Product of Array Except Self LeetCode Solution

Problem – Product of Array Except Self

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 product of any prefix or suffix of nums 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.)

Product of Array Except Self LeetCode Solution in Java

public class Solution {
public int[] productExceptSelf(int[] nums) {
    int n = nums.length;
    int[] res = new int[n];
    res[0] = 1;
    for (int i = 1; i < n; i++) {
        res[i] = res[i - 1] * nums[i - 1];
    }
    int right = 1;
    for (int i = n - 1; i >= 0; i--) {
        res[i] *= right;
        right *= nums[i];
    }
    return res;
}

Product of Array Except Self LeetCode Solution in C++

class Solution {
public:
    vector<int> productExceptSelf(vector<int>& nums) {
        int prod = 1, zeroCnt = count(begin(nums), end(nums), 0);
        if(zeroCnt > 1) return vector<int>(size(nums));               // Case-1
        for(auto c : nums) 
            if(c) prod *= c;                                          // calculate product of all elements except 0
        for(auto& c : nums)
            if(zeroCnt) c = c ? 0 : prod;                             // Case-2
            else c = prod / c;                                        // Case-3
        return nums;
    }
};

Product of Array Except Self LeetCode Solution in Python

class Solution:
    def productExceptSelf(self, nums):
        prod, zero_cnt = reduce(lambda a, b: a*(b if b else 1), nums, 1), nums.count(0)
        if zero_cnt > 1: return [0]*len(nums)
        for i, c in enumerate(nums):
            if zero_cnt: nums[i] = 0 if c else prod
            else: nums[i] = prod // c
        return nums

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