How Many Numbers Are Smaller Than the Current Number - Array - Easy - LeetCode
💻 coding

How Many Numbers Are Smaller Than the Current Number - Array - Easy - LeetCode

2 min read 360 words
2 min read
ShareWhatsAppPost on X
  • 1The problem requires counting how many numbers in an array are smaller than each element at a given index.
  • 2The solution involves a nested loop, resulting in a time complexity of O(n^2) and space complexity of O(n).
  • 3Example outputs demonstrate the function's correctness, showing the counts of smaller numbers for various input arrays.

AI-generated summary · May not capture all nuances

Key Insight
AskGif

"The problem requires counting how many numbers in an array are smaller than each element at a given index."

How Many Numbers Are Smaller Than the Current Number - Array - Easy - LeetCode

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:

using System;
using System.Collections.Generic;
using System.Text;

namespace LeetCode.AskGif.Easy.Array
{
 public class SmallerNumbersThanCurrentSoln
 {
 public int[] SmallerNumbersThanCurrent(int[] nums)
 {
 var res = new int[nums.Length];
 for (int i = 0; i < nums.Length; i++)
 {
 res[i] = 0;
 for (int j = 0; j < nums.Length; j++)
 {
 if (nums[i] > nums[j])
 {
 res[i]++;
 }
 }
 }

 return res;
 }
 }
}

Time Complexity: O(n^2)

Space Complexity: O(n) To Store Result

Unit Tests:

using LeetCode.AskGif.Easy.Array;
using Microsoft.VisualStudio.TestTools.UnitTesting;
using System;
using System.Collections.Generic;
using System.Text;

namespace CodingUnitTest.Easy.Array
{
 [TestClass]
 public class SmallerNumbersThanCurrentSolnTests
 {
 [TestMethod]
 public void SmallerNumbersThanCurrentSoln_First()
 {
 var nums = new int[] { 8, 1, 2, 2, 3 };
 var output = new int[]{4, 0, 1, 1, 3};
 var res = new SmallerNumbersThanCurrentSoln().SmallerNumbersThanCurrent(nums);

 AreEqual(res, output);
 }

 [TestMethod]
 public void SmallerNumbersThanCurrentSoln_Second()
 {
 var nums = new int[] { 6, 5, 4, 8 };
 var output = new int[] { 2, 1, 0, 3 };
 var res = new SmallerNumbersThanCurrentSoln().SmallerNumbersThanCurrent(nums);

 AreEqual(res, output);
 }

 [TestMethod]
 public void SmallerNumbersThanCurrentSoln_Third()
 {
 var nums = new int[] { 7, 7, 7, 7 };
 var output = new int[] { 0, 0, 0, 0 };
 var res = new SmallerNumbersThanCurrentSoln().SmallerNumbersThanCurrent(nums);

 AreEqual(res, output);
 }
 private void AreEqual(int[] res, int[] output)
 {
 Assert.AreEqual(res.Length, output.Length);
 for (int i = 0; i < res.Length; i++)
 {
 Assert.AreEqual(res[i], output[i]);
 }
 }
 }
}

Enjoyed this article?

Share it with someone who'd find it useful.

ShareWhatsAppPost on X

AskGif

Published on 7 June 2020 · 2 min read · 360 words

Part of AskGif Blog · coding

You might also like