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

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 →

Insert a Node at the Tail of a Linked List

You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Input Format: You have to complete the SinglyLink

View Solution →