Here I would be sharing the code I wrote for the problems posted on LeetCode.

Most problems took time for me to write the working code & optimize that. So I am giving those here as I solve Hard/Medium problems on LeetCode. I think, like me, many people would be able to understand the problems & come up with possible solution quickly, but during implementation are lost in the middle. So check the code given here & get the idea & write your own code your own way...Cheers!

238. Product of Array Except Self(Medium)

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

442. Find All Duplicates in an Array(Medium)

class Solution {
    public List<Integer> findDuplicates(int[] nums) {
       List<Integer> res = new ArrayList();
        int[] map = new int[nums.length + 1];
        for (int e : nums) {
            map[e]++;
        }
        for (int i = 1; i <= nums.length; i++) {
            if (map[i] == 2) {
                res.add(i);
            }
        }
        return res;
   }
}

262. Trips and Users(Hard)

select total.request_at Day,
case 
when cancel.cancels is null
then 0.00
else round(cancel.cancels/total.total, 2) 
end "Cancellation Rate"
from
(select filter.request_at,count(*) cancels from
(select t.client_id,t.status,t.request_at from trips t inner join users u 
 on u.Banned='No' and u.users_id = t.client_id and (status = 'cancelled_by_driver' or status = 'cancelled_by_client')
 and t.request_at between '2013-10-01' and'2013-10-03') filter
group by filter.request_at) cancel right outer join
(select filter.request_at,count(*) Total from
(select t.client_id,t.status,t.request_at from trips t, users u where u.Banned='No' and u.users_id = t.client_id  and t.request_at between '2013-10-01' and'2013-10-03') filter
group by filter.request_at) total
on total.request_at = cancel.request_at
order by total.request_at​

1250. Check If It Is a Good Array(Hard)

import java.util.ArrayList;
class Solution {
    public boolean isGoodArray(int[] nums) {
        int x= nums[0],y;
        for (int num : nums) {
            while (num>0){
                 y=x%num;
                 x=num;
                 num=y;
             }
            if(x==1){
               return true;
            }
       }
      return false;
   }
}

239. Sliding Window Maximum(Hard)

class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
int len = nums.length;
if(k == 0) {
return new int[0];
} else if(k == 1)
return nums;
int maxes = 0;
int i = 1;
for(; i < k; i++) {
if(nums[maxes] <= nums[i])
maxes = i;
}
nums[0] = nums[maxes];
for(; i < len; i++) {
if(maxes > i-k) {
if(nums[maxes] <= nums[i])
maxes = i;
nums[i-k+1] = nums[maxes];
} else {
maxes = i-k+1;
for(int m = maxes+1; m <= i; m++) {
if(nums[maxes] <= nums[m])
maxes = m;
}
nums[i-k+1] = nums[maxes];
}
}
return Arrays.copyOf(nums, len-k+1);
}
}​