238. Product of Array Except Self

// 나의 접근 방법. O(n2)

function productExceptSelf(nums: number[]): number[] {
    let answerArr: number[] = []

    for(let i = 0; i < nums.length; i++) {  // ← for로 수정
        if(i === 0) {  // ← 인덱스 체크로 수정
            let answer = 1;
            for(let j = i + 1; j < nums.length; j++) {
                answer *= nums[j]
            }
            answerArr.push(answer);
            
        } else if(i === nums.length - 1) {  // ← 마지막 인덱스 체크
            let answer = 1;
            for(let j = i - 1; j >= 0; j--) {  // ← >= 0으로 수정
                answer *= nums[j]
            }
            answerArr.push(answer);
        } else {  // ← 중간 요소들 처리 추가 필요
            let answer = 1;
            // 왼쪽 곱
            for(let j = 0; j < i; j++) {
                answer *= nums[j];
            }
            // 오른쪽 곱
            for(let j = i + 1; j < nums.length; j++) {
                answer *= nums[j];
            }
            answerArr.push(answer);
        }
    }

    return answerArr  // ← 오타 수정
}​

Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i].

The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.

You must write an algorithm that runs in O(n) time and without using the division operation.

 

Example 1:

Input: nums = [1,2,3,4]
Output: [24,12,8,6]
Example 2:

Input: nums = [-1,1,0,-3,3]
Output: [0,0,9,0,0]
 

Constraints:

2 <= nums.length <= 105
-30 <= nums[i] <= 30
The input is generated such that answer[i] is guaranteed to fit in a 32-bit integer.
 

Follow up: Can you solve the problem in O(1) extra space complexity? (The output array does not count as extra space for space complexity analysis.)

 

 

 

function productExceptSelf(nums) {
    const n = nums.length;
    const answer = new Array(n);
    
    // 1. 왼쪽에서 오른쪽으로 prefix product 계산
    const leftProducts = new Array(n);
    leftProducts[0] = 1;
    for (let i = 1; i < n; i++) {
        leftProducts[i] = leftProducts[i - 1] * nums[i - 1];
    }
    
    // 2. 오른쪽에서 왼쪽으로 suffix product 계산
    const rightProducts = new Array(n);
    rightProducts[n - 1] = 1;
    for (let i = n - 2; i >= 0; i--) {
        rightProducts[i] = rightProducts[i + 1] * nums[i + 1];
    }
    
    // 3. 두 값을 곱함
    for (let i = 0; i < n; i++) {
        answer[i] = leftProducts[i] * rightProducts[i];
    }
    
    return answer;
}

풀이 2: O(1) 공간 (최적화)

결과 배열을 재사용해서 추가 공간을 줄이기

 

function productExceptSelf(nums) {
    const n = nums.length;
    const answer = new Array(n);
    
    // 1. answer에 왼쪽 곱을 먼저 저장
    answer[0] = 1;
    for (let i = 1; i < n; i++) {
        answer[i] = answer[i - 1] * nums[i - 1];
    }
    
    // 2. 오른쪽에서 오면서 곱해줌
    let rightProduct = 1;
    for (let i = n - 1; i >= 0; i--) {
        answer[i] = answer[i] * rightProduct;
        rightProduct *= nums[i];
    }
    
    return answer;
}
```

## 동작 과정 (nums = [1,2,3,4])

**Step 1: 왼쪽 곱 계산**
```
answer = [1, 1, 1×2, 1×2×3]
       = [1, 1, 2, 6]
```

**Step 2: 오른쪽에서 곱하기**
```
i=3: answer[3] = 6 × 1 = 6,  rightProduct = 1×4 = 4
i=2: answer[2] = 2 × 4 = 8,  rightProduct = 4×3 = 12
i=1: answer[1] = 1 × 12 = 12, rightProduct = 12×2 = 24
i=0: answer[0] = 1 × 24 = 24, rightProduct = 24×1 = 24

answer = [24, 12, 8, 6] ✓