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.


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

Example 1


nums = [1, 20, 2, 11]




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


nums = [2, 2]




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


nums = [4, 10, 10]



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) {
        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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