# Leetcode Product of Array Except Self problem solution

In this Leetcode Product of Array Except Self problem solution we have 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.

## Problem solution in Python.

class Solution(object):
def productExceptSelf(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
left = [1] * len(nums)
right = [1] * len(nums)
res = []
for i in range (1, len(nums)):
left[i] = left[i-1] * nums[i-1]
for i in range (len(nums)-2, -1, -1):
right[i] = right[i+1] * nums[i+1]
for i in range (0, len(nums)):
res.append(left[i] * right[i])
return res

## Problem solution in Java.

class Solution {
public int[] productExceptSelf(int[] nums) {
int tmp = 1;
int n  =nums.length;
int[] left = new int[nums.length];
int[] right = new int[n];
left[0] = 1;
for (int i = 1; i < n; i++){
tmp *= nums[i-1];
left[i] = tmp;
}
right[n-1] = 1;
tmp = 1;
for (int i = n - 2; i >= 0; i--){
tmp *= nums[i+1];
right[i] = tmp;
}
for (int i = 0; i < n; i++){
nums[i] = left[i] * right[i];
}
return nums;
}
}

## Problem solution in C++.

class Solution {
public:
vector<int> productExceptSelf(vector<int>& nums) {
int cnt=0;
int n=nums.size();
vector<int>pref(n),suf(n);
for(int i=0;i<n;i++){
pref[i]=nums[i];
if(i) pref[i]*=pref[i-1];
}
for(int i=n-1;~i;i--){
suf[i]=nums[i];
if(i<n-1) suf[i]*=suf[i+1];
}
for(int i=0;i<n;i++){
if(i==0){
nums[i]=suf[i+1];
}else if(i==n-1){
nums[i]=pref[i-1];
}else{
nums[i]=(pref[i-1]*suf[i+1]);
}
}

return nums;

}
};

## Problem solution in C.

int* productExceptSelf(int* nums, int numsSize, int* returnSize) {
int sum = 1;
int flag = 0;
for (int i = 0; i < numsSize && flag < 2; i++){
if (!nums[i]) {flag++; continue;}
sum = sum * nums[i];
}
int* result = (int*)malloc(sizeof(int) * numsSize);
memset(result, 0, sizeof(int)*numsSize);
*returnSize = numsSize;
if (flag>1) return result;
for (int i = 0; i < numsSize; i++){
if (!nums[i]) {result[i] = sum; continue;}
if (flag > 0) {result[i] = 0; continue;}
result[i] = sum / nums[i];
}
return result;
}