1365. How Many Numbers Are Smaller Than the Current Number.md

1365. How Many Numbers Are Smaller Than the Current Number

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leetcode-cn Daily Challenge on October 26th, 2020.

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Difficulty : Easy

Related Topics : Array、HashTable

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Given the array nums, for each nums[i] find out how many numbers in the array are smaller than it. That is, for each nums[i] you have to count the number of valid j's such that j != i and nums[j] < nums[i].

Return the answer in an array.

Example 1:

Input: nums = [8,1,2,2,3]
Output: [4,0,1,1,3]
Explanation:
For nums[0]=8 there exist four smaller numbers than it (1, 2, 2 and 3).
For nums[1]=1 does not exist any smaller number than it.
For nums[2]=2 there exist one smaller number than it (1).
For nums[3]=2 there exist one smaller number than it (1).
For nums[4]=3 there exist three smaller numbers than it (1, 2 and 2).

Example 2:

Input: nums = [6,5,4,8]
Output: [2,1,0,3]

Example 3:

Input: nums = [7,7,7,7]
Output: [0,0,0,0]

Constraints:

• 2 <= nums.length <= 500
• 0 <= nums[i] <= 100

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Solution

• mine
  • Java

    • Runtime: 4 ms, faster than 63.52%, Memory Usage: 39 MB, less than 10.26% of Java online submissions

      // O(N * logN)time
      // O(N)space
      public int[] smallerNumbersThanCurrent(int[] nums) {
          int n = nums.length;
          Map<Integer, Integer> map = new HashMap<>();
          for(int i = 0; i < n; i++){
              map.put(i, nums[i]);
          }
          Arrays.sort(nums);
          Map<Integer, Integer> map2 = new HashMap<>();
          map2.put(nums[0], 0);
          int cur = 0;
          for(int i = 1; i < n; i++){
              if(nums[i] > nums[i - 1]){
                  cur = i;
              }
              map2.put(nums[i], cur);
          }
          int[] res = new int[n];
          for(int i = 0; i < n; i++){
              res[i] = map2.get(map.get(i));
          }
          return res;
      }
      

    • BucketSort Runtime: 1 ms, faster than 98.69%, Memory Usage: 39.2 MB, less than 10.26% of Java online submissions

      // O(N)time
      // O(N)space
      public int[] smallerNumbersThanCurrent(int[] nums) {
          int n = nums.length;
          int[] arr = new int[101];
          for(int num : nums){
              arr[num]++;
          }
          for(int i = 1; i < 101; i++){
              arr[i] += arr[i - 1];
          }
          int[] res = new int[n];
          for(int i = 0; i < n; i++){
              res[i] =nums[i] == 0 ? 0 : arr[nums[i] - 1];
          }
          return res;
      }
      

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