**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[0] and nums[0]) 2 and 2 (nums[0] and nums[1]) 2 and 2 (nums[1] and nums[1]) 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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