Count of Smaller Numbers After Self leetCode

 Given an integer array nums, return an integer array counts where counts[i] is the number of smaller elements to the right of nums[i].


 


Example 1:


Input: nums = [5,2,6,1]

Output: [2,1,1,0]

Explanation:

To the right of 5 there are 2 smaller elements (2 and 1).

To the right of 2 there is only 1 smaller element (1).

To the right of 6 there is 1 smaller element (1).

To the right of 1 there is 0 smaller element.

Example 2:


Input: nums = [-1]

Output: [0]

Example 3:


Input: nums = [-1,-1]

Output: [0,0]



BruteForce :

class Solution {
    public List<Integer> countSmaller(int[] nums) {
       
List<Integer> result= new ArrayList<>();
int n=nums.length;
for(int i=0;i<n;i++)
{
   int count=0;
    for(int j=i+1;j<n;j++)
    {
         
          if(nums[i]>nums[j])
          {
            count++;
         
          }
    }
      result.add(count);
 
}
     return result;
    }      
   
}


Optimal Approach:

class Solution {
    public List<Integer> countSmaller(int[] nums) {
        int n = nums.length;
        int[] result = new int[n];
        int[][] pairs = new int[n][2]; // [value, index]
       
        for (int i = 0; i < n; i++) {
            pairs[i][0] = nums[i];
            pairs[i][1] = i;
        }

        mergeSort(pairs, 0, n - 1, result);
       
        List<Integer> resList = new ArrayList<>();
        for (int count : result) {
            resList.add(count);
        }
        return resList;
    }

    private void mergeSort(int[][] arr, int left, int right, int[] result) {
        if (left >= right) return;

        int mid = (left + right) / 2;
        mergeSort(arr, left, mid, result);
        mergeSort(arr, mid + 1, right, result);
        merge(arr, left, mid, right, result);
    }

    private void merge(int[][] arr, int left, int mid, int right, int[] result) {
        List<int[]> temp = new ArrayList<>();
        int i = left, j = mid + 1;
        int rightCount = 0;

        while (i <= mid && j <= right) {
            if (arr[i][0] <= arr[j][0]) {
                result[arr[i][1]] += rightCount;
                temp.add(arr[i++]);
            } else {
                rightCount++;
                temp.add(arr[j++]);
            }
        }

        while (i <= mid) {
            result[arr[i][1]] += rightCount;
            temp.add(arr[i++]);
        }

        while (j <= right) {
            temp.add(arr[j++]);
        }

        for (int k = left; k <= right; k++) {
            arr[k] = temp.get(k - left);
        }
    }
}

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