Leetcode-1365 How Many Numbers Are Smaller Than the Current Number Solution in Java | Hindi Coding Community

 Given the array of integers nums, for each value in the arraynums[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].

Answer should be in the form of 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

Java Code :

class Solution {

public int[] smallerNumbersThanCurrent(int[] nums) {

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;

}

}