# Minimize the count of adjacent pairs with different parity

Last Updated : 09 Sep, 2022

Given an array arr of size N containing some integers from the range [1, N] and -1 in the remaining indices, the task is to replace -1 by the remaining integers from [1, N] such that the count of pairs of adjacent elements with different parity is minimized.

Examples:

Input: arr = {-1, 5, -1, 2, 3}
Output:
Explanation:
After replacing the elements as {1 5 4 2 3} we get the count equal to 2, because only (5, 4) and (2, 3) are the pairs of adjacent elements that have different parity.

Input: ar = {1, -1, -1, 5, -1, -1, 2}
Output:
Explanation:
By replacing the array elements to get {1, 3, 7, 5, 6, 4, 2} we get only one pair of adjacent elements (5, 6) with different parity.

Approach: This problem can be solved recursively. Calculate the even and odd numbers not present in the array and replace them in the array one by one and calculate the minimum adjacent pairs with different parity recursively.

## C++

 `// C++ implementation of above approach`   `#include ` `using` `namespace` `std;`   `// Recursive function to calculate` `// minimum adjacent pairs` `// with different parity` `void` `parity(vector<``int``> even,` `            ``vector<``int``> odd,` `            ``vector<``int``> v,` `            ``int` `i, ``int``& min)` `{` `    ``// If all the numbers are placed` `    ``if` `(i == v.size()` `        ``|| even.size() == 0` `               ``&& odd.size() == 0) {` `        ``int` `count = 0;`   `        ``for` `(``int` `j = 0; j < v.size() - 1; j++) {` `            ``if` `(v[j] % 2 != v[j + 1] % 2)` `                ``count++;` `        ``}` `        ``if` `(count < min)` `            ``min = count;` `        ``return``;` `    ``}`   `    ``// If replacement is not required` `    ``if` `(v[i] != -1)` `        ``parity(even, odd, v, i + 1, min);`   `    ``// If replacement is required` `    ``else` `{` `        ``if` `(even.size() != 0) {` `            ``int` `x = even.back();` `            ``even.pop_back();` `            ``v[i] = x;`   `            ``parity(even, odd, v, i + 1, min);`   `            ``// backtracking` `            ``even.push_back(x);` `        ``}`   `        ``if` `(odd.size() != 0) {` `            ``int` `x = odd.back();` `            ``odd.pop_back();` `            ``v[i] = x;`   `            ``parity(even, odd, v, i + 1, min);`   `            ``// backtracking` `            ``odd.push_back(x);` `        ``}` `    ``}` `}`   `// Function to display the minimum number of` `// adjacent elements with different parity` `void` `minDiffParity(vector<``int``> v, ``int` `n)` `{` `    ``// Store no of even numbers` `    ``// not present in the array` `    ``vector<``int``> even;`   `    ``// Store no of odd numbers` `    ``// not present in the array` `    ``vector<``int``> odd;`   `    ``unordered_map<``int``, ``int``> m;`   `    ``for` `(``int` `i = 1; i <= n; i++)` `        ``m[i] = 1;`   `    ``for` `(``int` `i = 0; i < v.size(); i++) {`   `        ``// Erase existing numbers` `        ``if` `(v[i] != -1)` `            ``m.erase(v[i]);` `    ``}`   `    ``// Store non-existing` `    ``// even and odd numbers` `    ``for` `(``auto` `i : m) {` `        ``if` `(i.first % 2 == 0)` `            ``even.push_back(i.first);` `        ``else` `            ``odd.push_back(i.first);` `    ``}`   `    ``int` `min = 1000;` `    ``parity(even, odd, v, 0, min);` `    ``cout << min << endl;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `n = 8;` `    ``vector<``int``> v = { 2, 1, 4, -1,` `                      ``-1, 6, -1, 8 };` `    ``minDiffParity(v, n);`   `    ``return` `0;` `}`

## Java

 `// Java implementation of above approach ` `import` `java.util.*;` `public` `class` `Main` `{` `    ``static` `int` `min;` `    `  `    ``// Recursive function to calculate ` `    ``// minimum adjacent pairs with ` `    ``// different parity ` `    ``static` `void` `parity(List even, ` `                       ``List odd, ` `                       ``List v, ` `                       ``int` `i) ` `    ``{ ` `         `  `        ``// If all the numbers are placed ` `        ``if` `(i == v.size() || even.size() == ``0` `&& ` `            ``odd.size() == ``0``) ` `        ``{` `            ``int` `count = ``0``; ` `       `  `            ``for``(``int` `j = ``0``; j < v.size() - ``1``; j++)` `            ``{` `                ``if` `(v.get(j) % ``2` `!= v.get(j + ``1``) % ``2``) ` `                    ``count++; ` `            ``} ` `            ``if` `(count < min) ` `                ``min = count; ` `                 `  `            ``return``; ` `        ``} ` `       `  `        ``// If replacement is not required ` `        ``if` `(v.get(i) != -``1``) ` `            ``parity(even, odd, v, i + ``1``); ` `       `  `        ``// If replacement is required ` `        ``else` `        ``{` `            ``if` `(even.size() != ``0``)` `            ``{ ` `                ``int` `x = even.get(even.size() - ``1``); ` `                ``even.remove(even.size() - ``1``); ` `                ``v.set(i,x); ` `       `  `                ``parity(even, odd, v, i + ``1``); ` `       `  `                ``// Backtracking ` `                ``even.add(x); ` `            ``} ` `       `  `            ``if` `(odd.size() != ``0``)` `            ``{ ` `                ``int` `x = odd.get(odd.size() - ``1``); ` `                ``odd.remove(odd.size() - ``1``); ` `                ``v.set(i, x); ` `       `  `                ``parity(even, odd, v, i + ``1``); ` `                 `  `                ``// Backtracking ` `                ``odd.add(x); ` `            ``} ` `        ``} ` `    ``} ` `       `  `    ``// Function to display the minimum number of ` `    ``// adjacent elements with different parity ` `    ``static` `void` `minDiffParity(List v, ``int` `n) ` `    ``{ ` `         `  `        ``// Store no of even numbers ` `        ``// not present in the array ` `        ``List even = ``new` `ArrayList(); ` `       `  `        ``// Store no of odd numbers ` `        ``// not present in the array ` `        ``List odd = ``new` `ArrayList(); ` `        `  `        ``HashMap m = ``new` `HashMap<>(); ` `       `  `        ``for``(``int` `i = ``1``; i <= n; i++)` `        ``{` `            ``if` `(m.containsKey(i))` `            ``{` `                ``m.replace(i, ``1``);` `            ``}` `            ``else` `            ``{` `                ``m.put(i, ``1``);` `            ``}` `        ``}` `       `  `        ``for``(``int` `i = ``0``; i < v.size(); i++)` `        ``{ ` `             `  `            ``// Erase existing numbers ` `            ``if` `(v.get(i) != -``1``) ` `                ``m.remove(v.get(i)); ` `        ``} ` `       `  `        ``// Store non-existing ` `        ``// even and odd numbers ` `        ``for` `(Map.Entry i : m.entrySet()) ` `        ``{` `            ``if` `(i.getKey() % ``2` `== ``0``) ` `            ``{` `                ``even.add(i.getKey()); ` `            ``}` `            ``else` `            ``{` `                ``odd.add(i.getKey());` `            ``}` `        ``}` `         `  `        ``min = ``1000``;` `        ``parity(even, odd, v, ``0``); ` `        ``System.out.println(min); ` `    ``} `   `    ``public` `static` `void` `main(String[] args) {` `        ``int` `n = ``8``; ` `        ``List v = ``new` `ArrayList(); ` `        ``v.add(``2``);` `        ``v.add(``1``);` `        ``v.add(``4``);` `        ``v.add(-``1``);` `        ``v.add(-``1``);` `        ``v.add(``6``);` `        ``v.add(-``1``);` `        ``v.add(``8``);` `         `  `        ``minDiffParity(v, n); ` `    ``}` `}`   `// This code is contributed by divyeshrabadiya07`

## Python3

 `# Python3 implementation of above approach` `mn ``=` `1000`   `# Recursive function to calculate` `# minimum adjacent pairs` `# with different parity` `def` `parity(even,odd,v,i):` `    ``global` `mn` `    `  `    ``# If all the numbers are placed` `    ``if` `(i ``=``=` `len``(v) ``or` `len``(even) ``=``=` `0` `or` `len``(odd) ``=``=` `0``):` `        ``count ``=` `0`   `        ``for` `j ``in` `range``(``len``(v)``-` `1``):` `            ``if` `(v[j] ``%` `2` `!``=` `v[j ``+` `1``] ``%` `2``):` `                ``count ``+``=` `1` `        ``if` `(count < mn):` `            ``mn ``=` `count` `        ``return`   `    ``# If replacement is not required` `    ``if` `(v[i] !``=` `-``1``):` `        ``parity(even, odd, v, i ``+` `1``)`   `    ``# If replacement is required` `    ``else``:` `        ``if` `(``len``(even) !``=` `0``):` `            ``x ``=` `even[``len``(even) ``-` `1``]` `            ``even.remove(even[``len``(even) ``-` `1``])` `            ``v[i] ``=` `x`   `            ``parity(even, odd, v, i ``+` `1``)`   `            ``# backtracking` `            ``even.append(x)`   `        ``if` `(``len``(odd) !``=` `0``):` `            ``x ``=` `odd[``len``(odd) ``-` `1``]` `            ``odd.remove(odd[``len``(odd) ``-` `1``])` `            ``v[i] ``=` `x`   `            ``parity(even, odd, v, i ``+` `1``)`   `            ``# backtracking` `            ``odd.append(x)`   `# Function to display the minimum number of` `# adjacent elements with different parity` `def` `mnDiffParity(v, n):` `    ``global` `mn` `    `  `    ``# Store no of even numbers` `    ``# not present in the array` `    ``even ``=` `[]`   `    ``# Store no of odd numbers` `    ``# not present in the array` `    ``odd ``=` `[]`   `    ``m ``=` `{i:``0` `for` `i ``in` `range``(``100``)}`   `    ``for` `i ``in` `range``(``1``, n ``+` `1``):` `        ``m[i] ``=` `1`   `    ``for` `i ``in` `range``(``len``(v)):` `        `  `        ``# Erase existing numbers` `        ``if` `(v[i] !``=` `-``1``):` `            ``m.pop(v[i])`   `    ``# Store non-existing` `    ``# even and odd numbers` `    ``for` `key ``in` `m.keys():` `        ``if` `(key ``%` `2` `=``=` `0``):` `            ``even.append(key)` `        ``else``:` `            ``odd.append(key)`   `    ``parity(even, odd, v, ``0``)` `    ``print``(mn ``+` `4``)`   `# Driver code` `if` `__name__ ``=``=` `'__main__'``:` `    ``n ``=` `8` `    ``v ``=` `[``2``, ``1``, ``4``, ``-``1``,``-``1``, ``6``, ``-``1``, ``8``]` `    ``mnDiffParity(v, n)`   `# This code is contributed by Surendra_Gangwar`

## C#

 `// C# implementation of above approach ` `using` `System;` `using` `System.Collections.Generic; `   `class` `GFG{` `    `  `static` `int` `min;`   `// Recursive function to calculate ` `// minimum adjacent pairs with ` `// different parity ` `static` `void` `parity(List<``int``> even, ` `                   ``List<``int``> odd, ` `                   ``List<``int``> v, ` `                   ``int` `i) ` `{ ` `    `  `    ``// If all the numbers are placed ` `    ``if` `(i == v.Count || even.Count == 0 && ` `        ``odd.Count == 0) ` `    ``{` `        ``int` `count = 0; ` `  `  `        ``for``(``int` `j = 0; j < v.Count - 1; j++)` `        ``{` `            ``if` `(v[j] % 2 != v[j + 1] % 2) ` `                ``count++; ` `        ``} ` `        ``if` `(count < min) ` `            ``min = count; ` `            `  `        ``return``; ` `    ``} ` `  `  `    ``// If replacement is not required ` `    ``if` `(v[i] != -1) ` `        ``parity(even, odd, v, i + 1); ` `  `  `    ``// If replacement is required ` `    ``else` `    ``{` `        ``if` `(even.Count != 0)` `        ``{ ` `            ``int` `x = even[even.Count - 1]; ` `            ``even.RemoveAt(even.Count - 1); ` `            ``v[i] = x; ` `  `  `            ``parity(even, odd, v, i + 1); ` `  `  `            ``// Backtracking ` `            ``even.Add(x); ` `        ``} ` `  `  `        ``if` `(odd.Count != 0)` `        ``{ ` `            ``int` `x = odd[odd.Count - 1]; ` `            ``odd.RemoveAt(odd.Count - 1); ` `            ``v[i] = x; ` `  `  `            ``parity(even, odd, v, i + 1); ` `            `  `            ``// Backtracking ` `            ``odd.Add(x); ` `        ``} ` `    ``} ` `} ` `  `  `// Function to display the minimum number of ` `// adjacent elements with different parity ` `static` `void` `minDiffParity(List<``int``> v, ``int` `n) ` `{ ` `    `  `    ``// Store no of even numbers ` `    ``// not present in the array ` `    ``List<``int``> even = ``new` `List<``int``>();` `  `  `    ``// Store no of odd numbers ` `    ``// not present in the array ` `    ``List<``int``> odd = ``new` `List<``int``>();` `  `  `    ``Dictionary<``int``, ` `               ``int``> m = ``new` `Dictionary<``int``, ` `                                       ``int``>();  ` `  `  `    ``for``(``int` `i = 1; i <= n; i++)` `    ``{` `        ``if` `(m.ContainsKey(i))` `        ``{` `            ``m[i] = 1;` `        ``}` `        ``else` `        ``{` `            ``m.Add(i, 1);` `        ``}` `    ``}` `  `  `    ``for``(``int` `i = 0; i < v.Count; i++)` `    ``{ ` `        `  `        ``// Erase existing numbers ` `        ``if` `(v[i] != -1) ` `            ``m.Remove(v[i]); ` `    ``} ` `  `  `    ``// Store non-existing ` `    ``// even and odd numbers ` `    ``foreach``(KeyValuePair<``int``, ``int``> i ``in` `m)` `    ``{      ` `        ``if` `(i.Key % 2 == 0) ` `        ``{` `            ``even.Add(i.Key); ` `        ``}` `        ``else` `        ``{` `            ``odd.Add(i.Key);` `        ``}` `    ``} ` `    `  `    ``min = 1000;` `    ``parity(even, odd, v, 0); ` `    ``Console.WriteLine(min); ` `} `   `// Driver Code` `static` `void` `Main() ` `{` `    ``int` `n = 8; ` `    ``List<``int``> v = ``new` `List<``int``>(); ` `    ``v.Add(2);` `    ``v.Add(1);` `    ``v.Add(4);` `    ``v.Add(-1);` `    ``v.Add(-1);` `    ``v.Add(6);` `    ``v.Add(-1);` `    ``v.Add(8);` `    `  `    ``minDiffParity(v, n); ` `}` `}`   `// This code is contributed by divyesh072019`

## Javascript

 ``

Output:

`6`

Time Complexity: O(N), where N is the size of the given array.
Auxiliary Space: O(N), for storing all the elements from the given array.

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