kninami

import java.util.ArrayList;

class Solution{
    public int solution(int N){
        ArrayList<Integer> array = new ArrayList<Integer>();
        
        int count = 0;
        int max = 0;
        boolean flag = false;
        

kninami

import java.util.*;
import java.lang.*;
import java.util.Arrays;

class Solution {
    public int solution(int N) {
        int N3 = 1;
        String bina = Integer.toBinaryString(N);
        String pro[] = bina.split("0");
        int[] score = new int[pro.length];

kninami

import java.util.*;

class Solution {
    public int solution(int[] A) {
        
        Map<Integer,Integer> matchMap = new HashMap<Integer,Integer>();
        int unPaired = 0;
        
        for(int i=0;i<A.length;i++){
            int value = matchMap.get(A[i])==null?0:matchMap.get(A[i]);

kninami

import java.util.HashMap;
import java.util.Iterator;
import java.util.Map.Entry;

class Solution {
    public int solution(int[] A) {
    		int result=0;
    		HashMap<Integer , Integer> map = new HashMap<Integer , Integer>();
    		
    		for(int i=0;i<A.length;i++){

kninami

function solution(A) {
    let solver = {
        makeCounterHash: (A) => {
            let counterMap = {};
            const ARR_LEN = A.length;
            const HALF_IDX = Math.floor(ARR_LEN / 2);

            for (let idx = 0; idx <= HALF_IDX; idx++) {
                const prenum = A[idx];
                if (counterMap.hasOwnProperty(prenum)) {

kninami

/*
node로 실행
풀이 방식
시간 복잡도를 O(n)으로 하기 위해 bit연산으로 하여 처리
10진수 값을 1과 AND 연산해서 2진수의 역순으로 하나씩 꺼내서 처리
*/
function solution(N) {
  let curNum = N;

  let maxZeroCnt = 0;

kninami

class Solution {
    public int[] solution(int[] A, int K) {
        int[] B = new int[A.length];
        
        for(int i=0; i<A.length; i++){
            B[(i+K)%A.length] = A[i];
        }
        
        return B;
    }

kninami

class Solution {
	public int[] solution(int[] A, int K) {
		int[] result = new int[A.length];
		int length = 0;
		
		if(K!=0 && A.length!=0){ length = K%A.length; }
		for(int i=0;i<A.length;i++){
			if(i+length<A.length){  
				result[i+length] = A[i]; 
			}else{

kninami

class Solution {
    public int solution(int[] A) {
        long N = A.length +1;
        long totalSum = (N*(N+1))/2;
        
        for(int i=0; i<A.length; i++){
            totalSum -= A[i];
        }
        
        return (int)totalSum;

kninami

function solution(A) {
    const sum = (A.length + 1) * (A.length + 1 + 1) / 2;
    return A.reduceRight((prev, cur) => prev - cur, sum);
}