arr, by updating values at each position of arr, with the value from map.
Below is the implementation of the above approach:
C++
// C++ program for above approach#include <bits/stdc++.h>using namespace std;// Function to replace all the// occurrences of a number with// another given number for Q queriesvoid update(vector<int>& A, int N, vector<vector<int> >& Q){ // Creating a hashmap map<int, vector<int> > hashmap; for (int i = 0; i < N; ++i) { hashmap[A[i]].push_back(i); } // Iterating with q in given queries for (auto q : Q) { if (hashmap.count(q[0])) { if (hashmap.count(q[1])) hashmap[q[1]].insert(hashmap[q[1]].end(), hashmap[q[0]].begin(), hashmap[q[0]].end()); else hashmap[q[1]] = hashmap[q[0]]; hashmap.erase(q[0]); } } // Creating map to store key value pairs map<int, int> new_map; for (auto it = hashmap.begin(); it != hashmap.end(); ++it) { for (auto index : it->second) new_map[index] = it->first; } // Updating the main array with final values for (auto it = new_map.begin(); it != new_map.end(); ++it) A[it->first] = it->second;}// Driver Codeint main(){ vector<int> arr = { 2, 2, 5, 1 }; int N = arr.size(); vector<vector<int> > query = { { 2, 4 }, { 5, 2 } }; update(arr, N, query); for (int i = 0; i < N; ++i) { cout << arr[i] << " "; } return 0;} // This code is contributed by rakeshsahni |
Java
// Java program for above approachimport java.io.*;import java.util.*;class GFG { // Function to replace all the // occurrences of a number with // another given number for Q queries static void update(List<Integer> A, int N, List<List<Integer> > Q) { // Creating a hashmap Map<Integer, List<Integer> > hashmap = new HashMap<>(); for (int i = 0; i < N; ++i) { List<Integer> indices = hashmap.getOrDefault( A.get(i), new ArrayList<>()); indices.add(i); hashmap.put(A.get(i), indices); } // Iterating with q in given queries for (List<Integer> q : Q) { if (hashmap.containsKey(q.get(0))) { if (hashmap.containsKey(q.get(1))) { hashmap.get(q.get(1)).addAll( hashmap.get(q.get(0))); } else { hashmap.put(q.get(1), hashmap.get(q.get(0))); } hashmap.remove(q.get(0)); } } // Creating map to store key value pairs Map<Integer, Integer> newMap = new HashMap<>(); for (Map.Entry<Integer, List<Integer> > entry : hashmap.entrySet()) { for (int index : entry.getValue()) { newMap.put(index, entry.getKey()); } } // Updating the main array with final values for (Map.Entry<Integer, Integer> entry : newMap.entrySet()) { A.set(entry.getKey(), entry.getValue()); } } public static void main(String[] args) { List<Integer> arr = Arrays.asList(2, 2, 5, 1); int N = arr.size(); List<List<Integer> > query = Arrays.asList( Arrays.asList(2, 4), Arrays.asList(5, 2)); update(arr, N, query); System.out.print(arr.toString()); }}// This code is contributed by lokesh. |
Python3
# Python program for above approach# Function to replace all the# occurrences of a number with# another given number for Q queriesdef update(A, N, Q): # Creating a hashmap hashmap = {a:[] for a in A} for i in range(N): hashmap[A[i]].append(i) # Iterating with q in given queries for q in Q: if q[0] in hashmap: if q[1] in hashmap: hashmap[q[1]].extend(hashmap[q[0]]) else: hashmap[q[1]] = hashmap[q[0]] del hashmap[q[0]] # Creating map to store key value pairs new_map = {} for key, value in hashmap.items(): for index in value: new_map[index] = key # Updating the main array with final values for key in new_map.keys(): A[key] = new_map[key] # Driver Codeif __name__ == '__main__': arr = [2, 2, 5, 1] N = len(arr) query = [[2, 4], [5, 2]] update(arr, N, query) print(arr) |
C#
// C# program for above approachusing System;using System.Collections.Generic;using System.Linq;class GFG { // Function to replace all the // occurrences of a number with // another given number for Q queries static void update(List<int> A, int N, List<List<int>> Q) { // Creating a hashmap var hashmap = new Dictionary<int, List<int>>(); for (int i = 0; i < N; ++i) { if (!hashmap.ContainsKey(A[i])) hashmap[A[i]] = new List<int>(); hashmap[A[i]].Add(i); } // Iterating with q in given queries for (int i = 0; i < Q.Count; i++) { if (hashmap.ContainsKey(Q[i][0])) { if (hashmap.ContainsKey(Q[i][1])) { hashmap[Q[i][1]].AddRange(hashmap[Q[i][0]]); } else { hashmap[Q[i][1]] = hashmap[Q[i][0]]; } hashmap.Remove(Q[i][0]); } } // Creating map to store key value pairs var newMap = new Dictionary<int, int>(); foreach (var entry in hashmap) { for (int j = 0; j < entry.Value.Count; j++) { newMap[entry.Value[j]] = entry.Key; } } // Updating the main array with final values for (int i = 0; i < A.Count; i++) { if (newMap.ContainsKey(i)) { A[i] = newMap[i]; } } } // Driver code public static void Main(string[] args) { List<int> arr = new List<int> { 2, 2, 5, 1 }; int N = arr.Count; List<List<int>> query = new List<List<int>> { new List<int> { 2, 4 }, new List<int> { 5, 2 } }; update(arr, N, query); Console.WriteLine(string.Join(" ", arr)); }} |
Javascript
<script>// Javascript program for above approach// Function to replace all the// occurrences of a number with// another given number for Q queriesfunction update(A, N, Q){ // Creating a hashmap let hashmap = new Map(); for(let i = 0; i < N; i++){ if(hashmap.has(A[i])){ let temp = hashmap.get(A[i]) temp.push(i) hashmap.set(A[i], temp) }else{ hashmap.set(A[i], [i]) } } // Iterating with q in given queries for(q of Q){ if(hashmap.has(q[0])){ if (hashmap.has(q[1])){ let temp = hashmap.get(q[1]); temp = [...temp, ...hashmap.get(q[0])] hashmap.set(q[1], temp); } else{ hashmap.set(q[1], hashmap.get(q[0])) } hashmap.delete(q[0]) } } // Creating map to store key value pairs let new_map = new Map() for(x of hashmap) for(index of x[1]) new_map.set(index, x[0]) // Updating the main array with final values for(key of new_map.keys()) A[key] = new_map.get(key) document.write(`[ ${A}] `)} // Driver Codelet arr = [2, 2, 5, 1]let N = arr.lengthlet query = [[2, 4], [5, 2]]update(arr, N, query)// This code is contributed by gfgking.</script> |
Output:
[4, 4, 2, 1]
Time Complexity: O(N+Q), where N is size of arr[] and Q is size of query[][].
Auxiliary Space: O(N)
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