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Missing NumbersGiven two arrays of integers, find which elements in the second array are missing from the first array. If a number occurs multiple times in the lists, you must ensure that the frequency of that number in both lists is the same. If that is

Bijoy Saha answered on October 6, 2022 Popularity 3/10 Helpfulness 1/10

answer Missing NumbersGiven two arrays of integers, find which elements in the second array are missing from the first array. If a number occurs multiple times in the lists, you must ensure that the frequency of that number in both lists is the same. If that is

Missing NumbersGiven two arrays of integers, find which elements in the second array are missing from the first array. If a number occurs multiple times in the lists, you must ensure that the frequency of that number in both lists is the same. If that is

Comment

import java.io.*;
import java.util.*;

public class Solution {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] A = new int[n];

        for (int i = 0; i < n; i++) {
            A[i] = in.nextInt();
        }

        int m = in.nextInt();
        int[] B = new int[m];

        for (int i = 0; i < m; i++) {
            B[i] = in.nextInt();
        }

        HashMap<Integer, Integer> freqs = new HashMap<Integer, Integer>();

        for (int i = 0; i < m; i++) {
            if (freqs.containsKey(B[i])) {
                int freq = freqs.get(B[i]);
                freqs.replace(B[i], freq + 1);
            } else {
                freqs.put(B[i], 1);
            }
        }

        for (int i = 0; i < n; i++) {
            if (freqs.containsKey(A[i])) {
                int freq = freqs.get(A[i]);
                if (freq == 1) {
                    freqs.remove(A[i]);
                } else {
                    freqs.replace(A[i], freq - 1);
                }
            } else {
                System.out.println("error");
            }
        }

        StringBuilder answer = new StringBuilder();
        Iterator it = freqs.entrySet().iterator();
        while (it.hasNext()) {
            Map.Entry pair = (Map.Entry)it.next();
            answer.append(pair.getKey());
            answer.append(" ");
        }

        System.out.println(answer.toString());
    }
}

Popularity 3/10 Helpfulness 1/10 Language java

Source: programs.programmingoneonone.com

Tags: arrays elements find frequency java

Link to this answer
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Contributed on Oct 06 2022

Bijoy Saha

0 Answers Avg Quality 2/10

Closely Related Answers

Missing NumbersGiven two arrays of integers, find which elements in the second array are missing from the first array. If a number occurs multiple times in the lists, you must ensure that the frequency of that number in both lists is the same. If that is

Comment

import java.io.*;
import java.util.*;

public class Solution {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] A = new int[n];

        for (int i = 0; i < n; i++) {
            A[i] = in.nextInt();
        }

        int m = in.nextInt();
        int[] B = new int[m];

        for (int i = 0; i < m; i++) {
            B[i] = in.nextInt();
        }

        HashMap<Integer, Integer> freqs = new HashMap<Integer, Integer>();

        for (int i = 0; i < m; i++) {
            if (freqs.containsKey(B[i])) {
                int freq = freqs.get(B[i]);
                freqs.replace(B[i], freq + 1);
            } else {
                freqs.put(B[i], 1);
            }
        }

        for (int i = 0; i < n; i++) {
            if (freqs.containsKey(A[i])) {
                int freq = freqs.get(A[i]);
                if (freq == 1) {
                    freqs.remove(A[i]);
                } else {
                    freqs.replace(A[i], freq - 1);
                }
            } else {
                System.out.println("error");
            }
        }

        StringBuilder answer = new StringBuilder();
        Iterator it = freqs.entrySet().iterator();
        while (it.hasNext()) {
            Map.Entry pair = (Map.Entry)it.next();
            answer.append(pair.getKey());
            answer.append(" ");
        }

        System.out.println(answer.toString());
    }
}

Popularity 3/10 Helpfulness 1/10 Language java

Source: programs.programmingoneonone.com

Tags: elements java frequency arrays find

Link to this answer
Share Copy Link

Contributed on Oct 06 2022

Bijoy Saha

0 Answers Avg Quality 2/10

Missing NumbersGiven two arrays of integers, find which elements in the second array are missing from the first array. If a number occurs multiple times in the lists, you must ensure that the frequency of that number in both lists is the same. If that is

Comment

import java.io.*;
import java.util.*;

public class Solution {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int[] A = new int[n];

        for (int i = 0; i < n; i++) {
            A[i] = in.nextInt();
        }

        int m = in.nextInt();
        int[] B = new int[m];

        for (int i = 0; i < m; i++) {
            B[i] = in.nextInt();
        }

        HashMap<Integer, Integer> freqs = new HashMap<Integer, Integer>();

        for (int i = 0; i < m; i++) {
            if (freqs.containsKey(B[i])) {
                int freq = freqs.get(B[i]);
                freqs.replace(B[i], freq + 1);
            } else {
                freqs.put(B[i], 1);
            }
        }

        for (int i = 0; i < n; i++) {
            if (freqs.containsKey(A[i])) {
                int freq = freqs.get(A[i]);
                if (freq == 1) {
                    freqs.remove(A[i]);
                } else {
                    freqs.replace(A[i], freq - 1);
                }
            } else {
                System.out.println("error");
            }
        }

        StringBuilder answer = new StringBuilder();
        Iterator it = freqs.entrySet().iterator();
        while (it.hasNext()) {
            Map.Entry pair = (Map.Entry)it.next();
            answer.append(pair.getKey());
            answer.append(" ");
        }

        System.out.println(answer.toString());
    }
}

Popularity 3/10 Helpfulness 1/10 Language java

Source: programs.programmingoneonone.com

Tags: frequency find java elements arrays

Link to this answer
Share Copy Link

Contributed on Oct 06 2022

Bijoy Saha

0 Answers Avg Quality 2/10

Missing NumbersGiven two arrays of integers, find which elements in the second array are missing from the first array. If a number occurs multiple times in the lists, you must ensure that the frequency of that number in both lists is the same. If that is

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Missing NumbersGiven two arrays of integers, find which elements in the second array are missing from the first array. If a number occurs multiple times in the lists, you must ensure that the frequency of that number in both lists is the same. If that is

Closely Related Answers

Missing NumbersGiven two arrays of integers, find which elements in the second array are missing from the first array. If a number occurs multiple times in the lists, you must ensure that the frequency of that number in both lists is the same. If that is

Missing NumbersGiven two arrays of integers, find which elements in the second array are missing from the first array. If a number occurs multiple times in the lists, you must ensure that the frequency of that number in both lists is the same. If that is

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