# 1394. Find Lucky Integer in an Array

## The Problem

Link to original problem on Leetcode.

Given an array of integers `arr`

, a lucky integer is an integer which has a frequency in the array equal to its value.

Return a *lucky integer* in the array. If there are multiple lucky integers return the **largest** of them. If there is no lucky integer return `-1`

.

## Examples

Example 1:

```
Input: arr = [2,2,3,4]
Output: 2
Explanation: The only lucky number in the array is 2 because frequency[2] == 2.
```

Example 2:

```
Input: arr = [1,2,2,3,3,3]
Output: 3
Explanation: 1, 2 and 3 are all lucky numbers, return the largest of them.
```

Example 3:

```
Input: arr = [2,2,2,3,3]
Output: -1
Explanation: There are no lucky numbers in the array.
```

Example 4:

```
Input: arr = [5]
Output: -1
```

Example 5:

```
Input: arr = [7,7,7,7,7,7,7]
Output: 7
```

## Constraints

`1 <= arr.length <= 500`

`1 <= arr[i] <= 500`

## My Solution

This solution creates a Map of the integer values in `arr`

along with their count. It then iterates over the Map, comparing keys and values for equality. An initial `result`

of `-1`

is then replaced if a lucky integer is found, and any larger lucky integer will in turn replace a smaller one. This takes $$O(n)$$ time.

`const findLucky = (arr) => {`

const integers = new Map();

let result = -1;

arr.forEach(i => {

if (!integers.has(i)) {

integers.set(i, 1);

} else {

integers.set(i, 1 + integers.get(i));

}

});

for ([key, value] of integers) {

if (key === value && key > result) {

result = key;

}

}

return result;

};