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.

LeetCode Problem - 1365

class Solution {
    // It calculates and returns the number of pairs of integers in 'nums' whose sum is less than 'target'.
    public int countPairs(List<Integer> nums, int target) {  
        // Initialize a variable 'count' to keep track of the number of pairs.
        int count = 0;
        // Iterate through each element at index 'i' in the 'nums' list.
        for(int i=0; i<nums.size(); i++){
            // Iterate through each element at index 'j' starting from 'i+1' to the end of the list.
            for(int j=i+1; j<nums.size(); j++){
                // Check if the sum of the elements at index 'i' and 'j' is less than 'target'.
                // If so, increment the 'count' as it represents a valid pair.
                if((nums.get(i) + nums.get(j)) < target){
                    count++;
                }
            }
        }

        // Return the total count of pairs whose sum is less than 'target'.
        return count;
    }
}