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[leetCode/JS] 2256. Minimum Average Difference 본문

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[leetCode/JS] 2256. Minimum Average Difference

쭘봉 2022. 12. 4. 09:56

난이도 [ 🤔 ] Medium 

 

문제 설명 

You are given a 0-indexed integer array nums of length n.

The average difference of the index i is the absolute difference between the average of the first i + 1 elements of nums and the average of the last n - i - 1 elements. Both averages should be rounded down to the nearest integer.

Return the index with the minimum average difference. If there are multiple such indices, return the smallest one.

Note:

  • The absolute difference of two numbers is the absolute value of their difference.
  • The average of n elements is the sum of the n elements divided (integer division) by n.
  • The average of 0 elements is considered to be 0.

 

숫자가 들어있는 numsindex 기준으로 좌우로 나눠서
각각 평균을 낸 후 각 평균의 차이가 가장 작은 index를 리턴하는 것이다.
example1 을 차근 차근 보면 이해가 쉽다.

 

입출력 예

Example 1:

Input: nums = [2,5,3,9,5,3]
Output: 3
Explanation:
- The average difference of index 0 is: |2 / 1 - (5 + 3 + 9 + 5 + 3) / 5| = |2 / 1 - 25 / 5| = |2 - 5| = 3.
- The average difference of index 1 is: |(2 + 5) / 2 - (3 + 9 + 5 + 3) / 4| = |7 / 2 - 20 / 4| = |3 - 5| = 2.
- The average difference of index 2 is: |(2 + 5 + 3) / 3 - (9 + 5 + 3) / 3| = |10 / 3 - 17 / 3| = |3 - 5| = 2.
- The average difference of index 3 is: |(2 + 5 + 3 + 9) / 4 - (5 + 3) / 2| = |19 / 4 - 8 / 2| = |4 - 4| = 0.
- The average difference of index 4 is: |(2 + 5 + 3 + 9 + 5) / 5 - 3 / 1| = |24 / 5 - 3 / 1| = |4 - 3| = 1.
- The average difference of index 5 is: |(2 + 5 + 3 + 9 + 5 + 3) / 6 - 0| = |27 / 6 - 0| = |4 - 0| = 4.
The average difference of index 3 is the minimum average difference so return 3.

 

Example 2:

Input: nums = [0]
Output: 0
Explanation:
The only index is 0 so return 0.
The average difference of index 0 is: |0 / 1 - 0| = |0 - 0| = 0.

 

Constraints

  • 1 <= nums.length <= 105
  • 0 <= nums[i] <= 105

 

 

 


내 솔루션

  • for문으로 순회한다.
  • 처음엔 좌우로 나눈 배열 a, b를 각각 sum / n 으로 평균을 구했는데, 느리다고 빠꾸.
  • 새로운 요소를 a엔 더하고 b엔  빼주면 된다는 로직을 통해 로직을 간소화함.
var minimumAverageDifference = function(nums) {
  let minimun = Infinity;
  let answer = null;
  let aSum = 0;
  let bSum = nums.reduce((acc, cur) => acc + cur, 0);
  for(let i = 0; i < nums.length; i++) {
    aSum += nums[i];
    bSum -= nums[i];
    const a = Math.floor(aSum / (i + 1));
    const b = Math.floor(bSum / (nums.length - i - 1) || 0);
    const min = Math.abs(a - b);
    if(minimun > min) {
      minimun = min;
      answer = i;
    }
  }
  return answer;
};


// 너무 느렸음.
// var minimumAverageDifference = function(nums) {
//   let minimun = Infinity;
//   let answer = null;
//   for(let i = 1; i < nums.length+1; i++) {
//     const a = Math.floor(nums.slice(0, i).reduce((acc, cur)=>acc+cur,0) / i);
//     const b = Math.floor(nums.slice(i).reduce((acc, cur)=>acc+cur,0) / (nums.length - i) || 0);
//     const min = Math.abs(a - b);
//     if(minimun > min) {
//       minimun = min;
//       answer = i - 1;
//     }
//   }
//   return answer;
// };

 

감상평

  • 오마이깟? 개빨랐다. 실행할 때, 서버 컨디션이 좋았나?
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