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Maximum number of operations required such that no pairs from a Matrix overlap
  • Last Updated : 18 Aug, 2020

Given an integer N followed by a matrix V[][] consisting of pairs {X, Y} in ascending order of X, the task is to for each pair given in ascending order of X, the following operations can be performed: 

  • Convert the pair {X, Y} to {X – Y, X}
  • Convert the pair {X, Y} to {X, X+Y}
  • Change the pair as {X, X}

The task is to find the count of addition and subtraction operations required such that no two pairs overlap.

Examples:

Input: N = 5, V[] = {{1, 2} {2, 1} {5, 10} {10, 9} {19, 1}} 
Output:
Explanation: 
{1, 2}: Operation 1 modifies pair to {-1, 1}. 
{2, 1}: Operation 2 modifies pair to {2, 3}. 
{5, 10}: Operation 3 modifies pair to {5, 5} 
{10, 9}: Operation 3 modifies pair to {10, 10} 
{19, 1}: Operation 2 modifies pair to {19, 20}. 
Therefore, none of the pairs overlap. Hence, the count of addition and subtraction operations required is 3.

Input: N = 3, V[][] = {{10, 20} {15, 10} {20, 16}} 
Output:
 



Approach: 
The main idea is to observe that, the answer, in any case, will not exceed N, since any of the three operations cannot be applied twice on a pair. Follow the steps below to solve the problem:

  • Always choose Operation 1 for the first pair, since X is minimum for the first pair.
  • Always choose Operation 2 for the last pair, since X is maximum for the last pair.
  • For the remaining pairs, check if applying Operation 1 violates rules or not. If it does not violate the rules then it will always maximize the result. Otherwise check for Operation 2. Increase count if any of the two operations is applicable.
  • If both the rules are not applicable, perform operation 3.
  • Finally, print count.

Below is the implementation of the above approach:

C++




// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find maximum count of operations
int find_max(vector<pair<int, int> > v, int n)
{
    // Initialize count by 0
    int count = 0;
  
    if (n >= 2)
        count = 2;
  
    else
        count = 1;
  
    // Iterate over remaining pairs
    for (int i = 1; i < n - 1; i++) {
  
        // Check if first operation
        // is applicable
        if (v[i - 1].first
            < (v[i].first - v[i].second))
            count++;
  
        // Check if 2nd operation is applicable
        else if (v[i + 1].first
                 > (v[i].first + v[i].second)) {
            count++;
            v[i].first = v[i].first + v[i].second;
        }
  
        // Otherwise
        else
            continue;
    }
  
    // Return  the count of operations
    return count;
}
  
// Driver Code
int main()
{
    int n = 3;
    vector<pair<int, int> > v;
  
    v.push_back({ 10, 20 });
    v.push_back({ 15, 10 });
    v.push_back({ 20, 16 });
  
    cout << find_max(v, n);
  
    return 0;
}

Java




// Java program to implement
// the above approach
import java.util.*;
  
class GFG{
      
static class pair
    int first, second; 
    public pair(int first, int second) 
    
        this.first = first; 
        this.second = second; 
    
  
// Function to find maximum count of operations
static int find_max(Vector<pair> v, int n)
{
    // Initialize count by 0
    int count = 0;
  
    if (n >= 2)
        count = 2;
    else
        count = 1;
  
    // Iterate over remaining pairs
    for(int i = 1; i < n - 1; i++)
    {
          
        // Check if first operation
        // is applicable
        if (v.get(i - 1).first < 
           (v.get(i).first - v.get(i).second))
            count++;
  
        // Check if 2nd operation is applicable
        else if (v.get(i + 1).first > 
                (v.get(i).first + v.get(i).second)) 
        {
            count++;
            v.get(i).first = v.get(i).first + 
                             v.get(i).second;
        }
  
        // Otherwise
        else
            continue;
    }
  
    // Return the count of operations
    return count;
}
  
// Driver Code
public static void main(String[] args)
{
    int n = 3;
    Vector<pair> v = new Vector<>();
  
    v.add(new pair(10, 20));
    v.add(new pair(15, 10));
    v.add(new pair(20, 16));
  
    System.out.print(find_max(v, n));
}
}
  
// This code is contributed by 29AjayKumar 

Python3




# Python3 program to implement
# the above approach
  
# Function to find maximum count of
# operations
def find_max(v, n):
  
    # Initialize count by 0
    count = 0
  
    if(n >= 2):
        count = 2
    else:
        count = 1
  
    # Iterate over remaining pairs
    for i in range(1, n - 1):
  
        # Check if first operation
        # is applicable
        if(v[i - 1][0] > (v[i][0] +
                          v[i][1])):
            count += 1
  
        # Check if 2nd operation is applicable
        elif(v[i + 1][0] > (v[i][0] +
                            v[i][1])):
            count += 1
            v[i][0] = v[i][0] + v[i][1]
  
        # Otherwise
        else:
            continue
  
    # Return the count of operations
    return count
  
# Driver Code
n = 3
v = []
  
v.append([ 10, 20 ])
v.append([ 15, 10 ])
v.append([ 20, 16 ])
  
print(find_max(v, n))
  
# This code is contributed by Shivam Singh

C#




// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
  
class GFG{
      
class pair
    public int first, second; 
    public pair(int first, int second) 
    
        this.first = first; 
        this.second = second; 
    
  
// Function to find maximum count of operations
static int find_max(List<pair> v, int n)
{
      
    // Initialize count by 0
    int count = 0;
  
    if (n >= 2)
        count = 2;
    else
        count = 1;
  
    // Iterate over remaining pairs
    for(int i = 1; i < n - 1; i++)
    {
          
        // Check if first operation
        // is applicable
        if (v[i - 1].first < 
           (v[i].first - v[i].second))
            count++;
  
        // Check if 2nd operation is applicable
        else if (v[i + 1].first > 
                (v[i].first + v[i].second)) 
        {
            count++;
            v[i].first = v[i].first + 
                         v[i].second;
        }
  
        // Otherwise
        else
            continue;
    }
  
    // Return the count of operations
    return count;
}
  
// Driver Code
public static void Main(String[] args)
{
    int n = 3;
    List<pair> v = new List<pair>();
  
    v.Add(new pair(10, 20));
    v.Add(new pair(15, 10));
    v.Add(new pair(20, 16));
  
    Console.Write(find_max(v, n));
}
}
  
// This code is contributed by 29AjayKumar
Output: 
2

Time Complexity: O(N) 
Auxiliary Space: O(1)
 

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