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 queriesvoid 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 Codeint 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 queriesimport java.util.*;class GFG{ // Function that print the sum of// the array for Q queriespublic 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 codepublic 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 queriesdef 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 Codearr = [ 1, 2, 1, 3, 2, ]X = [ 2, 3, 5 ]Y = [ 3, 1, 2 ]N = len(arr)Q = len(X)# Function callsumOfTheArrayForQuery(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 queriesusing System;using System.Collections.Generic;class GFG{ // Function that print the sum of// the array for Q queriespublic 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 codepublic 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 queriesfunction 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 Codevar 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 callsumOfTheArrayForQuery(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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