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

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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; ` `    ``} ` ` `  `    ``// return the result ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``// initialise array ` `    ``int` `Q[] = { 3, 1, 2, 1 }; ` `    ``int` `n = ``sizeof``(Q) / ``sizeof``(Q); ` ` `  `    ``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 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 ans = ``new` `Vector(); ` `  `  `    ``// 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 ans = ``new` `Vector(); ` `  `  `    ``// 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#

 `// 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; ` `    ``} ` ` `  `    ``// 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 `

Output:

```2 1 2 1
```

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.

1. Push element in the front
2. Find the index of a number
3. Update the index’s 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 is update 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 postion by finding the number of set elements lesser than the given query, once done set it’s 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 ` `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; ` `} `

## 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 `

Output:

```3 1 2 0
```

Time complexity : O(nlogn)
space complexity : O(2n)

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