Given two arrays A[] and B[] consisting of n and m elements respectively. Find the minimum number of elements to remove from each array such that no common element exist in both.
Examples:
Input : A[] = { 1, 2, 3, 4}
B[] = { 2, 3, 4, 5, 8 }
Output : 3
We need to remove 2, 3 and 4 from any array.
Input : A[] = { 4, 2, 4, 4}
B[] = { 4, 3 }
Output : 1
We need to remove 4 from B[]
Input : A[] = { 1, 2, 3, 4 }
B[] = { 5, 6, 7 }
Output : 0
There is no common element in both.
Count occurrence of each number in both arrays. If there is a number in both array remove number from array in which it appears less number of times add it to the result.
Implementation:
C++
// CPP program to find minimum element// to remove so no common element// exist in both array#include <bits/stdc++.h>using namespace std;// To find no elements to remove// so no common element existint minRemove(int a[], int b[], int n, int m){ // To store count of array element unordered_map<int, int> countA, countB; // Count elements of a for (int i = 0; i < n; i++) countA[a[i]]++; // Count elements of b for (int i = 0; i < m; i++) countB[b[i]]++; // Traverse through all common element, and // pick minimum occurrence from two arrays int res = 0; for (auto x : countA) if (countB.find(x.first) != countB.end()) res += min(x.second, countB[x.first]); // To return count of minimum elements return res;}// Driver program to test minRemove()int main(){ int a[] = { 1, 2, 3, 4 }; int b[] = { 2, 3, 4, 5, 8 }; int n = sizeof(a) / sizeof(a[0]); int m = sizeof(b) / sizeof(b[0]); cout << minRemove(a, b, n, m); return 0;} |
Java
// JAVA Code to Remove minimum number of elements// such that no common element exist in both arrayimport java.util.*;class GFG { // To find no elements to remove // so no common element exist public static int minRemove(int a[], int b[], int n, int m) { // To store count of array element HashMap<Integer, Integer> countA = new HashMap< Integer, Integer>(); HashMap<Integer, Integer> countB = new HashMap< Integer, Integer>(); // Count elements of a for (int i = 0; i < n; i++){ if (countA.containsKey(a[i])) countA.put(a[i], countA.get(a[i]) + 1); else countA.put(a[i], 1); } // Count elements of b for (int i = 0; i < m; i++){ if (countB.containsKey(b[i])) countB.put(b[i], countB.get(b[i]) + 1); else countB.put(b[i], 1); } // Traverse through all common element, and // pick minimum occurrence from two arrays int res = 0; Set<Integer> s = countA.keySet(); for (int x : s) if(countB.containsKey(x)) res += Math.min(countB.get(x), countA.get(x)); // To return count of minimum elements return res; } /* Driver program to test above function */ public static void main(String[] args) { int a[] = { 1, 2, 3, 4 }; int b[] = { 2, 3, 4, 5, 8 }; int n = a.length; int m = b.length; System.out.println(minRemove(a, b, n, m)); }} // This code is contributed by Arnav Kr. Mandal. |
Python3
# Python3 program to find minimum # element to remove so no common # element exist in both array# To find no elements to remove# so no common element existdef minRemove(a, b, n, m): # To store count of array element countA = dict() countB = dict() # Count elements of a for i in range(n): countA[a[i]] = countA.get(a[i], 0) + 1 # Count elements of b for i in range(n): countB[b[i]] = countB.get(b[i], 0) + 1 # Traverse through all common # element, and pick minimum # occurrence from two arrays res = 0 for x in countA: if x in countB.keys(): res += min(countA[x],countB[x]) # To return count of # minimum elements return res# Driver Codea = [ 1, 2, 3, 4 ]b = [2, 3, 4, 5, 8 ]n = len(a)m = len(b)print(minRemove(a, b, n, m))# This code is contributed # by mohit kumar |
C#
// C# Code to Remove minimum number of elements// such that no common element exist in both arrayusing System;using System.Collections.Generic;class GFG{ // To find no elements to remove // so no common element exist public static int minRemove(int []a, int []b, int n, int m) { // To store count of array element Dictionary<int,int> countA = new Dictionary<int,int>(); Dictionary<int,int>countB = new Dictionary<int,int>(); // Count elements of a for (int i = 0; i < n; i++) { if (countA.ContainsKey(a[i])) { var v = countA[a[i]]; countA.Remove(countA[a[i]]); countA.Add(a[i], v + 1); } else countA.Add(a[i], 1); } // Count elements of b for (int i = 0; i < m; i++) { if (countB.ContainsKey(b[i])) { var v = countB[b[i]]; countB.Remove(countB[b[i]]); countB.Add(b[i], v + 1); } else countB.Add(b[i], 1); } // Traverse through all common element, and // pick minimum occurrence from two arrays int res = 0; foreach (int x in countA.Keys) if(countB.ContainsKey(x)) res += Math.Min(countB[x], countA[x]); // To return count of minimum elements return res; } /* Driver code */ public static void Main(String[] args) { int []a = { 1, 2, 3, 4 }; int []b = { 2, 3, 4, 5, 8 }; int n = a.Length; int m = b.Length; Console.WriteLine(minRemove(a, b, n, m)); }}// This code has been contributed by 29AjayKumar |
Javascript
<script>// JavaScript Code to Remove // minimum number of elements// such that no common element// exist in both array // To find no elements to remove // so no common element exist function minRemove(a,b,n,m) { // To store count of array element let countA = new Map(); let countB = new Map(); // Count elements of a for (let i = 0; i < n; i++){ if (countA.has(a[i])) countA.set(a[i], countA.get(a[i]) + 1); else countA.set(a[i], 1); } // Count elements of b for (let i = 0; i < m; i++){ if (countB.has(b[i])) countB.set(b[i], countB.get(b[i]) + 1); else countB.set(b[i], 1); } // Traverse through all // common element, and // pick minimum occurrence // from two arrays let res = 0; for (let x of countA.keys()) if(countB.has(x)) res += Math.min(countB.get(x), countA.get(x)); // To return count of minimum elements return res; } /* Driver program to test above function */ let a=[1, 2, 3, 4 ]; let b=[2, 3, 4, 5, 8]; let n = a.length; let m = b.length; document.write(minRemove(a, b, n, m)); // This code is contributed by unknown2108</script> |
Output
3
Time Complexity: O(n+m)
Auxiliary Space: O(n+m)
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