LeetCode 1365: How Many Numbers Are Smaller Than the Current Number Solution

Master LeetCode problem 1365 (How Many Numbers Are Smaller Than the Current Number), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1365. How Many Numbers Are Smaller Than the Current Number

Easy

Array Hash Table Sorting Counting Sort

Problem Explanation

Explanation

Algorithmic Idea:
- We can solve this problem using a simple brute force approach.
- For each element in the array, we compare it with all other elements to count the numbers smaller than it.
Step-by-Step Iterations:
- Iterate over each element in the array.
- For each element, compare it with all other elements to count the numbers smaller than it.
- Store the count for each element in the result array.
Time Complexity: O(n^2) where n is the number of elements in the array.
Space Complexity: O(n) where n is the number of elements in the array.

Solution Code

class Solution {
    public int[] smallerNumbersThanCurrent(int[] nums) {
        int n = nums.length;
        int[] result = new int[n];
        
        for (int i = 0; i < n; i++) {
            int count = 0;
            for (int j = 0; j < n; j++) {
                if (i != j && nums[j] < nums[i]) {
                    count++;
                }
            }
            result[i] = count;
        }
        
        return result;
    }
}

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Frequently Asked Questions

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What is the time complexity of LeetCode 1365 (How Many Numbers Are Smaller Than the Current Number)?

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