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Want to solve **Get encrypted number Amazon OA?, **if yes then this article is for you.

Make sure you check the List of Amazon OA questions in the year of 2021 and 2022. Today, we are going to see another problem from amazon oa called *Get encrypted number* which was asked in 2022, this questions might be still active in some of the **Amazon OA** so save all the solutions, publishing questions directly is violation of amazon DCMA policy so we try our best to provide similar questions now lets see how can we solve this problem.

**Get encrypted number Amazon OA Solution**

You are given a **0-indexed** integer array `nums`

, where `nums[i]`

is a digit between `0`

and `9`

(**inclusive**).

The **triangular sum** of `nums`

is the value of the only element present in `nums`

after the following process terminates:

- Let
`nums`

comprise of`n`

elements. If`n == 1`

,**end**the process. Otherwise,**create**a new**0-indexed**integer array`newNums`

of length`n - 1`

. - For each index
`i`

, where`0 <= i < n - 1`

,**assign**the value of`newNums[i]`

as`(nums[i] + nums[i+1]) % 10`

, where`%`

denotes modulo operator. **Replace**the array`nums`

with`newNums`

.**Repeat**the entire process starting from step 1.

Return *the triangular sum of* `nums`

.

**Example 1:**

**Input:** nums = [1,2,3,4,5]

**Output:** 8

**Explanation:** The above diagram depicts the process from which we obtain the triangular sum of the array.

**Example 2:**

**Input:** nums = [5]

**Output:** 5

**Explanation:** Since there is only one element in nums, the triangular sum is the value of that element itself.

**Constraints:**

`1 <= nums.length <= 1000`

`0 <= nums[i] <= 9`

**Related: Google Online Assessment 2022 Questions List**

**SOLUTION**

**Program: Get encrypted number Amazon OA Solution** in Python

```
class Solution:
def triangularSum(self, nums: List[int]) -> int:
m = len(nums) - 1
mCk = 1
result = 0
for k, num in enumerate(nums):
result = (result + mCk * num) % 10
# print("r",result)
mCk *= m - k
# print("mc",mCk)
mCk //= k + 1
# print("mc",mCk)
return (result)
```

**Program: Get encrypted number Amazon OA Solution** in Java

```
public int triangularSum(int[] nums) {
if (nums.length == 1)
return nums[0];
int run = nums.length - 1; //store the number of times to run the loop
for (int i = 0; i < run; i++) {
for (int j = 0; j < nums.length - 1 - i; j++) {
nums[j] = (nums[j] + nums[j + 1]) % 10; //replace in the orginal array as you move along
}
}
return nums[0]; //first index will have answer
}
```

**Program: Get encrypted number Amazon OA Solution** in C++

```
class Solution {
public:
int triangularSum(vector<int>& n) {
int l=n.size();
if(l==1){
return n[0];
}else{
vector<int>arr(l-1);
for(int i=0; i<l-1; i++){
arr[i]=(n[i]+n[i+1])%10;
}
return triangularSum(arr);
}
}
};
```