# Find sum of product of every number and its frequency in given range

• Difficulty Level : Hard
• Last Updated : 10 Jun, 2021

Given an array arr[] of integers and an array of queries, the task is to find the sum of product of every number and its frequency in given range [L, R] where each ranges are given in the array of queries.

Examples:

Input: arr[] = [1, 2, 1], Queries: [{1, 2}, {1, 3}]
Output: [3, 4]
Explanation:
For query [1, 2], freq[1] = 1, freq[2] = 1, ans = (1 * freq[1]) + (2 * freq[2]) => ans = (1 * 1 + 2 * 1) = 3
For query [1, 3], freq[1] = 2, freq[2] = 1; ans = (1 * freq[1]) + (2 * freq[2]) => ans = (1 * 2) + (2 * 1) = 4

Input: arr[] = [1, 1, 2, 2, 1, 3, 1, 1], Queries: [{2, 7}, {1, 6}]
Output: [10, 10]
Explanation:
For query (2, 7), freq[1] = 3, freq[2] = 2, freq[3] = 3;
ans = (1 * freq[1]) + (2 * freq[2] ) + (3 * freq[3])
ans = (1 * 3) + (2 * 2) + (3 * 1) = 10

Naive Approach:
To solve the problem mentioned above the naive method is to iterate over the subarray given in the query. Maintain a map for the frequency of each number in the subarray and iterate over the map and compute the answer.

Below is the implementation of the above approach:

## C++

 // C++ implementation to find// sum of product of every number// and square of its frequency// in the given range #include using namespace std; // Function to solve queriesvoid answerQueries(    int arr[], int n,    vector >& queries){     for (int i = 0; i < queries.size(); i++) {         // Calculating answer        // for every query        int ans = 0;         // The end points        // of the ith query        int l = queries[i].first - 1;        int r = queries[i].second - 1;         // map for storing frequency        map<int, int> freq;        for (int j = l; j <= r; j++) {             // Iterating over the given            // subarray and storing            // frequency in a map             // Incrementing the frequency            freq[arr[j]]++;        }         // Iterating over map to find answer        for (auto& i : freq) {             // adding the contribution            // of ith number            ans += (i.first                    * i.second);        }         // print answer        cout << ans << endl;    }} // Driver codeint main(){     int arr[] = { 1, 2, 1 };    int n = sizeof(arr) / sizeof(arr[0]);     vector > queries        = { { 1, 2 },            { 1, 3 } };    answerQueries(arr, n, queries);}

## Java

 // Java program to find sum of// product of every number and// square of its frequency in// the given rangeimport java.util.*; class GFG{ // Function to solve queries    public static void answerQueries(int[] arr,                                 int n,                                 int[][] queries){    for(int i = 0; i < queries.length; i++)    {                 // Calculating answer        // for every query            int ans = 0;         // The end points        // of the ith query        int l = queries[i][0] - 1;        int r = queries[i][1] - 1;         // Hashmap for storing frequency        Map freq = new HashMap<>();         for(int j = l; j < r + 1; j++)        {                         // Iterating over the given            // subarray and storing            // frequency in a map             // Incrementing the frequency            freq.put(arr[j],                     freq.getOrDefault(arr[j], 0) + 1);        }        for(int k: freq.keySet())        {                         // Adding the contribution            // of ith number            ans += k * freq.get(k);        }             // Print answer    System.out.println(ans);    }} // Driver codepublic static void main(String args[] ){    int[] arr = { 1, 2, 1 };    int n = arr.length;    int[][] queries = { { 1, 2 },                        { 1, 3 } };                             // Calling function    answerQueries(arr, n, queries);}} // This code contributed by dadi madhav

## Python3

 # Python3 implementation to find# sum of product of every number# and square of its frequency# in the given range # Function to solve queriesdef answerQueries(arr, n, queries):         for i in range(len(queries)):         # Calculating answer        # for every query        ans = 0         # The end points        # of the ith query        l = queries[i][0] - 1        r = queries[i][1] - 1         # Map for storing frequency        freq = dict()        for j in range(l, r + 1):             # Iterating over the given            # subarray and storing            # frequency in a map             # Incrementing the frequency            freq[arr[j]] = freq.get(arr[j], 0) + 1         # Iterating over map to find answer        for i in freq:             # Adding the contribution            # of ith number            ans += (i * freq[i])         # Print answer        print(ans) # Driver codeif __name__ == '__main__':         arr = [ 1, 2, 1 ]    n = len(arr)     queries = [ [ 1, 2 ],                [ 1, 3 ] ]                     answerQueries(arr, n, queries) # This code is contributed by mohit kumar 29

## C#

 // C# program to find sum of// product of every number and // square of its frequency in // the given rangeusing System;using System.Collections.Generic; class GFG{     // Function to solve queriespublic static void answerQueries(int[] arr, int n,                                 int[,] queries){    for(int i = 0; i < queries.GetLength(0); i++)    {                 // Calculating answer         // for every query         int ans = 0;                 // The end points         // of the ith query        int l = queries[i, 0] - 1;        int r = queries[i, 1] - 1;                 // Hashmap for storing frequency        Dictionary<int,                   int> freq = new Dictionary<int,                                              int>();        for(int j = l; j < r+ 1; j++)        {                         // Iterating over the given             // subarray and storing             // frequency in a map              // Incrementing the frequency            freq[arr[j]] = freq.GetValueOrDefault(arr[j], 0) + 1;        }        foreach(int k in freq.Keys)        {                         // Adding the contribution             // of ith number            ans += k * freq[k];        }                 // Print answer        Console.WriteLine(ans);    }} // Driver codestatic public void Main(){    int[] arr = { 1, 2, 1 };    int n = arr.Length;    int[,] queries = { { 1, 2 }, { 1, 3 } };         // Calling function    answerQueries(arr, n, queries);}} // This code is contributed by avanitrachhadiya2155

## Javascript

 
Output:
3
4

Time Complexity: O(Q * N)
Auxiliary Space Complexity: O(N)

Efficient Approach:
To optimize the above method we will try to implement the problem using Mo’s Algorithm.

• Sort the queries first according to their block of size using custom comparator and also store the indexes of each query in a map for printing in order.
• Now, we will maintain two-pointer L and R which we iterate over the array for answering the queries. As we move the pointers, if we are adding some number in our range, we’ll first remove the contribution of its previous freq from the answer, then increment the frequency and finally add the contribution of new frequency in answer.
• And if we remove some element from the range, we’ll do the same, remove the contribution of existing freq of this number, decrement the freq, add the contribution of its new frequency.

Below is the implementation of the above approach:

## C++

 // C++ implementation to find sum// of product of every number// and square of its frequency// in the given range #include using namespace std; // Stores frequencyconst int N = 1e5 + 5; // Frequency arrayvector<int> freq(N); int sq; // Function for comparatorbool comparator(    pair<int, int>& a,    pair<int, int>& b){    // comparator for sorting    // according to the which query    // lies in the which block;    if (a.first / sq != b.first / sq)        return a.first < b.first;     // if same block,    // return which query end first    return a.second < b.second;} // Function to add numbers in rangevoid add(int x, int& ans, int arr[]){    // removing contribution of    // old frequency from answer    ans -= arr[x]           * freq[arr[x]];     // incrementing the frequency    freq[arr[x]]++;     // adding contribution of    // new frequency to answer    ans += arr[x]           * freq[arr[x]];} void remove(int x, int& ans, int arr[]){    // removing contribution of    // old frequency from answer    ans -= arr[x]           * freq[arr[x]];     // Decrement the frequency    freq[arr[x]]--;     // adding contribution of    // new frequency to answer    ans += arr[x]           * freq[arr[x]];} // Function to answer the queriesvoid answerQueries(    int arr[], int n,    vector >& queries){     sq = sqrt(n) + 1;     vector<int> answer(        int(queries.size()));     // map for storing the    // index of each query    map, int> idx;     // Store the index of queries    for (int i = 0; i < queries.size(); i++)        idx[queries[i]] = i;     // Sort the queries    sort(queries.begin(),         queries.end(),         comparator);     int ans = 0;     // pointers for iterating    // over the array    int x = 0, y = -1;    for (auto& i : queries) {         // iterating over all        // the queries        int l = i.first - 1, r = i.second - 1;        int id = idx[i];         while (x > l) {            // decrementing the left            // pointer and adding the            // xth number's contribution            x--;            add(x, ans, arr);        }        while (y < r) {             // incrementing the right            // pointer and adding the            // yth number's contribution            y++;            add(y, ans, arr);        }        while (x < l) {             // incrementing the left pointer            // and removing the            // xth number's contribution            remove(x, ans, arr);            x++;        }        while (y > r) {             // decrementing the right            // pointer and removing the            // yth number's contribution            remove(y, ans, arr);            y--;        }        answer[id] = ans;    }     // printing the answer of queries    for (int i = 0; i < queries.size(); i++)        cout << answer[i] << endl;} // Driver Codeint main(){     int arr[] = { 1, 1, 2, 2, 1, 3, 1, 1 };    int n = sizeof(arr) / sizeof(arr[0]);     vector > queries        = { { 2, 7 },            { 1, 6 } };    answerQueries(arr, n, queries);}
Output:
10
10

Time Complexity: O(N * sqrt{N})
Auxiliary Space Complexity: O(N)

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