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

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 the 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 an 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   ` `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);  ` ` `  `    ``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 `

Output:

```Yes
```

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

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Improved By : offbeat, sanjoy_62