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: 2
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: 1
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 <bits/stdc++.h>using namespace std;// Recursive function to calculate// minimum adjacent pairs// with different parityvoid 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 parityvoid 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 codeint 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<Integer> even, List<Integer> odd, List<Integer> 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<Integer> v, int n) { // Store no of even numbers // not present in the array List<Integer> even = new ArrayList<Integer>(); // Store no of odd numbers // not present in the array List<Integer> odd = new ArrayList<Integer>(); HashMap<Integer, Integer> 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<Integer, Integer> 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<Integer> v = new ArrayList<Integer>(); 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 approachmn = 1000# Recursive function to calculate# minimum adjacent pairs# with different paritydef 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 paritydef 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 codeif __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 Codestatic 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
<script>// Javascript implementation of above approachvar min = 10000;// Recursive function to calculate// minimum adjacent pairs// with different parityfunction parity(even, odd, v, i){ // If all the numbers are placed if (i == v.length || even.length == 0 && odd.length == 0) { var count = 0; for (var j = 0; j < v.length - 1; j++) { if (v[j] % 2 != v[j + 1] % 2) count++; } if (count < min) min = count; return min; } // If replacement is not required if (v[i] != -1) min = parity(even, odd, v, i + 1); // If replacement is required else { if (even.length != 0) { var x = even.back(); even.pop(); v[i] = x; min = parity(even, odd, v, i + 1); // backtracking even.push(x); } if (odd.length != 0) { var x = odd[odd.length-1]; odd.pop(); v[i] = x; min = parity(even, odd, v, i + 1); // backtracking odd.push(x); } } return min;}// Function to display the minimum number of// adjacent elements with different parityfunction minDiffParity(v, n){ // Store no of even numbers // not present in the array var even = []; // Store no of odd numbers // not present in the array var odd = []; var m = new Map(); for (var i = 1; i <= n; i++) m.set(i, 1); for (var i = 0; i < v.length; i++) { // Erase existing numbers if (v[i] != -1) m.delete(v[i]); } // Store non-existing // even and odd numbers m.forEach((value, key) => { if (i.first % 2 == 0) even.push(key); else odd.push(key); }); min = parity(even, odd, v, 0); document.write( min );}// Driver codevar n = 8;var v = [2, 1, 4, -1, -1, 6, -1, 8];minDiffParity(v, n);// This code is contributed by rutvik_56.</script> |
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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