# Minimum operations required to modify the array such that parity of adjacent elements is different

Given an array A[], the task is to find the minimum number of operations required to convert the array into B[] such that for every index in B (except the last) parity(b[i]) != parity(b[i + 1]) where parity(x) = x % 3.

Below is the operation to be performed:

Any element from the set {1, 2} can be added to any element of the array.

Examples:

Input: A[] = {2, 1, 3, 0}
Output: 1
1 can be added to 0 in a single operation and the array becomes {2, 1, 3, 1}.

Input: A[] = {2, 2, 2, 2}
Output: 2

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: An optimal approach is to update the element which is in between two other elements so that it can be updated such that its parity can be different from the element previous to it and the element next to it in a single operation (as the number of operations needs to be minimized).
So, start traversing the array from A[1] to A[n – 2] and for every element A[i] such that parity(A[i]) == parity(A[i – 1]) or parity(A[i]) == parity(A[i + 1]), update A[i] such that its parity is different from both the numbers surrounding it and keep a track of the count of operations performed. Print the count in the end.

Below is the implementation of the above approach:

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the parity of a number ` `int` `parity(``int` `a) ` `{ ` `    ``return` `a % 3; ` `} ` ` `  `// Function to return the minimum ` `// number of operations required ` `int` `solve(``int` `array[], ``int` `size) ` `{ ` `    ``int` `operations = 0; ` `    ``for` `(``int` `i = 0; i < size - 1; i++) { ` ` `  `        ``// Operation needs to be performed ` `        ``if` `(parity(array[i]) == parity(array[i + 1])) { ` ` `  `            ``operations++; ` `            ``if` `(i + 2 < size) { ` ` `  `                ``// Parity of previous element ` `                ``int` `pari1 = parity(array[i]); ` ` `  `                ``// Parity of next element ` `                ``int` `pari2 = parity(array[i + 2]); ` ` `  `                ``// Update parity of current element to be other than ` `                ``// the parities of the pervious and the next number ` `                ``if` `(pari1 == pari2) { ` `                    ``if` `(pari1 == 0) ` `                        ``array[i + 1] = 1; ` `                    ``else` `if` `(pari1 == 1) ` `                        ``array[i + 1] = 0; ` `                    ``else` `                        ``array[i + 1] = 1; ` `                ``} ` `                ``else` `{ ` `                    ``if` `((pari1 == 0 && pari2 == 1) ` `                        ``|| (pari1 == 1 && pari2 == 0)) ` `                        ``array[i + 1] = 2; ` `                    ``if` `((pari1 == 1 && pari2 == 2) ` `                        ``|| (pari1 == 2 && pari2 == 1)) ` `                        ``array[i + 1] = 0; ` `                    ``if` `((pari1 == 2 && pari2 == 0) ` `                        ``|| (pari1 == 0 && pari2 == 2)) ` `                        ``array[i + 1] = 1; ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``return` `operations; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `array[] = { 2, 1, 3, 0 }; ` `    ``int` `size = ``sizeof``(array) / ``sizeof``(array[0]); ` `    ``cout << solve(array, size) << endl; ` ` `  `  ``return` `0; ` `} `

 `// Java implementation of the approach  ` `class` `GFG  ` `{  ` `     `  `    ``// Function to return the parity of a number  ` `    ``static` `int` `parity(``int` `a)  ` `    ``{  ` `        ``return` `a % ``3``;  ` `    ``}  ` ` `  `    ``// Function to return the minimum  ` `    ``// number of operations required  ` `    ``static` `int` `solve(``int` `[]array, ``int` `size)  ` `    ``{  ` `        ``int` `operations = ``0``;  ` `        ``for` `(``int` `i = ``0``; i < size - ``1``; i++)  ` `        ``{  ` `     `  `            ``// Operation needs to be performed  ` `            ``if` `(parity(array[i]) == parity(array[i + ``1``]))  ` `            ``{  ` `     `  `                ``operations++;  ` `                ``if` `(i + ``2` `< size)  ` `                ``{  ` `     `  `                    ``// Parity of previous element  ` `                    ``int` `pari1 = parity(array[i]);  ` `     `  `                    ``// Parity of next element  ` `                    ``int` `pari2 = parity(array[i + ``2``]);  ` `     `  `                    ``// Update parity of current  ` `                    ``// element to be other than  ` `                    ``// the parities of the pervious  ` `                    ``// and the next number  ` `                    ``if` `(pari1 == pari2)  ` `                    ``{  ` `                        ``if` `(pari1 == ``0``)  ` `                            ``array[i + ``1``] = ``1``;  ` `                        ``else` `if` `(pari1 == ``1``)  ` `                            ``array[i + ``1``] = ``0``;  ` `                        ``else` `                            ``array[i + ``1``] = ``1``;  ` `                    ``}  ` `                    ``else` `                    ``{  ` `                        ``if` `((pari1 == ``0` `&& pari2 == ``1``) ||  ` `                            ``(pari1 == ``1` `&& pari2 == ``0``))  ` `                            ``array[i + ``1``] = ``2``;  ` `                        ``if` `((pari1 == ``1` `&& pari2 == ``2``)  ` `                            ``|| (pari1 == ``2` `&& pari2 == ``1``))  ` `                            ``array[i + ``1``] = ``0``;  ` `                        ``if` `((pari1 == ``2` `&& pari2 == ``0``)  ` `                            ``|| (pari1 == ``0` `&& pari2 == ``2``))  ` `                            ``array[i + ``1``] = ``1``;  ` `                    ``}  ` `                ``}  ` `            ``}  ` `        ``}  ` `     `  `        ``return` `operations;  ` `    ``}  ` ` `  `    ``// Driver Code  ` `    ``public` `static` `void` `main (String arg[])  ` `    ``{  ` `        ``int` `[]array = { ``2``, ``1``, ``3``, ``0` `};  ` `        ``int` `size = array.length;  ` `        ``System.out.println(solve(array, size));  ` `    ``}  ` `}  ` ` `  `// This code is contributed by Ryuga `

 `# Python3 implementation of the approach  ` ` `  `# Function to return the parity  ` `# of a number  ` `def` `parity(a): ` `    ``return` `a ``%` `3` ` `  `# Function to return the minimum  ` `# number of operations required  ` `def` `solve(array, size):  ` ` `  `    ``operations ``=` `0` `    ``for` `i ``in` `range``(``0``, size ``-` `1``):  ` ` `  `        ``# Operation needs to be performed  ` `        ``if` `parity(array[i]) ``=``=` `parity(array[i ``+` `1``]):  ` ` `  `            ``operations ``+``=` `1` `            ``if` `i ``+` `2` `< size:  ` ` `  `                ``# Parity of previous element  ` `                ``pari1 ``=` `parity(array[i])  ` ` `  `                ``# Parity of next element  ` `                ``pari2 ``=` `parity(array[i ``+` `2``])  ` ` `  `                ``# Update parity of current element to ` `                ``# be other than the parities of the ` `                ``# previous and the next number  ` `                ``if` `pari1 ``=``=` `pari2:  ` `                    ``if` `pari1 ``=``=` `0``:  ` `                        ``array[i ``+` `1``] ``=` `1` `                    ``elif` `pari1 ``=``=` `1``:  ` `                        ``array[i ``+` `1``] ``=` `0` `                    ``else``: ` `                        ``array[i ``+` `1``] ``=` `1` `                 `  `                ``else``: ` `                    ``if` `((pari1 ``=``=` `0` `and` `pari2 ``=``=` `1``) ``or`  `                        ``(pari1 ``=``=` `1` `and` `pari2 ``=``=` `0``)):  ` `                        ``array[i ``+` `1``] ``=` `2` `                    ``if` `((pari1 ``=``=` `1` `and` `pari2 ``=``=` `2``) ``or`  `                        ``(pari1 ``=``=` `2` `and` `pari2 ``=``=` `1``)):  ` `                        ``array[i ``+` `1``] ``=` `0` `                    ``if` `((pari1 ``=``=` `2` `and` `pari2 ``=``=` `0``) ``and`  `                        ``(pari1 ``=``=` `0` `and` `pari2 ``=``=` `2``)): ` `                        ``array[i ``+` `1``] ``=` `1` ` `  `    ``return` `operations  ` ` `  `# Driver Code  ` `if` `__name__ ``=``=` `"__main__"``:  ` ` `  `    ``array ``=` `[``2``, ``1``, ``3``, ``0``]  ` `    ``size ``=` `len``(array)  ` `    ``print``(solve(array, size))  ` ` `  `# This code is contributed by Rituraj Jain `

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// Function to return the parity of a number ` `static` `int` `parity(``int` `a) ` `{ ` `    ``return` `a % 3; ` `} ` ` `  `// Function to return the minimum ` `// number of operations required ` `static` `int` `solve(``int` `[]array, ``int` `size) ` `{ ` `    ``int` `operations = 0; ` `    ``for` `(``int` `i = 0; i < size - 1; i++) ` `    ``{ ` ` `  `        ``// Operation needs to be performed ` `        ``if` `(parity(array[i]) == parity(array[i + 1])) ` `        ``{ ` ` `  `            ``operations++; ` `            ``if` `(i + 2 < size) ` `            ``{ ` ` `  `                ``// Parity of previous element ` `                ``int` `pari1 = parity(array[i]); ` ` `  `                ``// Parity of next element ` `                ``int` `pari2 = parity(array[i + 2]); ` ` `  `                ``// Update parity of current  ` `                ``// element to be other than ` `                ``// the parities of the pervious  ` `                ``// and the next number ` `                ``if` `(pari1 == pari2)  ` `                ``{ ` `                    ``if` `(pari1 == 0) ` `                        ``array[i + 1] = 1; ` `                    ``else` `if` `(pari1 == 1) ` `                        ``array[i + 1] = 0; ` `                    ``else` `                        ``array[i + 1] = 1; ` `                ``} ` `                ``else` `                ``{ ` `                    ``if` `((pari1 == 0 && pari2 == 1) || ` `                        ``(pari1 == 1 && pari2 == 0)) ` `                        ``array[i + 1] = 2; ` `                    ``if` `((pari1 == 1 && pari2 == 2) ` `                        ``|| (pari1 == 2 && pari2 == 1)) ` `                        ``array[i + 1] = 0; ` `                    ``if` `((pari1 == 2 && pari2 == 0) ` `                        ``|| (pari1 == 0 && pari2 == 2)) ` `                        ``array[i + 1] = 1; ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``return` `operations; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `[]array = { 2, 1, 3, 0 }; ` `    ``int` `size = array.Length; ` `    ``Console.WriteLine(solve(array, size)); ` `} ` `} ` ` `  `// This code is contributed  ` `// by Akanksha Rai `

 ` `

Output:
```1
```

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