Given a sorted array arr[] of size N and an array Q[][] having queries in the form of {x, y}. In each query {x, y}, update the given array by incrementing the value arr[x] by y. The task is to find the minimum number of swaps required to sort the array obtained after performing each query in the given array individually.
Examples:
Input: arr[] = {2, 3, 4, 5, 6}, Q[][] = {{2, 8}, {3, 1}}
Output: 2 0
Explanation:
Following are the number of swaps required for each query:
Query 1: Increment arr[2] by 8. Therefore, arr[] = {2, 3, 12, 5, 6}.
To make this array sorted, shift 12 right by 2 positions.
Now arr[] = {2, 3, 5, 6, 12}. Hence, it requires 2 swaps.
Query 2: Increment arr[3] by 1. Therefore, arr[] = {2, 3, 4, 6, 6}.
The array is still sorted. Hence, it requires 0 swaps.Input: arr[] = {2, 3, 4, 5, 6}, Q[][] = {{0, -1}, {4, -11}};
Output: 0 4
Explanation:
Following are the number of swaps required for each query:
Query 1: Increment arr[0] by -1. Therefore, arr[] = {1, 3, 4, 5, 6}.
The array is still sorted. Hence, it requires 0 swaps.
Query 2: Increment arr[4] by -11. Therefore, arr[] = {2, 3, 4, 5, -5}.
To make this array sorted, shift -5 left by 4 positions.
Now arr[] = {-5, 2, 3, 4, 5}. Hence, it requires 4 swaps.
Naive Approach: The simplest approach is to update the given array by incrementing the value arr[x] by y for each query {x, y}. After that, traverse the updated array and swap arr[x] to the right while arr[x] is greater than arr[x+1], incrementing x each time then swap arr[x] to the left while arr[x] is smaller than arr[x-1], decrementing x each time. Print the absolute difference between the initial and the final value of x.
Time Complexity: O(Q*N2) where N is the length of the given array and Q is the total number of queries.
Auxiliary Space: O(N)
Efficient Approach: The idea is to use Binary Search to find the minimum number of swaps required to make the given array sorted after each query. Follow the below steps to solve the problem:
- For each query {x, y}, store the value arr[x]+y in a variable newElement.
- Using Binary Search, find the index of the value present in the given array that is just smaller than or equal to the value newElement.
- If no such value can be found, print x, otherwise let that value be at index j.
- Print the absolute difference between the index i and the index j.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to return the position of// the given value using binary searchint computePos(int arr[], int n, int value){ // Return 0 if every value is // greater than the given value if (value < arr[0]) return 0; // Return N-1 if every value is // smaller than the given value if (value > arr[n - 1]) return n - 1; // Perform Binary Search int start = 0; int end = n - 1; // Iterate till start < end while (start < end) { // Find the mid int mid = (start + end + 1) / 2; // Update start and end if (arr[mid] >= value) end = mid - 1; else start = mid; } // Return the position of // the given value return start;}// Function to return the number of// make the array sortedvoid countShift(int arr[], int n, vector<vector<int> >& queries){ for (auto q : queries) { // Index x to update int index = q[0]; // Increment value by y int update = q[1]; // Set newElement equals // to x + y int newElement = arr[index] + update; // Compute the new index int newIndex = computePos( arr, n, newElement); // Print the minimum number // of swaps cout << abs(newIndex - index) << " "; }}// Driver Codeint main(){ // Given array arr[] int arr[] = { 2, 3, 4, 5, 6 }; int N = sizeof(arr) / sizeof(arr[0]); // Given Queries vector<vector<int> > queries = { { 0, -1 }, { 4, -11 } }; // Function Call countShift(arr, N, queries); return 0;} |
Java
// Java program for the // above approachimport java.util.*;class GFG{// Function to return the position of// the given value using binary searchstatic int computePos(int arr[], int n, int value){ // Return 0 if every value is // greater than the given value if (value < arr[0]) return 0; // Return N-1 if every value is // smaller than the given value if (value > arr[n - 1]) return n - 1; // Perform Binary Search int start = 0; int end = n - 1; // Iterate till start < end while (start < end) { // Find the mid int mid = (start + end + 1) / 2; // Update start and end if (arr[mid] >= value) end = mid - 1; else start = mid; } // Return the position of // the given value return start;}// Function to return the number of// make the array sortedstatic void countShift(int arr[], int n, Vector<Vector<Integer> > queries){ for (Vector<Integer> q : queries) { // Index x to update int index = q.get(0); // Increment value by y int update = q.get(1); // Set newElement equals // to x + y int newElement = arr[index] + update; // Compute the new index int newIndex = computePos(arr, n, newElement); // Print the minimum number // of swaps System.out.print(Math.abs(newIndex - index) + " "); }}// Driver Codepublic static void main(String[] args){ // Given array arr[] int arr[] = {2, 3, 4, 5, 6}; int N = arr.length; // Given Queries Vector<Vector<Integer> > queries = new Vector<>(); Vector<Integer> v = new Vector<>(); Vector<Integer> v1 = new Vector<>(); v.add(0); v.add(-1); queries.add(v); v1.add(4); v1.add(-11); queries.add(v1); // Function Call countShift(arr, N, queries);}}// This code is contributed by Princi Singh |
Python3
# Python3 program for the above approach# Function to return the position of# the given value using binary searchdef computePos(arr, n, value): # Return 0 if every value is # greater than the given value if (value < arr[0]): return 0 # Return N-1 if every value is # smaller than the given value if (value > arr[n - 1]): return n - 1 # Perform Binary Search start = 0 end = n - 1 # Iterate till start < end while (start < end): # Find the mid mid = (start + end + 1) // 2 # Update start and end if (arr[mid] >= value): end = mid - 1 else: start = mid # Return the position of # the given value return start# Function to return the number of# make the array sorteddef countShift(arr, n, queries): for q in queries: # Index x to update index = q[0] # Increment value by y update = q[1] # Set newElement equals # to x + y newElement = arr[index] + update # Compute the new index newIndex = computePos(arr, n, newElement) # Print the minimum number # of swaps print(abs(newIndex - index), end = " ")# Driver Codeif __name__ == '__main__': # Given array arr[] arr = [ 2, 3, 4, 5, 6 ] N = len(arr) # Given Queries queries = [ [ 0, -1 ], [4, -11 ] ] # Function Call countShift(arr, N, queries)# This code is contributed by mohit kumar 29 |
C#
// C# program for the // above approachusing System;using System.Collections.Generic;class GFG{// Function to return the position of// the given value using binary searchstatic int computePos(int []arr, int n, int value){ // Return 0 if every value is // greater than the given value if (value < arr[0]) return 0; // Return N-1 if every value is // smaller than the given value if (value > arr[n - 1]) return n - 1; // Perform Binary Search int start = 0; int end = n - 1; // Iterate till start // < end while (start < end) { // Find the mid int mid = (start + end + 1) / 2; // Update start and end if (arr[mid] >= value) end = mid - 1; else start = mid; } // Return the position of // the given value return start;}// Function to return the number of// make the array sortedstatic void countShift(int []arr, int n, List<List<int> > queries){ foreach (List<int> q in queries) { // Index x to update int index = q[0]; // Increment value by y int update = q[1]; // Set newElement equals // to x + y int newElement = arr[index] + update; // Compute the new index int newIndex = computePos(arr, n, newElement); // Print the minimum number // of swaps Console.Write(Math.Abs(newIndex - index) + " "); }}// Driver Codepublic static void Main(String[] args){ // Given array []arr int []arr = {2, 3, 4, 5, 6}; int N = arr.Length; // Given Queries List<List<int> > queries = new List<List<int>>(); List<int> v = new List<int>(); List<int> v1 = new List<int>(); v.Add(0); v.Add(-1); queries.Add(v); v1.Add(4); v1.Add(-11); queries.Add(v1); // Function Call countShift(arr, N, queries);}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program for the above approach// Function to return the position of// the given value using binary searchfunction computePos(arr, n, value){ // Return 0 if every value is // greater than the given value if (value < arr[0]) return 0; // Return N-1 if every value is // smaller than the given value if (value > arr[n - 1]) return n - 1; // Perform Binary Search var start = 0; var end = n - 1; // Iterate till start < end while (start < end) { // Find the mid var mid = parseInt((start + end + 1) / 2); // Update start and end if (arr[mid] >= value) end = mid - 1; else start = mid; } // Return the position of // the given value return start;}// Function to return the number of// make the array sortedfunction countShift(arr, n, queries){ for(var i =0; i< queries.length; i++) { // Index x to update var index = queries[i][0]; // Increment value by y var update = queries[i][1]; // Set newElement equals // to x + y var newElement = arr[index] + update; // Compute the new index var newIndex = computePos( arr, n, newElement); // Print the minimum number // of swaps document.write( Math.abs(newIndex - index) + " "); }}// Driver Code// Given array arr[]var arr = [ 2, 3, 4, 5, 6 ];var N = arr.length;// Given Queriesvar queries = [[ 0, -1 ], [ 4, -11 ]];// Function CallcountShift(arr, N, queries);</script> |
0 4
Time Complexity: O(Q*N*log N)
Auxiliary Space: O(N)
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