HowManyNumbersAreSmallerThantheCurrentNumber

 
 class HowManyNumbersAreSmallerThantheCurrentNumber {
    public int[] smallerNumbersThanCurrent() {
        int[] nums={8,1,2,2,3};
        int[] arr=new int[nums.length];
        
        for (int i = 0; i < nums.length; i++) {
            int count=0;
            for (int j = 0; j < nums.length; j++) {
                if(nums[i]>nums[j])
                {
                    count++;
                }
            }
            arr[i]=count;
        }
        return arr;
    }
}
//     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).