Print the Array of size N containing values in range [0, M) after Q query updates
Given array arr[] of size N containing cyclic variables having states from 0 to (M – 1) (i.e when incremented from M-1 it goes to 0). The task is to fulfill the Q queries of which are of any of the following two types:
- 1st Type: 1 L R K – increment all the values in range of indices [L, R], K times cyclically.
- 2nd Type: 2 L R – print the updated values in range of indices [L, R].
Examples:
Input: arr[] = {2, 2, 7, 2, 5}, Q = 5, M = 8
queries[][] = {{1, 0, 3, 4},
{1, 4, 4, 2},
{1, 0, 0, 7},
{2, 1, 3},
{2, 3, 3}}
Output: {6, 3, 6}, {6}
Explanation: The states after performing each operation are:
After 1st: {6, 6, 3, 6, 5}
After 2nd: {6, 6, 3, 6, 7}
After 3rd: {5, 6, 3, 6, 7}
So in 4th query elements from index 1 to 3 and in 5th query element in index 3 are printed.Input: arr[] = [2, 3, 4, 5], Q = 3, M = 6
queries[][] = {{1, 0, 0, 3},
{1, 1, 2, 2},
{1, 3, 3, 1},
{2, 0, 3}}
Output: {5, 5, 1, 1}
Approach: The problem can be solved using greedy approach. Follow the steps mentioned below to implement the approach:
- When a query is of first type update all the values in range [L, R], K times.
- When a query is of second type then print the value in range [L, R].
Below is the implementation of the above approach.
C++14
// C++ code to implement above approach#include <bits/stdc++.h>using namespace std;// Function to implement the queriesvoid update(int arr[], int N, int M, int Q, vector<vector<int> >& queries){ // Loop to implement the queries for (int i = 0; i < Q; i++) { if (queries[i][0] == 1) { for (int j = queries[i][1]; j <= queries[i][2]; j++) arr[j] = (arr[j] + queries[i][3]) % M; } else if (queries[i][0] == 2) { for (int j = queries[i][1]; j <= queries[i][2]; j++) cout << arr[j] << " "; cout << endl; } }}// Driver's codeint main(){ int N = 5, M = 8, Q = 5; int arr[] = { 2, 2, 7, 2, 5 }; vector<vector<int> > queries(Q); queries[0] = { 1, 0, 3, 4 }; queries[1] = { 1, 4, 4, 2 }; queries[2] = { 1, 0, 0, 7 }; queries[3] = { 2, 1, 3 }; queries[4] = { 2, 3, 3 }; update(arr, N, M, Q, queries); return 0;} |
Java
// Java code to implement the above approachimport java.io.*;class GFG { // Function to implement the queries public static void update(int[] arr, int N, int M, int Q, int[][] queries) { // Loop to implement the queries for (int i = 0; i < Q; i++) { if (queries[i][0] == 1) { for (int j = queries[i][1]; j <= queries[i][2]; j++) arr[j] = (arr[j] + queries[i][3]) % M; } else if (queries[i][0] == 2) { for (int j = queries[i][1]; j <= queries[i][2]; j++) System.out.print(arr[j]+ " "); System.out.println(); } } } // Driver's code public static void main (String[] args) { int N = 5, M = 8, Q = 5; int[] arr = { 2, 2, 7, 2, 5 }; int[][] queries = new int[][] { new int[] { 1, 0, 3, 4 }, new int[] { 1, 4, 4, 2 }, new int[] { 1, 0, 0, 7 }, new int[] { 2, 1, 3 }, new int[] { 2, 3, 3 } }; update(arr, N, M, Q, queries); }}// This code is contributed by Shubham Singh |
Python3
# Python3 code to implement above approach# Function to implement the queriesdef update(arr, N, M, Q, queries): # Loop to implement the queries for i in range(Q): if (queries[i][0] == 1): for j in range(queries[i][1], queries[i][2] + 1): arr[j] = (arr[j] + queries[i][3]) % M elif (queries[i][0] == 2): for j in range(queries[i][1], queries[i][2] + 1): print(arr[j], end = " ") print()# Driver codeif __name__ == "__main__": N = 5 M = 8 Q = 5 arr = [2, 2, 7, 2, 5] queries = [] queries.append([1, 0, 3, 4]) queries.append([1, 4, 4, 2]) queries.append([1, 0, 0, 7]) queries.append([2, 1, 3]) queries.append([2, 3, 3]) update(arr, N, M, Q, queries) # This code is contributed by ukasp |
C#
// C# code to implement the above approachusing System;public class GFG{ // Function to implement the queries public static void update(int[] arr, int N, int M, int Q, int[][] queries) { // Loop to implement the queries for (int i = 0; i < Q; i++) { if (queries[i][0] == 1) { for (int j = queries[i][1]; j <= queries[i][2]; j++) arr[j] = (arr[j] + queries[i][3]) % M; } else if (queries[i][0] == 2) { for (int j = queries[i][1]; j <= queries[i][2]; j++) Console.Write(arr[j]+ " "); Console.WriteLine(); } } } // Driver's code public static void Main() { int N = 5, M = 8, Q = 5; int[] arr = { 2, 2, 7, 2, 5 }; int[][] queries = new int[][] { new int[] { 1, 0, 3, 4 }, new int[] { 1, 4, 4, 2 }, new int[] { 1, 0, 0, 7 }, new int[] { 2, 1, 3 }, new int[] { 2, 3, 3 } }; update(arr, N, M, Q, queries); }}// This code is contributed by Shubham Singh |
Javascript
<script> // JavaScript code for the above approach // Function to implement the queries function update(arr, N, M, Q, queries) { // Loop to implement the queries for (let i = 0; i < Q; i++) { if (queries[i][0] == 1) { for (let j = queries[i][1]; j <= queries[i][2]; j++) arr[j] = (arr[j] + queries[i][3]) % M; } else if (queries[i][0] == 2) { for (let j = queries[i][1]; j <= queries[i][2]; j++) document.write(arr[j] + " "); document.write('<br>') } } } // Driver's code let N = 5, M = 8, Q = 5; let arr = [2, 2, 7, 2, 5]; let queries = new Array(Q); queries[0] = [1, 0, 3, 4]; queries[1] = [1, 4, 4, 2]; queries[2] = [1, 0, 0, 7]; queries[3] = [2, 1, 3]; queries[4] = [2, 3, 3]; update(arr, N, M, Q, queries); // This code is contributed by Potta Lokesh </script> |
6 3 6 6
Time Complexity: O(Q*N)
Auxiliary Space: O(1)
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