package com.fishercoder.solutions;

import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

/**
 * 1356. Sort Integers by The Number of 1 Bits
 *
 * Given an integer array arr. You have to sort the integers in the array in ascending order by the number of 1's in
 * their binary representation and in case of two or more integers have the same number of 1's you have to sort them in ascending order.
 *
 * Return the sorted array.
 *
 * Example 1:
 * Input: arr = [0,1,2,3,4,5,6,7,8]
 * Output: [0,1,2,4,8,3,5,6,7]
 * Explantion: [0] is the only integer with 0 bits.
 * [1,2,4,8] all have 1 bit.
 * [3,5,6] have 2 bits.
 * [7] has 3 bits.
 * The sorted array by bits is [0,1,2,4,8,3,5,6,7]
 *
 * Example 2:
 * Input: arr = [1024,512,256,128,64,32,16,8,4,2,1]
 * Output: [1,2,4,8,16,32,64,128,256,512,1024]
 * Explantion: All integers have 1 bit in the binary representation, you should just sort them in ascending order.
 *
 * Example 3:
 * Input: arr = [10000,10000]
 * Output: [10000,10000]
 *
 * Example 4:
 * Input: arr = [2,3,5,7,11,13,17,19]
 * Output: [2,3,5,17,7,11,13,19]
 *
 * Example 5:
 * Input: arr = [10,100,1000,10000]
 * Output: [10,100,10000,1000]
 *
 * Constraints:
 * 1 <= arr.length <= 500
 * 0 <= arr[i] <= 10^4
 * */
public class _1356 {
    public static class Solution1 {
        public int[] sortByBits(int[] arr) {
            Map<Integer, List<Integer>> map = new HashMap<>();
            for (int num : arr) {
                int count = Integer.bitCount(num);
                if (!map.containsKey(count)) {
                    map.put(count, new ArrayList<>());
                }
                map.get(count).add(num);
            }
            int[] result = new int[arr.length];
            int i = 0;
            for (int count : map.keySet()) {
                List<Integer> list = map.get(count);
                Collections.sort(list);
                for (int num : list) {
                    result[i++] = num;
                }
            }
            return result;
        }
    }
}

@@ -163,1155 +163,8 @@
 p;%0A%0A
-/**%0A * 1356. Sort Integers by The Number of 1 Bits%0A *%0A * Given an integer array arr. You have to sort the integers in the array in ascending order by the number of 1's in%0A * their binary representation and in case of two or more integers have the same number of 1's you have to sort them in ascending order.%0A *%0A * Return the sorted array.%0A *%0A * Example 1:%0A * Input: arr = %5B0,1,2,3,4,5,6,7,8%5D%0A * Output: %5B0,1,2,4,8,3,5,6,7%5D%0A * Explantion: %5B0%5D is the only integer with 0 bits.%0A * %5B1,2,4,8%5D all have 1 bit.%0A * %5B3,5,6%5D have 2 bits.%0A * %5B7%5D has 3 bits.%0A * The sorted array by bits is %5B0,1,2,4,8,3,5,6,7%5D%0A *%0A * Example 2:%0A * Input: arr = %5B1024,512,256,128,64,32,16,8,4,2,1%5D%0A * Output: %5B1,2,4,8,16,32,64,128,256,512,1024%5D%0A * Explantion: All integers have 1 bit in the binary representation, you should just sort them in ascending order.%0A *%0A * Example 3:%0A * Input: arr = %5B10000,10000%5D%0A * Output: %5B10000,10000%5D%0A *%0A * Example 4:%0A * Input: arr = %5B2,3,5,7,11,13,17,19%5D%0A * Output: %5B2,3,5,17,7,11,13,19%5D%0A *%0A * Example 5:%0A * Input: arr = %5B10,100,1000,10000%5D%0A * Output: %5B10,100,10000,1000%5D%0A *%0A * Constraints:%0A * 1 %3C= arr.length %3C= 500%0A * 0 %3C= arr%5Bi%5D %3C= 10%5E4%0A * */%0A
 publ

refactor 1356