### How Many Numbers Are Smaller Than the Current Number in an Array - The Coding Shala

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In this post, we will learn how to check How Many Numbers Are Smaller Than the Current Number in an Array using Java Program.

## How Many Numbers Are Smaller Than the Current Number in an Array Java Program

Given the array nums, for each num [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].

**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).

### Java Solution

**Approach 1**

Using brute force(Two loops).

**Java Program:**

class Solution { public int[] smallerNumbersThanCurrent(int[] nums) { int[] ans = new int[nums.length]; for(int i=0; i<nums.length; i++){ int tmp = 0; for(int j=0; j<nums.length; j++) { if(i!=j){ if(nums[i] > nums[j]) tmp++; } } ans[i] = tmp; } return ans; } }

**Approach 2**

Here All the given numbers are between 0 and 100 so we can create an array of length 101 and put all the numbers in this array with the count at the corresponding index.

**Java Program: **

class Solution { public int[] smallerNumbersThanCurrent(int[] nums) { int[] count = new int[101]; for(int i=0;i<nums.length;i++) { count[nums[i]]++; } for(int i=1;i<101;i++){ count[i] += count[i-1]; } int[] ans = new int[nums.length]; for(int i=0;i<nums.length;i++){ if(nums[i] == 0) ans[i] = 0; else ans[i] = count[nums[i]-1]; } return ans; } }

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