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Check if the count of inversions of two given types on an Array are equal or not

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  • Last Updated : 13 May, 2021
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Given an array a[] on which, the following two types of inversions are performed:

  • Count of pairs of indices (i, j) such that A[i] > A[j] and i < j
  • Count of pairs of indices (i, j) such that A[i] > A[j] and j = i + 1

The task is to check if the count of both the inversions is equal or not. If they are equal, print “Yes”. Otherwise, print “No”.

Examples:

Input: a[] = {1, 0, 2} 
Output: Yes 
Explanation: 
Count of inversion of Type 1 = 1 [(i, j) : (0, 1)] 
Count of inversion of Type 2 = 1 [(i, j) : (0, 1)]
Input: a[] = {1, 2, 0} 
Output: No 
Explanation: 
Count of inversion of Type 1 = 2 [(i, j) : (0, 2);(1, 2)] 
Count of inversion of Type 2 = 1 [(i, j) : (1, 2)]

Approach: 
To solve the problem, the difference between the two inversions need to be understood:

  • For Type 2, if j = 5, then i can only be 4 as j = i + 1
  • For Type 1, if j = 5, then i can be from 0 to 4, as i is less than j.
  • Therefore, the inversion of Type 1 is basically an inversion of Type 2 summed up with all pair of indices(i, j) with i which is less than (j – 1) and a[i] > a[j].
  • So, for any index j, the task is to check if there is index i, which is less than j – 1 and a[i] > a[j]. If any such pair of indices (i, j) is found, then print “No“. Otherwise, print “Yes“.

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 check if the
// count of inversion of two
// types are same or not
bool solve(int a[], int n)
{
    int mx = INT_MIN;
 
    for (int j = 1; j < n; j++) {
 
        // If maximum value is found
        // to be greater than a[j],
        // then that pair of indices
        // (i, j) will add extra value
        // to inversion of Type 1
        if (mx > a[j])
 
            return false;
 
        // Update max
        mx = max(mx, a[j - 1]);
    }
 
    return true;
}
 
// Driver code
int main()
{
 
    int a[] = { 1, 0, 2 };
 
    int n = sizeof(a) / sizeof(a[0]);
 
    bool possible = solve(a, n);
 
    if (possible)
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
 
    return 0;
}

Java




// Java program to implement
// the above approach
import java.io.*;
 
class GFG{
     
// Function to check if the
// count of inversion of two
// types are same or not
static boolean solve(int a[], int n)
{
    int mx = Integer.MIN_VALUE;
 
    for(int j = 1; j < n; j++)
    {
         
        // If maximum value is found
        // to be greater than a[j],
        // then that pair of indices
        // (i, j) will add extra value
        // to inversion of Type 1
        if (mx > a[j])
            return false;
 
        // Update max
        mx = Math.max(mx, a[j - 1]);
    }
    return true;
}
     
// Driver code
public static void main (String[] args)
{
    int a[] = { 1, 0, 2 };
 
    int n = a.length;
     
    boolean possible = solve(a, n);
     
    if (possible)
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by offbeat

Python3




# Python3 program to implement 
# the above approach
import sys
 
# Function to check if the
# count of inversion of two
# types are same or not
def solve(a, n):
     
    mx = -sys.maxsize - 1
 
    for j in range(1, n):
 
        # If maximum value is found
        # to be greater than a[j],
        # then that pair of indices
        # (i, j) will add extra value
        # to inversion of Type 1
        if (mx > a[j]):
            return False
 
        # Update max
        mx = max(mx, a[j - 1])
     
    return True
 
# Driver code
a = [ 1, 0, 2 ]
 
n = len(a)
 
possible = solve(a, n)
 
if (possible != 0):
    print("Yes")
else:
    print("No")
 
# This code is contributed by sanjoy_62

C#




// C# program to implement 
// the above approach
using System;
     
class GFG{
     
// Function to check if the
// count of inversion of two
// types are same or not
static bool solve(int[] a, int n)
{
    int mx = Int32.MinValue;
     
    for(int j = 1; j < n; j++)
    {
             
        // If maximum value is found
        // to be greater than a[j],
        // then that pair of indices
        // (i, j) will add extra value
        // to inversion of Type 1
        if (mx > a[j])
            return false;
     
        // Update max
        mx = Math.Max(mx, a[j - 1]);
    }
    return true;
}
         
// Driver code
public static void Main ()
{
    int[] a = { 1, 0, 2 };
    int n = a.Length;
         
    bool possible = solve(a, n);
         
    if (possible)
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed by sanjoy_62

Javascript




<script>
// Javascript implementation of the above approach
 
// Function to check if the
// count of inversion of two
// types are same or not
function solve(a, n)
{
    let mx = 0;
 
    for(let j = 1; j < n; j++)
    {
         
        // If maximum value is found
        // to be greater than a[j],
        // then that pair of indices
        // (i, j) will add extra value
        // to inversion of Type 1
        if (mx > a[j])
            return false;
 
        // Update max
        mx = Math.max(mx, a[j - 1]);
    }
    return true;
}
   
  // Driver Code
     
    let a = [ 1, 0, 2 ];
 
    let n = a.length;
     
    let possible = solve(a, n);
     
    if (possible)
        document.write("Yes");
    else
        document.write("No");
 
</script>

Output: 

Yes

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


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