# Reverse Equivalent Pairs - Google Top Interview Questions

### Problem Statement :

```Consider a function rev(num) which reverses the digits of num. For example, rev(123) = 321 and rev(100) = 1.

Given a list of positive integers nums, return the number of pairs (i, j) where i ≤ j and nums[i] + rev(nums[j]) = nums[j] + rev(nums[i]). Mod the result by 10 ** 9 + 7.

Constraints

n ≤ 100,000 where n is the length of nums

Example 1

Input

nums = [1, 20, 2, 11]

Output

7

Explanation

We can have the following pairs:

1 and 1

1 and 2

1 and 11

20 and 20

2 and 2

2 and 11

11 and 11

Example 2

Input

nums = [2, 2]

Output

3

Explanation

We can have the following pairs:

2 and 2 (nums and nums)

2 and 2 (nums and nums)

2 and 2 (nums and nums)

Example 3

Input

nums = [4, 10, 10]

Output

4```

### Solution :

```                        ```Solution in C++ :

int rev_num(int n) {
int res = 0;

while (n > 0) {
res = res * 10 + n % 10;
n /= 10;
}

return res;
}

int solve(vector<int>& nums) {
unordered_map<int, int> mp;
int cnt = 0;

for (int& i : nums) cnt = (cnt + ++mp[i - rev_num(i)]) % (int)(1e9 + 7);

return cnt;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
private static final int MOD = (int) 1e9 + 7;
public int solve(int[] nums) {
Map<Integer, Integer> map = new HashMap<>();
int res = 0;
for (int i : nums) {
int my = i - getReverse(i);
int prev = map.getOrDefault(my, 0);
res = (res + prev) % MOD;
map.put(my, (prev + 1) % MOD);
}
return res + nums.length;
}

private StringBuilder sb = new StringBuilder();

private int getReverse(int num) {
sb.setLength(0);
return Integer.parseInt(sb.append(num).reverse().toString());
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, nums):
freq = defaultdict(int)
ans = 0
for val in nums:
signature = val - int(str(val)[::-1])
freq[signature] += 1
ans += freq[signature]
return ans % (10 ** 9 + 7)```
```

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