Queries to search an element in an array and move it to front after every query

Last Updated : 27 Sep, 2022

Given an integer M which represents an array initially having numbers 1 to M. Also given is a Query array. For every query, search the number in the initial array and bring it to the front of the array. The task is to return the indexes of the searched element in the given array for every query.
Examples:

Input : Q[] = {3, 1, 2, 1}, M = 5
Output : [2, 1, 2, 1]
Explanations :
Since m = 5 the initial array is [1, 2, 3, 4, 5].
Query1: Search for 3 in the [1, 2, 3, 4, 5] and move it in the beginning. After moving, the array looks like [3, 1, 2, 4, 5]. 3 is at index 2.
Query2: Move 1 from [3, 1, 2, 4, 5] to the beginning of the array to make the array look like [1, 3, 2, 4, 5]. 1 is present at index 1.
Query3: Move 2 from [1, 3, 2, 4, 5] to the beginning of the array to make the array look like [2, 1, 3, 2, 4, 5]. 2 is present at index 2.
Query4: Move 1 from [2, 1, 3, 4, 5] to the beginning of the array to make the array look like [1, 2, 3, 4, 5]. 1 is present at index 1.

Input : Q[] = {4, 1, 2, 2}, M = 4
Output : 3, 1, 2, 0

Naive approach: The naive approach is to use a hash table to search for the element and linearly do shifts by performing swaps. The time complexity will be quadratic in nature for this approach.
Below is the naive implementation of the above approach:

C++

// C++ program to search the element // in an array for every query and // move the searched element to // the front after every query #include <bits/stdc++.h> using namespace std; // Function to find the indices vector<int> processQueries(int Q[], int m, int n) {  int a[m + 1], pos[m + 1];  for (int i = 1; i <= m; i++) {  a[i - 1] = i;  pos[i] = i - 1;  }  vector<int> ans;  // iterate in the query array  for (int i = 0; i < n; i++) {  int q = Q[i];  // store current element  int p = pos[q];  ans.push_back(p);  for (int i = p; i > 0; i--) {  // swap positions of the element  swap(a[i], a[i - 1]);  pos[a[i]] = i;  }  pos[a[0]] = 0;  }  // return the result  return ans; } // Driver code int main() {  // initialise array  int Q[] = { 3, 1, 2, 1 };  int n = sizeof(Q) / sizeof(Q[0]);  int m = 5;  vector<int> ans;  // Function call  ans = processQueries(Q, m, n);  // Print answers to queries  for (int i = 0; i < ans.size(); i++)  cout << ans[i] << " ";  return 0; }

Java

// Java program to search the element // in an array for every query and // move the searched element to // the front after every query import java.util.*; class GFG{   // Function to find the indices static Vector<Integer> processQueries(int Q[], int m, int n) {  int []a = new int[m + 1];  int []pos = new int[m + 1];    for (int i = 1; i <= m; i++) {  a[i - 1] = i;  pos[i] = i - 1;  }    Vector<Integer> ans = new Vector<Integer>();    // iterate in the query array  for (int i = 0; i < n; i++) {  int q = Q[i];    // store current element  int p = pos[q];    ans.add(p);    for (int j = p; j > 0; j--) {    // swap positions of the element  a[j] = a[j] + a[j - 1];  a[j - 1] = a[j] - a[j - 1];  a[j] = a[j] - a[j - 1];    pos[a[j]] = j;  }    pos[a[0]] = 0;  }    // return the result  return ans; }   // Driver code public static void main(String[] args) {  // initialise array  int Q[] = { 3, 1, 2, 1 };  int n = Q.length;    int m = 5;    Vector<Integer> ans = new Vector<Integer>();    // Function call  ans = processQueries(Q, m, n);    // Print answers to queries  for (int i = 0; i < ans.size(); i++)  System.out.print(ans.get(i)+ " ");   } } // This code is contributed by sapnasingh4991

Python3

# Python3 program to search the element  # in an array for every query and  # move the searched element to  # the front after every query # Function to find the indices  def processQueries(Q, m, n) : a = [0]*(m + 1); pos = [0]*(m + 1); for i in range(1, m + 1) : a[i - 1] = i; pos[i] = i - 1; ans = []; # iterate in the query array  for i in range(n) : q = Q[i]; # store current element  p = pos[q]; ans.append(p); for i in range(p,0,-1) : # swap positions of the element  a[i], a[i - 1] = a[i - 1],a[i]; pos[a[i]] = i; pos[a[0]] = 0; # return the result  return ans; # Driver code  if __name__ == "__main__" : # initialise array  Q = [ 3, 1, 2, 1 ]; n = len(Q); m = 5; ans = []; # Function call  ans = processQueries(Q, m, n); # Print answers to queries  for i in range(len(ans)) : print(ans[i],end=" "); # This code is contributed by Yash_R

// C# program to search the element // in an array for every query and // move the searched element to // the front after every query using System; using System.Collections.Generic; public class GFG{ // Function to find the indices static List<int> processQueries(int []Q, int m, int n) {  int []a = new int[m + 1];  int []pos = new int[m + 1];  for(int i = 1; i <= m; i++)  {  a[i - 1] = i;  pos[i] = i - 1;  }    List<int> ans = new List<int>();  // Iterate in the query array  for(int i = 0; i < n; i++)  {  int q = Q[i];  // Store current element  int p = pos[q];  ans.Add(p);  for(int j = p; j > 0; j--)  {    // Swap positions of the element  a[j] = a[j] + a[j - 1];  a[j - 1] = a[j] - a[j - 1];  a[j] = a[j] - a[j - 1];  pos[a[j]] = j;  }  pos[a[0]] = 0;  }  // Return the result  return ans; } // Driver code public static void Main(String[] args) {  // Initialise array  int []Q = { 3, 1, 2, 1 };  int n = Q.Length;  int m = 5;  List<int> ans = new List<int>();  // Function call  ans = processQueries(Q, m, n);  // Print answers to queries  for(int i = 0; i < ans.Count; i++)  Console.Write(ans[i] + " "); } } // This code is contributed by sapnasingh4991

JavaScript

<script> // JavaScript program to search the element // in an array for every query and // move the searched element to // the front after every query // Function to find the indices function processQueries(Q,m,n) {  let a = new Array(m + 1);  let pos = new Array(m + 1);    for (let i = 1; i <= m; i++) {  a[i - 1] = i;  pos[i] = i - 1;  }    let ans = [];    // iterate in the query array  for (let i = 0; i < n; i++) {  let q = Q[i];    // store current element  let p = pos[q];    ans.push(p);    for (let j = p; j > 0; j--) {    // swap positions of the element  a[j] = a[j] + a[j - 1];  a[j - 1] = a[j] - a[j - 1];  a[j] = a[j] - a[j - 1];    pos[a[j]] = j;  }    pos[a[0]] = 0;  }    // return the result  return ans; } // Driver code // initialise array let Q=[3, 1, 2, 1]; let n = Q.length; let m = 5; let ans=[]; // Function call ans = processQueries(Q, m, n); // Print answers to queries for (let i = 0; i < ans.length; i++)  document.write(ans[i]+ " ");   // This code is contributed by rag2127 </script>

Output:

2 1 2 1

Time complexity: O(n*p)
Auxiliary space: O(m)

Efficient Approach: An efficient method to solve the above problem is to use Fenwick Tree. Using the below 3 operations, the problem can be solved.

Push element in the front
Find the index of a number
Update the indexes of the rest of the elements.

Keep the elements in a sorted manner using the set data structure, and then follow the below-mentioned points:

Instead of pushing the element to the front i.e assigning an index 0 for every query.
We assign -1 for the first query, -2 for the second query, -3 for the third, and so on till -m.
Doing so, the range of index’s updates to [-m, m]
Perform a right shift for all values [-m, m] by a value of m, so our new range is [0, 2m]
Initialize a Fenwick tree of size 2m and set all the values from [1…m] i.e [m..2m]
For every query, find its position by finding the number of set elements lesser than the given query, once done set its position to 0 in the Fenwick tree.

Below is the implementation of the above approach:

C++

// C++ program to search the element // in an array for every query and // move the searched element to // the front after every query #include <bits/stdc++.h> using namespace std; // Function to update the fenwick tree void update(vector<int>& tree, int i, int val) {  // update the next element  while (i < tree.size()) {  tree[i] += val;  // move to the next  i += (i & (-i));  } } // Function to get the // sum from the fenwick tree int getSum(vector<int>& tree, int i) {  int s = 0;  // keep adding till we have not  // reached the root of the tree  while (i > 0) {  s += tree[i];  // move to the parent  i -= (i & (-i));  }  return s; } // function to process the queries vector<int> processQueries(vector<int>& queries, int m) {  vector<int> res, tree((2 * m) + 1, 0);  // Hash-table  unordered_map<int, int> hmap;  // Iterate and increase the frequency  // count in the fenwick tree  for (int i = 1; i <= m; ++i) {  hmap[i] = i + m;  update(tree, i + m, 1);  }  // Traverse for all queries  for (int querie : queries) {  // Get the sum from the fenwick tree  res.push_back(getSum(tree, hmap[querie]) - 1);  // remove it from the fenwick tree  update(tree, hmap[querie], -1);  // Add it back at the first index  update(tree, m, 1);  hmap[querie] = m;  m--;  }  // return the final result  return res; } // Driver code int main() {  // initialise the Queries  vector<int> Queries = { 4, 1, 2, 2 };  // initialise M  int m = 4;  vector<int> ans;  ans = processQueries(Queries, m);  for (int i = 0; i < ans.size(); i++)  cout << ans[i] << " ";  return 0; }

Java

// Java program to search the element // in an array for every query and // move the searched element to // the front after every query import java.util.*; class GFG{ // Function to update the fenwick tree static void update(int []tree, int i, int val) {  // update the next element  while (i < tree.length)   {  tree[i] += val;  // move to the next  i += (i & (-i));  } } // Function to get the // sum from the fenwick tree static int getSum(int []tree, int i) {  int s = 0;  // keep adding till we have not  // reached the root of the tree  while (i > 0)  {  s += tree[i];  // move to the parent  i -= (i & (-i));  }  return s; } // function to process the queries static Vector<Integer> processQueries(int []queries,   int m) {  Vector<Integer>res = new Vector<>();  int []tree = new int[(2 * m) + 1];  // Hash-table  HashMap<Integer,Integer> hmap = new HashMap<>();  // Iterate and increase the frequency  // count in the fenwick tree  for (int i = 1; i <= m; ++i)  {  hmap.put(i, i+m);  update(tree, i + m, 1);  }    // Traverse for all queries  for (int querie : queries)   {  // Get the sum from the fenwick tree  res.add(getSum(tree, hmap.get(querie) - 1));  // remove it from the fenwick tree  update(tree, hmap.get(querie), -1);  // Add it back at the first index  update(tree, m, 1);  hmap.put(querie, m);  m--;  }  // return the final result  return res; } // Driver code public static void main(String[] args) {  // initialise the Queries  int []Queries = { 4, 1, 2, 2 };  // initialise M  int m = 4;  Vector<Integer> ans;  ans = processQueries(Queries, m);  System.out.print(ans); } } // This code is contributed by Rohit_ranjan

Python3

# Python3 program to search the element # in an array for every query and # move the searched element to # the front after every query # Function to update the fenwick tree def update(tree, i, val): # Update the next element while (i < len(tree)): tree[i] += val # Move to the next i += (i & (-i)) # Function to get the # sum from the fenwick tree def getSum(tree, i): s = 0 # Keep adding till we have not # reached the root of the tree while (i > 0): s += tree[i] # Move to the parent i -= (i & (-i)) return s # Function to process the queries def processQueries(queries, m): res = [] tree = [0] * (2 * m + 1) # Hash-table hmap = {} # Iterate and increase the frequency # count in the fenwick tree for i in range(1, m + 1): hmap[i] = i + m update(tree, i + m, 1) # Traverse for all queries for querie in queries: # Get the sum from the fenwick tree res.append(getSum(tree, hmap[querie]) - 1) # Remove it from the fenwick tree update(tree, hmap[querie], -1) # Add it back at the first index update(tree, m, 1) hmap[querie] = m m -= 1 # Return the final result return res # Driver code if __name__ == "__main__": # Initialise the Queries Queries = [ 4, 1, 2, 2 ] # Initialise M m = 4 ans = processQueries(Queries, m) for i in range(len(ans)): print(ans[i], end = " ") # This code is contributed by chitranayal

// C# program to search the element // in an array for every query and // move the searched element to // the front after every query using System; using System.Collections.Generic; class GFG{ // Function to update the fenwick tree static void update(int []tree,   int i, int val) {  // update the next element  while (i < tree.Length)   {  tree[i] += val;  // move to the next  i += (i & (-i));  } } // Function to get the // sum from the fenwick tree static int getSum(int []tree, int i) {  int s = 0;  // keep adding till we have not  // reached the root of the tree  while (i > 0)  {  s += tree[i];  // move to the parent  i -= (i & (-i));  }  return s; } // function to process the queries static List<int> processQueries(int []queries,   int m) {  List<int>res = new List<int>();  int []tree = new int[(2 * m) + 1];  // Hash-table  Dictionary<int,  int> hmap = new Dictionary<int,  int>();  // Iterate and increase the frequency  // count in the fenwick tree  for (int i = 1; i <= m; ++i)  {  hmap.Add(i, i+m);  update(tree, i + m, 1);  }    // Traverse for all queries  foreach (int querie in queries)   {  // Get the sum from the fenwick tree  res.Add(getSum(tree, hmap[querie] - 1));  // remove it from the fenwick tree  update(tree, hmap[querie], -1);  // Add it back at the first index  update(tree, m, 1);  if(hmap.ContainsKey(querie))  hmap[querie] = m;  else  hmap.Add(querie, m);  m--;  }  // return the readonly result  return res; } // Driver code public static void Main(String[] args) {  // initialise the Queries  int []Queries = {4, 1, 2, 2};  // initialise M  int m = 4;  List<int> ans;  ans = processQueries(Queries, m);  foreach (int i in ans)  Console.Write(i + " "); } } // This code is contributed by shikhasingrajput

JavaScript

<script> // Javascript program to search the element // in an array for every query and // move the searched element to // the front after every query    // Function to update the fenwick tree  function update(tree,i,val)  {  // update the next element  while (i < tree.length)  {  tree[i] += val;    // move to the next  i += (i & (-i));  }  }    // Function to get the // sum from the fenwick tree  function getSum(tree,i)  {  let s = 0;    // keep adding till we have not  // reached the root of the tree  while (i > 0)  {  s += tree[i];    // move to the parent  i -= (i & (-i));  }  return s;  }    // function to process the queries  function processQueries(queries,m)  {  let res = [];  let tree = new Array((2 * m) + 1);  for(let i=0;i<tree.length;i++)  {  tree[i]=0;  }  // Hash-table  let hmap = new Map();    // Iterate and increase the frequency  // count in the fenwick tree  for (let i = 1; i <= m; ++i)  {  hmap.set(i, i+m);  update(tree, i + m, 1);  }    // Traverse for all queries  for (let querie=0;querie< queries.length;querie++)  {    // Get the sum from the fenwick tree  res.push(getSum(tree, hmap.get(queries[querie]) - 1));    // remove it from the fenwick tree  update(tree, hmap.get(queries[querie]), -1);    // Add it back at the first index  update(tree, m, 1);    hmap.set(queries[querie], m);    m--;  }    // return the final result  return res;  }    // Driver code  let Queries=[4, 1, 2, 2];  // initialise M  let m = 4;    let ans;    ans = processQueries(Queries, m);    document.write(ans.join(" "));     // This code is contributed by avanitrachhadiya2155 </script>