easy How Many Numbers Are Smaller Than the Current Number

Problem Statement

Given an integer array nums, return the answer array where answer[i] is the count of smaller elements than nums[i].

Examples

• Example 1:

  • Input: nums = [5, 3, 6, 1, 4]
  • Expected Output: [3, 1, 4, 0, 2]
  • Justification: For the first element (5), there are three elements (3,1 and 4) that are smaller. For the second element (3), there is one element (1) smaller than it. For (6), there are four elements smaller than it. For (1), there are no smaller elements, and for (4), there are two elements (3 and 1) that are smaller.

• Example 2:

  • Input: nums = [6, 6, 6, 6, 6]
  • Expected Output: [0, 0, 0, 0, 0]
  • Justification: Each element is equal to the others, so there are no elements smaller than any given element.

• Example 3:

  • Input: nums = [1, 2, 3, 4, 5]
  • Expected Output: [0, 1, 2, 3, 4]
  • Justification: This is a strictly increasing sequence. Each number has all the previous numbers in the sequence smaller than itself.

Solution

To solve this problem, we can use a brute force approach by comparing each element with every other element to count how many numbers are smaller than it. However, a more efficient approach involves first sorting the array and using the sorted positions to determine how many elements are smaller. By sorting, we map each element to its position in the sorted order, which directly correlates to the number of elements smaller than it, considering duplicates.

This approach is effective because sorting gives us an ordered sequence where the position of each element naturally indicates the count of numbers less than it, except for handling duplicates. To handle duplicates effectively, we maintain a mapping of each unique number to its index in the sorted array. This ensures that each unique number gets the lowest possible rank, which corresponds to the count of numbers less than it.

Step-by-step Algorithm

1. Sort the Input Array: Create a sorted copy of the input array nums. This sorted array will help us identify the relative position of each number, which indirectly tells us how many numbers are smaller than each element.

2. Initialize an Empty HashMap (index_map): This hashmap will store each unique number from the sorted array as keys, and their first occurrence index as values. This index represents the count of unique numbers that are smaller than the key number.

3. Populate the HashMap: Iterate over the sorted array sorted_nums. For each number num in sorted_nums, if num is not already a key in index_map, add num as a key with its index i as its value. If num is already a key, skip to the next number. This step ensures each number is mapped to the number of unique elements that are smaller than it.

4. Generate the Result Array: Iterate over the original array nums, and for each element num, use index_map to find the count of numbers smaller than num. Place this count in the corresponding position of a new result array.

5. Return the Result Array: The result array, now filled with the counts of smaller numbers for each element in nums, is returned as the final output.

Algorithm Walkthrough

• Input Array: [5, 3, 6, 1, 4]

• Step 1: Create a copy of the array and sort it.

  • Sorted Array: [1, 3, 4, 5, 6]

• Step 2: Initialize an empty hashmap index_map.

• Step 3: Populate the hashmap.

  • Iteration 1: num = 1, index_map becomes {1: 0}
  • Iteration 2: num = 3, index_map becomes {1: 0, 3: 1}
  • Iteration 3: num = 4, index_map becomes {1: 0, 3: 1, 4: 2}
  • Iteration 4: num = 5, index_map becomes {1: 0, 3: 1, 4: 2, 5: 3}
  • Iteration 5: num = 6, index_map becomes {1: 0, 3: 1, 4: 2, 5: 3, 6: 4}

• Step 4: Generate the result array by iterating over the original array and using index_map.

  • Original Array Element 5: 3 numbers are smaller (index_map[5]), so first element in result array is 3.
  • Original Array Element 3: 1 number is smaller (index_map[3]), so second element in result array is 1.
  • Original Array Element 6: 4 numbers are smaller (index_map[6]), so third element in result array is 4.
  • Original Array Element 1: 0 numbers are smaller (index_map[1]), so fourth element in result array is 0.
  • Original Array Element 4: 2 numbers are smaller (index_map[4]), so fifth element in result array is 2.

• Step 5: The final result array is [3, 1, 4, 0, 2], which is returned as the output.

Code

import java.util.Arrays;
import java.util.HashMap;

public class Solution {

  public int[] smallerNumbersThanCurrent(int[] nums) {
    int[] sortedNums = nums.clone();
    Arrays.sort(sortedNums);
    HashMap<Integer, Integer> indexMap = new HashMap<>();

    // Map each number to its first occurrence index in the sorted array
    for (int i = 0; i < sortedNums.length; i++) {
      indexMap.putIfAbsent(sortedNums[i], i);
    }

    // Replace each element with its corresponding value from the map
    int[] result = new int[nums.length];
    for (int i = 0; i < nums.length; i++) {
      result[i] = indexMap.get(nums[i]);
    }

    return result;
  }

  public static void main(String[] args) {
    Solution solution = new Solution();

    // Example 1
    int[] example1 = solution.smallerNumbersThanCurrent(
      new int[] { 5, 3, 6, 1, 4 }
    );
    System.out.println(Arrays.toString(example1)); // Expected: [3, 1, 4, 0, 2]

    // Example 2
    int[] example2 = solution.smallerNumbersThanCurrent(
      new int[] { 6, 6, 6, 6, 6 }
    );
    System.out.println(Arrays.toString(example2)); // Expected: [0, 0, 0, 0, 0]

    // Example 3
    int[] example3 = solution.smallerNumbersThanCurrent(
      new int[] { 1, 2, 3, 4, 5 }
    );
    System.out.println(Arrays.toString(example3)); // Expected: [0, 1, 2, 3, 4]
  }
}

Complexity Analysis

Time Complexity

1. Sorting: The initial sorting of the array takes time, where (n) is the number of elements in the input array.
2. Mapping: Creating the map from the sorted array takes time since we iterate through the sorted array once.
3. Building the Result Array: Iterating through the original array to form the result array, using the map for lookup, also takes time.

Therefore, the overall time complexity of the algorithm is due to the sorting step, which dominates the linear iterations.

Space Complexity

1. Sorted Array: We create a sorted copy of the input array, which takes space.
2. Mapping/Indexing Structure: The space taken by the map or dictionary is in the worst case when all elements are unique.
3. Result Array: The output array also takes space.

Thus, the total space complexity is , considering the additional structures used besides the input.