How Many Numbers Are Smaller Than the Current Number
Description
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(javascript)
/**
* @param {number[]} nums
* @return {number[]}
*/
var smallerNumbersThanCurrent = function(nums) {
let count = 0;
let array = [];
for (let i = 0; i < nums.length; i++) {
for (let j = 0; j < nums.length; j++) {
if (i !== j && nums[j] < nums[i]) {
count++;
}
}
array.push(count);
count = 0
}
return array;
};