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

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[] is to be replaced by Y.
• After each query 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 1st Query, Replace 2 with 3, arr[] = { 1, 3, 1, 3, 3 }, Sum = 11.
After 2nd Query, Replace 3 with 1, arr[] = { 1, 1, 1, 1, 1 }, Sum = 5.
After 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 in 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++ implementation to find the sum` `// of the array for the given Q queries`   `#include ` `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 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 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 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# 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`

Output:
```11 5 5

```

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

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