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1636. Sort Array by Increasing Frequency

Description

Given an array of integers nums, sort the array in increasing order based on the frequency of the values. If multiple values have the same frequency, sort them in decreasing order.

Return the sorted array.

 

Example 1:

Input: nums = [1,1,2,2,2,3]
Output: [3,1,1,2,2,2]
Explanation: '3' has a frequency of 1, '1' has a frequency of 2, and '2' has a frequency of 3.

Example 2:

Input: nums = [2,3,1,3,2]
Output: [1,3,3,2,2]
Explanation: '2' and '3' both have a frequency of 2, so they are sorted in decreasing order.

Example 3:

Input: nums = [-1,1,-6,4,5,-6,1,4,1]
Output: [5,-1,4,4,-6,-6,1,1,1]

 

Constraints:

• 1 <= nums.length <= 100
• -100 <= nums[i] <= 100

Solutions

• Java
• C++
• Python
• Go
• TypeScript
• Javascript
• RenderScript

• class Solution {
      public int[] frequencySort(int[] nums) {
          int[] cnt = new int[201];
          List<Integer> t = new ArrayList<>();
          for (int v : nums) {
              v += 100;
              ++cnt[v];
              t.add(v);
          }
          t.sort((a, b) -> cnt[a] == cnt[b] ? b - a : cnt[a] - cnt[b]);
          int[] ans = new int[nums.length];
          int i = 0;
          for (int v : t) {
              ans[i++] = v - 100;
          }
          return ans;
      }
  }
  

• class Solution {
  public:
      vector<int> frequencySort(vector<int>& nums) {
          vector<int> cnt(201);
          for (int v : nums) {
              ++cnt[v + 100];
          }
          sort(nums.begin(), nums.end(), [&](const int a, const int b) {
              if (cnt[a + 100] == cnt[b + 100]) return a > b;
              return cnt[a + 100] < cnt[b + 100];
          });
          return nums;
      }
  };
  

• class Solution:
      def frequencySort(self, nums: List[int]) -> List[int]:
          cnt = Counter(nums)
          return sorted(nums, key=lambda x: (cnt[x], -x))
  
  

• func frequencySort(nums []int) []int {
  	cnt := make([]int, 201)
  	for _, v := range nums {
  		cnt[v+100]++
  	}
  	sort.Slice(nums, func(i, j int) bool {
  		a, b := nums[i]+100, nums[j]+100
  		return cnt[a] < cnt[b] || cnt[a] == cnt[b] && a > b
  	})
  	return nums
  }
  

• function frequencySort(nums: number[]): number[] {
      const map = new Map<number, number>();
      for (const num of nums) {
          map.set(num, (map.get(num) ?? 0) + 1);
      }
      return nums.sort((a, b) => map.get(a) - map.get(b) || b - a);
  }
  
  

• /**
   * @param {number[]} nums
   * @return {number[]}
   */
  var frequencySort = function (nums) {
      const m = new Map();
      for (let i = 0; i < nums.length; i++) {
          m.set(nums[i], (m.get(nums[i]) || 0) + 1);
      }
      nums.sort((a, b) => (m.get(a) != m.get(b) ? m.get(a) - m.get(b) : b - a));
      return nums;
  };
  
  

• use std::collections::HashMap;
  impl Solution {
      pub fn frequency_sort(mut nums: Vec<i32>) -> Vec<i32> {
          let n = nums.len();
          let mut map = HashMap::new();
          for &num in nums.iter() {
              *map.entry(num).or_insert(0) += 1;
          }
          nums.sort_by(|a, b| {
              if map.get(a) == map.get(b) {
                  return b.cmp(a);
              }
              map.get(a).cmp(&map.get(b))
          });
          nums
      }
  }
  
  

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