Given an array, arr[] of N positive integers and M queries of the form {a, b, val, f}. The task is to print the array after performing each query to increment array elements in the range [a, b] by a value val f number of times.
Examples:
Input: arr[] = {1, 2, 3}, M=3, Q[][] = {{1, 2, 1, 4}, {1, 3, 2, 3}, {2, 3, 4, 5}}
Output: 11 32 29
Explanation:
After applying 1st Query 4 times,
Array will be: 5 6 3
After applying 2nd Query 3 times,
Array will be: 11 12 9
After applying 3rd Query 5 times,
Array will be: 11 32 29
Therefore, the final array will be {11, 32, 29}.Input: arr[] = {1}, M = 1, Q[][] = {{1, 1, 1, 1}}
Output: 2
Explanation:
After applying 1st and the only query 1 time only.
Array will be: 2
Naive Approach: The simplest approach is to perform each query on the given array i.e., for each query {a, b, val, f} traverse the array over the range [a, b] and increase each element by value val to f number of times. Print the array after performing each query.
Time Complexity: O(N * M * max(Freq))
Auxiliary Space: O(1)
Better Approach: The idea is based on the difference array which can be used in Range Update operations. Below are the steps:
- Find the difference array D[] of a given array A[] is defined as D[i] = (A[i] – A[i – 1]) (0 < i < N) and D[0] = A[0] considering 0 based indexing.
- Add val to D[a – 1] and subtract it from D[(b – 1) + 1], i.e., D[a – 1] += val, D[(b – 1) + 1] -= val. Perform this operation Freq number of times.
- Now update the given array using the difference array. Update A[0] to D[0] and print it. For rest of the elements, do A[i] = A[i-1] + D[i].
- Print the resultant array after the above steps.
Time Complexity: O(N + M * max(Freq))
Auxiliary Space: O(N) Extra space for Difference Array
Efficient Approach: This approach is similar to the previous approach but an extended version of the application of a difference array. Previously the task was to update values from indices a to b by val, f number of times. Here instead of calling the range update function f number of times, call it only once for each query:
- Update values from indices a to b by val*f, only 1 time for each query.
- Add val*f to D[a – 1] and subtract it from D[(b – 1) + 1], i.e., increase D[a – 1] by val*f, and decrease D[b] by val*f.
- Now update the main array using the difference array. Update A[0] to D[0] and print it.
- For rest of the elements, Update A[i] by (A[i-1] + D[i]).
- Print the resultant array after the above steps.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function that creates a difference// array D[] for A[]vector<int> initializeDiffArray( vector<int>& A){ int N = A.size(); // Stores the difference array vector<int> D(N + 1); D[0] = A[0], D[N] = 0; // Update difference array D[] for (int i = 1; i < N; i++) D[i] = A[i] - A[i - 1]; // Return difference array return D;}// Function that performs the range// update queriesvoid update(vector<int>& D, int l, int r, int x){ // Update the ends of the range D[l] += x; D[r + 1] -= x;}// Function that perform all query// once with modified update Callvoid UpdateDiffArray(vector<int>& DiffArray, int Start, int End, int Val, int Freq){ // For range update, difference // array is modified update(DiffArray, Start - 1, End - 1, Val * Freq);}// Function to take queriesvoid queriesInput(vector<int>& DiffArray, int Q[][4], int M){ // Traverse the query for (int i = 0; i < M; i++) { // Function Call for updates UpdateDiffArray(DiffArray, Q[i][0], Q[i][1], Q[i][2], Q[i][3]); }}// Function to updates the array// using the difference arrayvoid UpdateArray(vector<int>& A, vector<int>& D){ // Traverse the array A[] for (int i = 0; i < A.size(); i++) { // 1st Element if (i == 0) { A[i] = D[i]; } // A[0] or D[0] decides values // of rest of the elements else { A[i] = D[i] + A[i - 1]; } }}// Function that prints the arrayvoid PrintArray(vector<int>& A){ // Print the element for (int i = 0; i < A.size(); i++) { cout << A[i] << " "; } return;}// Function that print the array// after performing all queriesvoid printAfterUpdate(vector<int>& A, int Q[][4], int M){ // Create and fill difference // array for range updates vector<int> DiffArray = initializeDiffArray(A); queriesInput(DiffArray, Q, M); // Now update Array A using // Difference Array UpdateArray(A, DiffArray); // Print updated Array A // after M queries PrintArray(A);}// Driver Codeint main(){ // N = Array size, M = Queries int N = 3, M = 3; // Given array A[] vector<int> A{ 1, 2, 3 }; // Queries int Q[][4] = { { 1, 2, 1, 4 }, { 1, 3, 2, 3 }, { 2, 3, 4, 5 } }; // Function Call printAfterUpdate(A, Q, M); return 0;} |
Java
// Java program for the // above approachimport java.util.*;class GFG{ // N = Array size, // M = Queriesstatic int N = 3, M = 3; static int []A = new int[N];//Stores the difference arraystatic int []D = new int[N + 1]; // Function that creates // a difference array D[] // for A[]static void initializeDiffArray(){ D[0] = A[0]; D[N] = 0; // Update difference array D[] for (int i = 1; i < N; i++) D[i] = A[i] - A[i - 1];}// Function that performs // the range update queriesstatic void update(int l, int r, int x){ // Update the ends // of the range D[l] += x; D[r + 1] -= x;}// Function that perform all query// once with modified update Callstatic void UpdateDiffArray(int Start, int End, int Val, int Freq){ // For range update, difference // array is modified update(Start - 1, End - 1, Val * Freq);}// Function to take queriesstatic void queriesInput( int Q[][]){ // Traverse the query for (int i = 0; i < M; i++) { // Function Call for updates UpdateDiffArray(Q[i][0], Q[i][1], Q[i][2], Q[i][3]); }}// Function to updates the array// using the difference arraystatic void UpdateArray(){ // Traverse the array A[] for (int i = 0; i < N; i++) { // 1st Element if (i == 0) { A[i] = D[i]; } // A[0] or D[0] decides // values of rest of // the elements else { A[i] = D[i] + A[i - 1]; } }}// Function that prints // the arraystatic void PrintArray(){ // Print the element for (int i = 0; i < N; i++) { System.out.print(A[i] + i + 1 + " "); } return;}// Function that print the array// after performing all queriesstatic void printAfterUpdate(int []A, int Q[][], int M){ // Create and fill difference // array for range updates initializeDiffArray(); queriesInput( Q); // Now update Array // A using Difference // Array UpdateArray(); // Print updated Array A // after M queries PrintArray();}// Driver Codepublic static void main(String[] args){ // Given array A[] int []A = {1, 2, 3}; // Queries int [][]Q = {{1, 2, 1, 4}, {1, 3, 2, 3}, {2, 3, 4, 5}}; // Function Call printAfterUpdate(A, Q, M);}}// This code is contributed by gauravrajput1 |
Python3
# Python3 program for the above approach# Function that creates a difference# array D[] for A[]def initializeDiffArray(A): N = len(A) # Stores the difference array D = [0] * (N + 1) D[0] = A[0] D[N] = 0 # Update difference array D[] for i in range(1, N): D[i] = A[i] - A[i - 1] # Return difference array return D# Function that performs the range# update queriesdef update(D, l, r, x): # Update the ends of the range D[l] += x D[r + 1] -= x# Function that perform all query# once with modified update Calldef UpdateDiffArray(DiffArray, Start, End, Val, Freq): # For range update, difference # array is modified update(DiffArray, Start - 1, End - 1, Val * Freq)# Function to take queriesdef queriesInput(DiffArray, Q, M): # Traverse the query for i in range(M): # Function Call for updates UpdateDiffArray(DiffArray, Q[i][0], Q[i][1], Q[i][2], Q[i][3])# Function to updates the array# using the difference arraydef UpdateArray(A, D): # Traverse the array A[] for i in range(len(A)): # 1st Element if (i == 0): A[i] = D[i] # A[0] or D[0] decides values # of rest of the elements else: A[i] = D[i] + A[i - 1]# Function that prints the arraydef PrintArray(A): # Print the element for i in range(len(A)): print(A[i], end = " ") return# Function that print the array# after performing all queriesdef printAfterUpdate(A, Q, M): # Create and fill difference # array for range updates DiffArray = initializeDiffArray(A) queriesInput(DiffArray, Q, M) # Now update Array A using # Difference Array UpdateArray(A, DiffArray) # Print updated Array A # after M queries PrintArray(A)# Driver Codeif __name__ == '__main__': # N = Array size, M = Queries N = 3 M = 3 # Given array A[] A = [ 1, 2, 3 ] # Queries Q = [ [ 1, 2, 1, 4 ], [ 1, 3, 2, 3 ], [ 2, 3, 4, 5 ] ] # Function call printAfterUpdate(A, Q, M)# This code is contributed by mohit kumar 29 |
C#
// C# program for the// above approachusing System;class GFG{// N = Array size,// M = Queriesstatic int N = 3, M = 3;static int[] A = new int[N];// Stores the difference arraystatic int[] D = new int[N + 1];// Function that creates// a difference array D[]// for A[]static void initializeDiffArray(){ D[0] = A[0]; D[N] = 0; // Update difference array D[] for (int i = 1; i < N; i++) D[i] = A[i] - A[i - 1];}// Function that performs// the range update queriesstatic void update(int l, int r, int x){ // Update the ends // of the range D[l] += x; D[r + 1] -= x;}// Function that perform all query// once with modified update Callstatic void UpdateDiffArray(int Start, int End, int Val, int Freq){ // For range update, difference // array is modified update(Start - 1, End - 1, Val * Freq);}// Function to take queriesstatic void queriesInput(int[, ] Q){ // Traverse the query for (int i = 0; i < M; i++) { // Function Call for updates UpdateDiffArray(Q[i, 0], Q[i, 1], Q[i, 2], Q[i, 3]); }}// Function to updates the array// using the difference arraystatic void UpdateArray(){ // Traverse the array A[] for (int i = 0; i < N; i++) { // 1st Element if (i == 0) { A[i] = D[i]; } // A[0] or D[0] decides // values of rest of // the elements else { A[i] = D[i] + A[i - 1]; } }}// Function that prints// the arraystatic void PrintArray(){ // Print the element for (int i = 0; i < N; i++) { Console.Write(A[i] + i + 1 + " "); } return;}// Function that print the array// after performing all queriesstatic void printAfterUpdate(int[] A, int[, ] Q, int M){ // Create and fill difference // array for range updates initializeDiffArray(); queriesInput(Q); // Now update Array // A using Difference // Array UpdateArray(); // Print updated Array A // after M queries PrintArray();}// Driver Codepublic static void Main(String[] args){ // Given array A[] int[] A = {1, 2, 3}; // Queries int[, ] Q = {{1, 2, 1, 4}, {1, 3, 2, 3}, {2, 3, 4, 5}}; // Function Call printAfterUpdate(A, Q, M);}// This code is contributed by Chitranayal |
Javascript
<script>// Javascript program to implement// the above approach// N = Array size, // M = Querieslet N = 3, M = 3; let A = new Array(N).fill(0);//Stores the difference arraylet D = new Array(N+1).fill(0); // Function that creates // a difference array D[] // for A[]function initializeDiffArray(){ D[0] = A[0]; D[N] = 0; // Update difference array D[] for (let i = 1; i < N; i++) D[i] = A[i] - A[i - 1];}// Function that performs // the range update queriesfunction update(l, r, x){ // Update the ends // of the range D[l] += x; D[r + 1] -= x;}// Function that perform all query// once with modified update Callfunction UpdateDiffArray(Start, End, Val, Freq){ // For range update, difference // array is modified update(Start - 1, End - 1, Val * Freq);}// Function to take queriesfunction queriesInput(Q){ // Traverse the query for (let i = 0; i < M; i++) { // Function Call for updates UpdateDiffArray(Q[i][0], Q[i][1], Q[i][2], Q[i][3]); }}// Function to updates the array// using the difference arrayfunction UpdateArray(){ // Traverse the array A[] for (let i = 0; i < N; i++) { // 1st Element if (i == 0) { A[i] = D[i]; } // A[0] or D[0] decides // values of rest of // the elements else { A[i] = D[i] + A[i - 1]; } }}// Function that print// the arrayfunction PrintArray(){ // Print the element for (let i = 0; i < N; i++) { document.write(A[i] + i + 1 + " "); } return;}// Function that print the array// after performing all queriesfunction printAfterUpdate(A, Q, M){ // Create and fill difference // array for range updates initializeDiffArray(); queriesInput( Q); // Now update Array // A using Difference // Array UpdateArray(); // Print updated Array A // after M queries PrintArray();}// Driver Code // Given array A[] let a = [1, 2, 3]; // Queries let Q = [[1, 2, 1, 4], [1, 3, 2, 3], [2, 3, 4, 5]]; // Function Call printAfterUpdate(a, Q, M); </script> |
Output
11 32 29
Time Complexity: O(N + M)
Auxiliary Space: O(N)
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