Array sum after replacing all occurrences of X by Y for Q queries

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Given an integer array arr[] and Q queries, the task is to find the sum of the array for each query of the following type:

Each query contains 2 integers X and Y, where all the occurrences of X in arr[] are to be replaced by Y.
After each query, they print the sum of the array.

Examples:

Input: arr[] = { 1, 2, 1, 3, 2}, X[] = { 2, 3, 5 }, Y[] = { 3, 1, 2 }
Output: 11 5 5
Explanation:
After the 1st query, replace 2 with 3, arr[] = { 1, 3, 1, 3, 3 }, Sum = 11.
After the 2nd query, replace 3 with 1, arr[] = { 1, 1, 1, 1, 1 }, Sum = 5.
After the 3rd query, replace 5 with 2, arr[] = { 1, 1, 1, 1, 1 }, Sum = 5.

Input: arr[] = { 12, 22, 11, 11, 2}, X[] = {2, 11, 22}, Y[] = {12, 222, 2}
Output: 68 490 470

Naive Approach:
The simplest approach to solve the problem mentioned above is to traverse through the array and replace all the instances of X with Y for each query and calculate the sum.

Time Complexity: O(N * Q)

Efficient Approach:
To optimize the above method, follow the steps given below:

Precompute and store the sum of the array in a variable S and store the frequencies of array elements in a Map count.
Then, do the following for each query:
- Find the frequency of X stored on the map.
- Subtract X * count[X] from S.
- Set count[Y] = count[X] and then count[X] = 0.
- Add Y * count[Y] to S.
- Print the updated value of S.

Below is the implementation of the above approach:

C++

// C++ implementation to find the sum
// of the array for the given Q queries
 
#include <bits/stdc++.h>
using namespace std;
 
// Function that print the sum of
// the array for Q queries
void sumOfTheArrayForQuery(int* A, int N,
                           int* X, int* Y,
                           int Q)
{
    int sum = 0;
 
    // Stores the frequencies
    // of array elements
    unordered_map<int, int> count;
 
    // Calculate the sum of
    // the initial array and
    // store the frequency of
    // each element in map
 
    for (int i = 0; i < N; i++) {
        sum += A[i];
        count[A[i]]++;
    }
 
    // Iterate for all the queries
 
    for (int i = 0; i < Q; i++) {
        // Store query values
        int x = X[i], y = Y[i];
 
        // Decrement the sum accordingly
        sum -= count[X[i]] * X[i];
 
        // Increment the sum accordingly
        sum += count[X[i]] * Y[i];
 
        // Set count of Y[i]
        count[Y[i]] += count[X[i]];
 
        // Reset count of X[i]
        count[X[i]] = 0;
 
        // Print the sum
        cout << sum << " ";
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 1, 3, 2 };
    int X[] = { 2, 3, 5 };
    int Y[] = { 3, 1, 2 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    int Q = sizeof(X) / sizeof(X[0]);
 
    // Function call
    sumOfTheArrayForQuery(arr, N, X, Y, Q);
 
    return 0;
}

Java

// Java implementation to 
// find the sum of the array 
// for the given Q queries
import java.util.*;
class GFG{
   
// Function that print the sum of
// the array for Q queries
public static void sumOfTheArrayForQuery(int[] A, int N, 
                                         int[] X, int[] Y, 
                                         int Q)
{
  int sum = 0;
 
  // Stores the frequencies
  // of array elements
  // Create an empty hash map 
  HashMap<Integer, 
          Integer> count = new HashMap<>(); 
 
  // Calculate the sum of
  // the initial array and
  // store the frequency of
  // each element in map
  for (int i = 0; i < N; i++) 
  {
    sum += A[i];
    if (count.containsKey(A[i])) 
    {
      count.replace(A[i], 
      count.get(A[i]) + 1);
    }
    else
    {
      count.put(A[i], 1);
    }
  }
 
  // Iterate for all the queries
  for (int i = 0; i < Q; i++) 
  {
    // Store query values
    int x = X[i], y = Y[i];
 
    if(count.containsKey(X[i]))
    {
      // Decrement the sum accordingly
      sum -= count.get(X[i]) * X[i];
      // Increment the sum accordingly
      sum += count.get(X[i]) * Y[i];
    }
 
    // Set count of Y[i]
    if(count.containsKey(Y[i]) && 
       count.containsKey(X[i]))
    {
      count.replace(Y[i], 
      count.get(Y[i]) + 
      count.get(X[i]));
    }
 
    // Reset count of X[i]
    if(count.containsKey(X[i]))
    {
      count.replace(X[i], 0);
    }
 
    // Print the sum
    System.out.print(sum + " ");
  }
}
 
// Driver code
public static void main(String[] args) 
{
  int arr[] = {1, 2, 1, 3, 2};
  int X[] = {2, 3, 5};
  int Y[] = {3, 1, 2};
 
  int N = arr.length;
  int Q = X.length;
 
  // Function call
  sumOfTheArrayForQuery(arr, N, 
                        X, Y, Q);
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 implementation to find the sum 
# of the array for the given Q queries 
 
# Function that print the sum of 
# the array for Q queries
def sumOfTheArrayForQuery(A, N, X, Y, Q):
     
    sum = 0
 
    # Stores the frequencies 
    # of array elements 
    count = {}
 
    # Calculate the sum of 
    # the initial array and 
    # store the frequency of 
    # each element in map 
    for i in range(N):
        sum += A[i]
         
        if A[i] in count:
            count[A[i]] += 1
        else:
            count[A[i]] = 1
 
    # Iterate for all the queries
    for i in range(Q):
 
        # Store query values
        x = X[i]
        y = Y[i]
         
        if X[i] not in count:
            count[X[i]] = 0
        if Y[i] not in count:
            count[Y[i]] = 0
 
        # Decrement the sum accordingly
        sum -= (count[X[i]] * X[i])
 
        # Increment the sum accordingly
        sum += count[X[i]] * Y[i]
 
        # Set count of Y[i]
        count[Y[i]] += count[X[i]]
 
        # Reset count of X[i] 
        count[X[i]] = 0
 
        # Print the sum
        print(sum, end = " ")
 
# Driver Code
arr = [ 1, 2, 1, 3, 2, ]
X = [ 2, 3, 5 ]
Y = [ 3, 1, 2 ]
N = len(arr)
Q = len(X)
 
# Function call
sumOfTheArrayForQuery(arr, N, X, Y, Q)
 
# This code is contributed by avanitrachhadiya2155

C#

// C# implementation to 
// find the sum of the array 
// for the given Q queries
using System;
using System.Collections.Generic;
class GFG{
   
// Function that print the sum of
// the array for Q queries
public static void sumOfTheArrayForQuery(int[] A, int N, 
                                         int[] X, int[] Y, 
                                         int Q)
{
  int sum = 0;
 
  // Stores the frequencies
  // of array elements
  // Create an empty hash map 
  Dictionary<int, 
             int> count = new Dictionary<int,
                                         int>(); 
 
  // Calculate the sum of
  // the initial array and
  // store the frequency of
  // each element in map
  for (int i = 0; i < N; i++) 
  {
    sum += A[i];
    if (count.ContainsKey(A[i])) 
    {
      count[A[i]]= count[A[i]] + 1;
    }
    else
    {
      count.Add(A[i], 1);
    }
  }
 
  // Iterate for all the queries
  for (int i = 0; i < Q; i++) 
  {
    // Store query values
    int x = X[i], y = Y[i];
 
    if(count.ContainsKey(X[i]))
    {
      // Decrement the sum accordingly
      sum -= count[X[i]] * X[i];
      // Increment the sum accordingly
      sum += count[X[i]] * Y[i];
    }
 
    // Set count of Y[i]
    if(count.ContainsKey(Y[i]) && 
       count.ContainsKey(X[i]))
    {
      count[Y[i]] = count[Y[i]] + 
                    count[X[i]];
    }
 
    // Reset count of X[i]
    if(count.ContainsKey(X[i]))
    {
      count[X[i]] = 0;
    }
 
    // Print the sum
    Console.Write(sum + " ");
  }
}
 
// Driver code
public static void Main(String[] args) 
{
  int []arr = {1, 2, 1, 3, 2};
  int []X = {2, 3, 5};
  int []Y = {3, 1, 2};
 
  int N = arr.Length;
  int Q = X.Length;
 
  // Function call
  sumOfTheArrayForQuery(arr, N, 
                        X, Y, Q);
}
}
 
// This code is contributed by Amit Katiyar

Javascript

<script>
 
 
// Javascript implementation to find the sum
// of the array for the given Q queries
 
// Function that print the sum of
// the array for Q queries
function sumOfTheArrayForQuery(A, N, X, Y, Q)
{
    var sum = 0;
 
    // Stores the frequencies
    // of array elements
    var count = new Map();
 
    // Calculate the sum of
    // the initial array and
    // store the frequency of
    // each element in map
 
    for (var i = 0; i < N; i++) {
        sum += A[i];
        if(count.has(A[i]))
            count.set(A[i], count.get(A[i])+1)
        else
            count.set(A[i], 1)
    }
 
    // Iterate for all the queries
 
    for (var i = 0; i < Q; i++) {
        // Store query values
        var x = X[i], y = Y[i];
 
        if(count.has(X[i]))
        {
            // Decrement the sum accordingly
            sum -= count.get(X[i]) * X[i];
 
                // Increment the sum accordingly
            sum += count.get(X[i]) * Y[i];
        }
 
        if(count.has(Y[i]))
        {
            // Set count of Y[i]
            count.set(Y[i], count.get(Y[i]) + count.get(X[i]));
        }
 
        // Reset count of X[i]
        count.set(X[i] , 0);
 
        // Print the sum
        document.write( sum + " ");
    }
}
 
// Driver Code
var arr = [1, 2, 1, 3, 2];
var X = [2, 3, 5 ];
var Y = [3, 1, 2 ];
var N = arr.length;
var Q = X.length;
// Function call
sumOfTheArrayForQuery(arr, N, X, Y, Q);
 
 
</script>

Output:

11 5 5

Time Complexity: O(N+Q), as each query has a computational complexity of O(1).
Auxiliary Space: O(N)

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