How Many Numbers Are Smaller Than the Current Number

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

Solution

  1. Using Hash table to solve the problem in O(n) time.
  2. In our given constraint, nums[i] will be 0 <= nums[i] <= 100,
  3. So declare hash[101] which holds the number from 0 to 100.

C# Program

public class Solution {
    public int[] SmallerNumbersThanCurrent(int[] nums) {
        int[] hash = new int[101];
        for(int i=0;i<nums.Length;i++)
        {
            hash[nums[i]]++;
        }
        for(int i=1;i<=100;i++)
        {
            hash[i] = hash[i] + hash[i-1];
        }
        for(int i=0;i<nums.Length;i++)
        {
            if(nums[i] != 0)
                nums[i] = hash[nums[i]-1];
            
        }
        return nums;
    }
}

Time Complexity: O(n)

Space Complexity: O(1)

One thought on “How Many Numbers Are Smaller Than the Current Number

  1. Pingback: 100 Days Leetcode Challenge – Passion of Programming

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