## Introduction:

The problem involves identifying pairs of indices in an array that satisfy certain conditions, with a specific mathematical operation. This blog will discuss the problem, the provided solution, and break down the code for a better understanding.

## Problem Overview:

The goal is to find the count of such nice pairs of indices. Due to the potential size of the result, it should be returned modulo (10^9 + 7).

## Provided Solution:

The solution is implemented in Java and involves the following steps:

First we will simplify the given expression i in one side and j on other side. Then we end up with following expression.

`nums[i] - rev(nums[i]) == nums[j] - rev(nums[j])`

- Calculate the difference between each element and its reverse and store them in an array
`subtracted`

. - Use a HashMap to store the frequency of each element in the
`subtracted`

array. - Iterate through the HashMap, calculate the count of nice pairs, and accumulate the result.
- Return the final count modulo (10^9 + 7).

**Code Breakdown:**

Let’s break down the key parts of the provided Java code:

```
class Solution {
public int countNicePairs(int[] nums) {
int subtracted[] = new int[nums.length];
for(int i=0;i<nums.length;i++) {
subtracted[i] = nums[i] - reverse(nums[i]);
}
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
for(int i=0;i<nums.length;i++) {
if(map.containsKey(subtracted[i])) {
map.put(subtracted[i], map.get(subtracted[i])+ 1);
} else {
map.put(subtracted[i], 1);
}
}
long mod = 1000000007;
long ans = 0;
for (Map.Entry<Integer,Integer> entry : map.entrySet()) {
long freq = entry.getValue();
int key = entry.getKey();
if (freq > 1) {
System.out.println(freq +" "+ key);
ans = (ans + ((freq * (freq -1))/2)%mod)%mod;
}
}
return (int)ans;
}
private int reverse(int i) {
StringBuilder sb = new StringBuilder();
sb.append(i);
return Integer.parseInt(sb.reverse().toString());
}
}
```

**Conclusion:**

This blog has discussed the problem of finding nice pairs in arrays, presented a Java solution, and explained the key components of the code. Understanding the provided solution is crucial for tackling similar mathematical and array manipulation problems.