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 :



title-img 




                        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)


View More Similar Problems

2D Array-DS

Given a 6*6 2D Array, arr: 1 1 1 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 An hourglass in A is a subset of values with indices falling in this pattern in arr's graphical representation: a b c d e f g There are 16 hourglasses in arr. An hourglass sum is the sum of an hourglass' values. Calculate the hourglass sum for every hourglass in arr, then print t

View Solution →

Dynamic Array

Create a list, seqList, of n empty sequences, where each sequence is indexed from 0 to n-1. The elements within each of the n sequences also use 0-indexing. Create an integer, lastAnswer, and initialize it to 0. There are 2 types of queries that can be performed on the list of sequences: 1. Query: 1 x y a. Find the sequence, seq, at index ((x xor lastAnswer)%n) in seqList.

View Solution →

Left Rotation

A left rotation operation on an array of size n shifts each of the array's elements 1 unit to the left. Given an integer, d, rotate the array that many steps left and return the result. Example: d=2 arr=[1,2,3,4,5] After 2 rotations, arr'=[3,4,5,1,2]. Function Description: Complete the rotateLeft function in the editor below. rotateLeft has the following parameters: 1. int d

View Solution →

Sparse Arrays

There is a collection of input strings and a collection of query strings. For each query string, determine how many times it occurs in the list of input strings. Return an array of the results. Example: strings=['ab', 'ab', 'abc'] queries=['ab', 'abc', 'bc'] There are instances of 'ab', 1 of 'abc' and 0 of 'bc'. For each query, add an element to the return array, results=[2,1,0]. Fun

View Solution →

Array Manipulation

Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each of the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in the array. Example: n=10 queries=[[1,5,3], [4,8,7], [6,9,1]] Queries are interpreted as follows: a b k 1 5 3 4 8 7 6 9 1 Add the valu

View Solution →

Print the Elements of a Linked List

This is an to practice traversing a linked list. Given a pointer to the head node of a linked list, print each node's data element, one per line. If the head pointer is null (indicating the list is empty), there is nothing to print. Function Description: Complete the printLinkedList function in the editor below. printLinkedList has the following parameter(s): 1.SinglyLinkedListNode

View Solution →