[Leet Code - eazy] 1365. 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.

 

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

 

나의 풀이

 

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var smallerNumbersThanCurrent = function(nums) {
    const answer = []
    for(let i=0; i<nums.length; i++){
        let count =0
        for(let j=0; j<nums.length; j++){
            if(i === j) continue
            if(nums[i] > nums[j]){
                count+=1
            }
        }
        answer.push(count)
    }
    return answer
};

 

문제를 풀며 느낀 점

 

- 배열의 요소를 돌면서 자기 자신을 제외한 배열 요소가 클 때마다 +1 해주고 count값을 차례대로 배열에 담아 return 해 주는 문제라서 이중 for문을 통해서 자기 자신일 때는 continue를 통해서 건너뛰어준 값 들을 배열에 넣어서 return 하