Queries to find minimum swaps required to sort given array with updates

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:

  1. For each query {x, y}, store the value arr[x]+y in a variable newElement.
  2. 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.
  3. If no such value can be found, print x, otherwise let that value be at index j.
  4. Print the absolute difference between the index i and the index j.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the position of
// the given value using binary search
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 sorted
void 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 Code
int 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;
}

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Java

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// Java program for the
// above approach
import java.util.*;
class GFG{
 
// Function to return the position of
// the given value using binary search
static 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 sorted
static 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 Code
public 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

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Python3

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# Python3 program for the above approach
 
# Function to return the position of
# the given value using binary search
def 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 sorted
def 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 Code
if __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

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C#

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// C# program for the
// above approach
using System;
using System.Collections.Generic;
class GFG{
 
// Function to return the position of
// the given value using binary search
static 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 sorted
static 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 Code
public 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

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Output: 

0 4








 

Time Complexity: O(Q*N*log N)
Auxiliary Space: O(N)

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