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_1356.java
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_1356.java
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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;
}
}
}